Lecture 13: Differentiation Derivatives of Trigonometric Functions

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Lecture 13: Differentiation Derivatives of Trigonometric Functions Derivatives of the Basic Trigonometric Functions Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions 1/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is cos (x ± y) 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin Recall that in Example 31(c) we guesse that sin x = cos x x by consiering the graphs of sin an cos. We will now prove this using the efinition of the erivative an some basic trigonometric ientities. First recall the sum an ifference formulas for sin sin (x ± y) = sin x cos y ± cos x sin y Though we on t nee it right away, the corresponing formula for cos is cos (x ± y) = cos x cos y sin x sin y 2/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f (x) 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h Recall that using the Squeeze Theorem we prove that sin x lim x 0 x 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue With f (x) = sin x, using Formula 3 we have f sin (x + h) sin x (x) = lim h 0 h sin x cos h + cos x sin h sin x = lim h 0 h sin x (cos h 1) + cos x sin h = lim h 0 h = sin x ( lim h 0 cos h 1 h ) ( + cos x lim h 0 ) sin h h Recall that using the Squeeze Theorem we prove that sin x lim = 1 x 0 x 3/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim h 0 h 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = x sin x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = sin x = sin x (0) + cos x (1) x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that cos h 1 lim = 0 h 0 h Thus we have f (x) = sin x = sin x (0) + cos x (1) = cos x x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x You use the sum formula for cos to prove the corresponing ifferentiation formula for cos x, which is x cos x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of sin continue Further, using the same approach as use in Example 13 we can show that Thus we have as we preicte. cos h 1 lim = 0 h 0 h f (x) = sin x = sin x (0) + cos x (1) = cos x x You use the sum formula for cos to prove the corresponing ifferentiation formula for cos x, which is cos x = sin x x 4/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) x ( π2 x ) 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) ( π2 x ) x = sin x( 1) 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of cos Using the Chain Rule We can prove the formula for the erivative of cos in a ifferent way. Two basic trigonometric ientities are sin ( π 2 x ) = cos x cos ( π 2 x ) = sin x since x an π 2 x are complementary angles in a right triangle. Thus, using the Chain Rule gives x cos x = x sin ( π 2 x ) = cos ( π 2 x ) ( π2 x ) x = sin x( 1) = sin x 5/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) cos 2 x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x cos 2 x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) cos 2 x = cos2 x + sin 2 x cos 2 x = 1 cos 2 x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x = ( 1 cos x cos 2 x ) 2 = 1 cos 2 x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule Recall that tan x = sin x cos x Thus, using the Quotient Rule gives x tan x = sin x x cos x ( x sin x = ) ( ) cos x sin x x cos x (cos x) 2 (cos x) cos x sin x ( sin x) = cos 2 x = cos2 x + sin 2 x = ( 1 cos x cos 2 x ) 2 = sec 2 x = 1 cos 2 x 6/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 x ) 2 = 1 + ( sin x cos x = cos2 x cos 2 x + sin2 x cos 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x = sec 2 x 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivative of tan Using the Quotient Rule continue An alternative way to simplify the previous expression is x tan x = cos2 x + sin 2 x cos 2 = cos2 x x cos 2 x + sin2 x cos 2 x ( ) sin x 2 = 1 + = 1 + tan 2 x cos x So that x tan x = sec2 x = 1 + tan 2 x The equality above can also be prove using the Pythagorean ientity 1 + tan 2 x = sec 2 x Most text books use the sec 2 x formula for the erivative of tan x, but Maple an other symbolic ifferentiating programs use the 1 + tan 2 x formula. 7/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: x sin u 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x cos u 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x u cos u = sin u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u u cos u = sin u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u cos u = sin u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x x sec u u cos u = sin u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x x csc u 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x u cos u = sin u x x u csc u = csc u cot u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x x cot u u cos u = sin u x x u csc u = csc u cot u x x 8/25

