Derivatives of Exponential, Logarithmic and Trigonometric Functions

Transcription

1 Calculus Lia Vas Derivatives of Exponential, Logaritmic and Trigonometric Functions Derivative of te inverse function. If f(x) is a one-to-one function (i.e. te grap of f(x) passes te orizontal line test), ten f(x) as te inverse function f (x). Recall tat f and f are related by te following formulas y = f (x) x = f(y). Also, recall tat te graps of f (x) and f(x) are symmetrical wit respect to line y = x. Some pairs of inverse functions you encountered before are given in te table below were n is a positive integer and a is a positive real number. f x 2 x n e x a x f x n x ln x log a x Wit y = f (x), dy denotes te derivative of f and since x = f(y), dy we ave tat of f. Since te reciprocal of dy is dy (f ) (x) = dy = dy = f (y). Tus, te derivative of te inverse function of f is reciprocal of te derivative of f. Anoter way to see tis is to consider relation f(f (x)) = x or f (f(x)) = x, denotes te derivative and to differentiate any of tese identities. For example, differentiating f (f(x)) = x and using te cain rule for te left and side produces (f ) (f(x)) f (x) = = (f ) (f(x)) = f (x). Grapically, tis rule means tat Te slope of te tangent to f (x) at point (b, a) is reciprocal to te slope of te tangent to f(x) at point (a, b). Example. If f(x) as te inverse, f(2) =, and f (2) = 3, find (f ) ().

2 Solution. Since f passes te point (2, ), f passes te point (,2). Te tangent to f(x) at x = 2 is 3 so te tangent to f at x = is te reciprocal 3. Hence (f ) () = 3. Exponential Functions and teir derivatives. In a pre-calculus course you ave encountered exponential function a x of any base a > 0 and teir inverse functions. All tese functions can be considered to be a composite of e u and x ln a since a x = e ln ax = e x ln a Tus, using te cain rule and formula for derivative of e x, you can obtain te formula for derivative of any a x. Te derivative of e x can be computed by de x = lim e x+ e x 0 = lim 0 e x e e x = e x lim 0 e e Before you see a proof of lim 0 = in iger calculus courses, you can convince yourself tat e te limit lim 0 is by evaluating te quotient e at several values of close to 0 as in te table below. Tis indicates tat lim 0 e e = and so dex = ex lim 0 e = e x. Tus, te derivative of e x is e x. Example 2. Find te derivative of te following functions (a) y = e 3x (b) y = x 2 e 3x Solution. (a) We can consider te function y = e 3x as a composite of te outer function e u and te inner function 3x. y = e 3x y = e 3x 3 derivative of derivative of derivative of te composite te outer, te inner keep te inner uncanged Tus te derivative is y = 3e 3x. (b) Te function is a product of f(x) = x 2 and g(x) = e 3x. By part (a), g (x) = 3e 3x. Since f (x) = 2x, te product rule produces y = f g + g f = 2xe 3x + x 2 (3)e 3x = (2x + 3x 2 )e 3x. Using te formula dex = ex we can obtain te derivative of a x. Recall tat y = a x = e ln ax = e x ln a. In te last step we used te rule log a (x r ) = r log a x. Tus te function y = a x = e x ln a can be consider as a composite of e u and u = x ln a. Since ln a is a constant u = ln a (just as in part (a) of te previous example we ad u = 3x u = 3). Tus, te derivative of y = a x is y = e x ln a ln a. Note tat e x ln a is te original function y = a x. Tus, y = a x ln a and we ave tat 2

3 te derivative of a x is a x ln a. Example 3. Find te derivative of te following functions (a) y = 2 5x+7 (b) y = 3x 3 x Solution. (a) Consider te function as a composite of 2 u and u = 5x + 7. Using te formula for a x wit a = 2 obtain 2 u ln 2 = 2 5x+7 ln 2 for te derivative of te outer. Since te derivative of te inner is 5, y = 5 ln 2 2 5x+7. (b) Note tat te function can be written as y = 2 (3x 3 x ). Te derivative of te first term in te parentesis is 3 x ln 3 by te formula for derivative of a x wit a = 3. Using te cain rule wit inner function x, te derivative of te second part 3 x is 3 x ln 3( ) = 3 x ln 3. Tus y = 2 (3x ln x ln 3) = ln 3 2 (3x + 3 x ). Logaritmic function and teir derivatives. Recall tat te function log a x is te inverse function of a x : tus log a x = y a y = x. If a = e, te notation ln x is sort for log e x and te function ln x is called te natural logaritm. Te derivative of y = ln x can be obtained from derivative of te inverse function x = e y. Note tat te derivative x of