Derivative of sin Derivative of cos Using the Chain Rule Derivative of tan Using the Quotient Rule Derivatives the Six Trigonometric Functions Derivatives the Six Trigonometric Functions Using basic ifferentiation rules as in the erivation of the erivative formula for tan we can fin erivative formulas for all of the other trigonometric functions. Also, recall that when we erive the General Rule for the Exponential Function we state that we woul give all erivative formulas in a general form using the Chain Rule. In this form we introuce an intermeiate variable u assume to represent some function of x. With this assumption the erivative rules for all six basic trigonometric functions are: u sin u = cos u x x x tan u = sec2 u u ( ) x = 1 + tan 2 u u x u sec u = sec u tan u x x x cot u = csc2 u u ( ) x = 1 + cot 2 u u x u cos u = sin u x x u csc u = csc u cot u x x 8/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Example 46 Differentiating with Trig Functions Fin an simplify the inicate erivative(s) of each function. (a) (b) (c) Fin f (x) an f (x) for f (x) = x 2 cos (3x). Fin s t for s = cos t sin t + cos t. ( Fin C (x) for C(x) = tan e ) 1+x 2. 9/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) = 2 cos (3x) 6x sin (3x) 6x sin (3x) 9x 2 cos (3x) 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(a) Using the Prouct Rule followe by the Chain Rule (for cos (3x)) gives f (x) = 2x cos (3x) + x 2 [ 3 sin (3x)] = 2x cos (3x) 3x 2 sin (3x) = x [2 cos (3x) 3x sin (3x)] For f (x) use the expression in the secon line. Again using the Prouct an Chain Rules gives f (x) = 2 cos (3x) 6x sin (3x) 6x sin (3x) 9x 2 cos (3x) ( = 2 9x 2) cos (3x) 12x sin (3x) 10/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s t 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 1 = (sin t + cos t) 2 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(b) Using the Quotient Rule gives s sin t (sin t + cos t) cos t (cos t sin t) = t (sin t + cos t) 2 = sin2 t sin t cos t cos 2 t + cos t sin t (sin t + cos t) 2 ( ) = sin 2 t + cos 2 t (sin t + cos t) 2 1 = (sin t + cos t) 2 This example illustrates the fact that when simplifying erivatives involving trig functions, you sometimes nee to use stanar trigonometric ientities. 11/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with C(x) = f (g (h (k(x)))) where f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) ( = sec 2 1+x 2 e ) ( ) ( ) ( e 1+x 2 1 2 1 + x 2) 1/2 (2x) 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 46(c) This is a composite function with where C(x) = f (g (h (k(x)))) f (x) = tan x, g(x) = e x, h(x) = x = x 1/2, k(x) = 1 + x 2 Using the Chain Rule three times gives C (x) = f (g (h (k(x)))) g (h (k(x))) h (k(x)) k (x) ( = sec 2 1+x 2 e ) ( ) ( ) ( e 1+x 2 1 2 1 + x 2) 1/2 (2x) ( = xe 1+x 2 sec 2 ) e 1+x 2 1 + x 2 12/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Example 47 Dampe Oscillations Consier the function q(t) = e 7t sin (24t) This function escribes ampe simple harmonic motion. It gives the position of a mass attache to a spring relative to the equilibrium (resting) position of the spring. A frictional force acts to graually slow the mass. (a) Fin q (t) an q (t) an explain their meaning in terms of the ampe oscillatory motion. (b) Note that q(0) = 0. This means that the initial position of the mass is at the equilibrium position of the spring. Fin the initial velocity of the mass. Also fin the velocity when the mass first returns to the equilibrium position. (c) Draw a graph of the function q(t). () Fin the first two times when the oscillating mass turns aroun. Show the corresponing points on the graph of q(t). (e) Show that the function q(t) satisfies the ifferential equation 2 q q + 14 t2 t + 625q = 0 13/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives q (t) 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) + 24 2 7 2) ] sin (24t) 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) + 24 2 7 2) ] sin (24t) = e 7t [336 cos (24t) + 527 sin (24t)] 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(a) Using the Prouct an Chain Rules gives q (t) = 7e 7t sin (24t) + e 7t [24 cos (24t)] = e 7t [24 cos (24t) 7 sin (24t)] From our interpretation of the erivative as a rate of change, we know that this is the velocity of the mass at time t. Taking the erivative of the expression above for q (t) gives [ ] q (t) = 7e 7t [24 cos (24t) 7 sin (24t)] + e 7t 24 2 sin (24t) 7(24) cos (24t) [ ( = e 7t 14(24) cos (24t) + 24 2 7 2) ] sin (24t) = e 7t [336 cos (24t) + 527 sin (24t)] The rate of change of velocity is acceleration, so this gives the acceleration of the mass at time t. 14/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when. 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/24 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/24 = 9.6 24 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(b) Substituting t = 0 into the expression for the velocity v(t) = q (t) gives v(0) = e 0 [24 cos (0) 7 sin (0)] = 24 The mass returns to the equilibrium position when q(t) = 0. The first time after t = 0 when this happens is when 24t = π t = π 24 The velocity at this time is ( π ) v = e 7π/24 [24 cos (π) 7 sin (π)] = 24e 7π/24 = 9.6 24 This velocity is less than the initial velocity an in the opposite irection. 15/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. 0.1 0.2 0.3 0.4 16/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. As we saw in the last part, not only oes the amplitue ecrease, but so oes the velocity. 0.1 0.2 0.3 0.4 16/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. As we saw in the last part, not only oes the amplitue ecrease, but so oes the velocity. Further, note that where the graph is concave own, q (t) < 0, the mass is ecelerating. The velocity is getting less positive or more negative. 0.1 0.2 0.3 0.4 16/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(c) The graph of the function q(t) looks like this. The amplitue of the motion ecreases following an envelope given by the ecaying exponential function e 7t, as shown. As we saw in the last part, not only oes the amplitue ecrease, but so oes the velocity. Further, note that where the graph is concave own, q (t) < 0, the mass is ecelerating. The velocity is getting less positive or more negative. An, where the graph is concave up, q (t) > 0, the mass is accelerating. The velocity is getting more positive or less negative. 0.1 0.2 0.3 0.4 16/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 0.1 0.2 0.3 0.4 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 0.1 0.2 0.3 0.4 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 24t = arctan ( 24 7 ) + kπ 0.1 0.2 0.3 0.4 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 24t = arctan ( 24 7 ) + kπ 0.1 0.2 0.3 0.4 where k is any integer. 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 24t = arctan ( 24 7 ) + kπ 0.1 0.2 0.3 0.4 where k is any integer. The first two positive t values are t 1 = 0.0536 an t 2 = 0.1845. 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 24t = arctan ( 24 7 ) + kπ 0.1 0.2 0.3 0.4 where k is any integer. The first two positive t values are t 1 = 0.0536 an t 2 = 0.1845. At these values the tangent line to the graph of q(t) is horizontal, as shown. 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47() The mass turns aroun when v(t) = q (t) = 0. This happens when 24 cos (24t) 7 sin (24t) = 0 tan (24t) = 24 7 24t = arctan ( 24 7 ) + kπ 0.1 0.2 0.3 0.4 where k is any integer. The first two positive t values are t 1 = 0.0536 an t 2 = 0.1845. At these values the tangent line to the graph of q(t) is horizontal, as shown. 17/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(e) From part (a) we have 2 q q + 14 t2 t + 625q 18/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(e) From part (a) we have 2 q q + 14 t2 t + 625q = e 7t [336 cos (24t) + 527 sin (24t)] 18/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(e) From part (a) we have 2 q q + 14 t2 t + 625q = e 7t [336 cos (24t) + 527 sin (24t)] + 14e 7t [24 cos (24t) 7 sin (24t)] + 625e 7t sin (24t) 18/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(e) From part (a) we have 2 q q + 14 t2 t + 625q = e 7t [336 cos (24t) + 527 sin (24t)] + 14e 7t [24 cos (24t) 7 sin (24t)] + 625e 7t sin (24t) = e 7t [( 336 + 14 24) cos (24t) + ( 527 14 7 + 625) sin (24t)] 18/25

Example 46 Differentiating with Trig Functions Example 47 Dampe Oscillations Solution: Example 47(e) From part (a) we have 2 q q + 14 t2 t + 625q = e 7t [336 cos (24t) + 527 sin (24t)] + 14e 7t [24 cos (24t) 7 sin (24t)] + 625e 7t sin (24t) = e 7t [( 336 + 14 24) cos (24t) + ( 527 14 7 + 625) sin (24t)] = 0 18/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one π 2 π 2 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one, an so oes not have an inverse function. π 2 π 2 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the tangent function, π 2 π 2 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the tangent function, as shown in the graph, to the interval π 2 π 2 π 2 < x < π 2 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the tangent function, as shown in the graph, to the interval π 2 π 2 π 2 < x < π 2 gives a one-to-one function. 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arctan Function Recall that the graph of the tangent function looks like this. From this graph we realize that the tangent function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the tangent function, as shown in the graph, to the interval π 2 π 2 π 2 < x < π 2 gives a one-to-one function. So the inverse of the tangent function is efine as arctan x = tan 1 x = the angle between π 2 an π 2 whose tangent is x 19/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives tan (arctan x) x 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives ( ) x tan (arctan x) = 1 + tan 2 (arctan x) x arctan x 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives ( x tan (arctan x) = = ) 1 + tan 2 (arctan x) ( 1 + x 2) x arctan x x arctan x 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives ( x tan (arctan x) = = ) 1 + tan 2 (arctan x) x arctan x ( 1 + x 2) x arctan x = x x = 1 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives ( x tan (arctan x) = Solving for x arctan x gives x arctan x = ) 1 + tan 2 (arctan x) x arctan x ( 1 + x 2) x arctan x = x x = 1 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arctan Function With this efinition of the inverse tangent function, we see that its omain is all real numbers an its range is ( π 2, π ) 2. Further, we have tan (arctan x) = x an arctan (tan x) = x for π 2 < x < π 2 Taking the erivative of the first of the formulas above, as we i to fin the erivative of the ln function, gives ( x tan (arctan x) = = ) 1 + tan 2 (arctan x) x arctan x ( 1 + x 2) x arctan x = x x = 1 Solving for x arctan x gives x arctan x = 1 1 + x 2 20/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one 1 π 2 π 2 1 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. π 2 1 π 2 1 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, π 2 1 π 2 1 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 gives a one-to-one function. 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 gives a one-to-one function. So the inverse of the sine function is efine as arcsin x = sin 1 x = the angle between π 2 an π 2 whose sin is x 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 gives a one-to-one function. So the inverse of the sine function is efine as arcsin x = sin 1 x = the angle between π 2 an π 2 whose sin is x The omain of inverse sine function, so efine, is 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 gives a one-to-one function. So the inverse of the sine function is efine as arcsin x = sin 1 x = the angle between π 2 an π 2 whose sin is x The omain of inverse sine function, so efine, is [ 1, 1] an its range is 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The arcsin Function Recall that the graph of the sine function looks like this. From this graph we realize that the sine function is not one-to-one, an so oes not have an inverse function. However, restricting the omain of the sine function, as shown in the graph, to the interval π 2 1 π 2 π 2 x π 2 1 gives a one-to-one function. So the inverse of the sine function is efine as arcsin x = sin 1 x = the angle between π 2 an π 2 whose sin is x [ The omain of inverse sine function, so efine, is [ 1, 1] an its range is π 2, π ] 2. 21/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. So that sin (arcsin x) x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. So that x sin (arcsin x) = 1 sin 2 (arcsin x) x arcsin x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. So that x sin (arcsin x) = 1 sin 2 (arcsin x) x arcsin x = 1 x 2 x arcsin x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. So that x sin (arcsin x) = 1 sin 2 (arcsin x) x arcsin x = 1 x 2 x arcsin x Solving for x arcsin x gives x arcsin x 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions The Derivative of the arcsin Function As with the arctan function, we have sin (arcsin x) = x an arcsin (sin x) = x for π 2 x π 2 Taking the erivative of the first of the formulas above, as we just i for the arctan, gives sin (arcsin x) = cos (arcsin x) x x arcsin x = x x = 1 Now for π 2 x π 2 we have cos x 0, an using the basic Pythagorean ientity cos 2 x + sin 2 x = 1, gives cos x = 1 sin 2 x. So that x sin (arcsin x) = 1 sin 2 (arcsin x) x arcsin x = 1 x 2 x arcsin x Solving for x arcsin x gives x arcsin x = 1 1 x 2 22/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Example 48 Differentiating with Inverse Trig Functions Fin an simplify the inicate erivative(s) of each function. (a) Fin f (x) an f (x) for f (x) = ( 1 + x 2) arctan x. (b) Fin y ( ) x for y = arcsin 1 x 2. 23/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives f (x) 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives ( f (x) = 2x arctan x + 1 + x 2) ( ) 1 1 + x 2 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives ( f (x) = 2x arctan x + 1 + x 2) ( ) 1 1 + x 2 = 2x arctan x + 1 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives ( f (x) = 2x arctan x + 1 + x 2) ( ) 1 1 + x 2 = 2x arctan x + 1 Taking the erivative of the expression above, using the Prouct Rule again, gives f (x) 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives ( f (x) = 2x arctan x + 1 + x 2) ( ) 1 1 + x 2 = 2x arctan x + 1 Taking the erivative of the expression above, using the Prouct Rule again, gives ( ) f 1 (x) = 2 arctan x + 2x 1 + x 2 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(a) Using the Prouct Rule gives ( f (x) = 2x arctan x + 1 + x 2) ( ) 1 1 + x 2 = 2x arctan x + 1 Taking the erivative of the expression above, using the Prouct Rule again, gives ( ) f 1 (x) = 2 arctan x + 2x 1 + x 2 = 2 arctan x + 2x 1 + x 2 24/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = 1 ( ) 2 1 1 x 2 ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) x 1 x 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) ( ) ( ) x 1 x = 1 x 2 x 2 1 x 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) ( ) ( ) x 1 x = 1 x 2 x 2 1 x 2 x = x 1 x 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) Thus, if 0 x < 1, then ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) ( ) ( ) x 1 x = 1 x 2 x 2 1 x 2 ( ) x arcsin 1 x 2 1 = 1 x 2 x = x 1 x 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) Thus, if 0 x < 1, then But you can show that ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) ( ) ( ) x 1 x = 1 x 2 x 2 1 x 2 ( ) x arcsin 1 x 2 1 = 1 x 2 x arccos x = 1 1 x 2 x = x 1 x 2 25/25

The arctan Function The arcsin Function Example 48 Differentiating with Inverse Trig Functions Solution: Example 48(b) Using the Chain Rule gives y x = = 1 ( ) 2 1 1 x 2 ( 1 1 (1 x 2 ) Thus, if 0 x < 1, then But you can show that ( ) 1 ( 1 x 2) 1/2 ( 2x) 2 ) ( ) ( ) ( ) x 1 x = 1 x 2 x 2 1 x 2 ( ) x arcsin 1 x 2 1 = 1 x 2 x arccos x = 1 1 x 2 ( ) So is it true that arccos x = arcsin 1 x 2? x = x 1 x 2 25/25

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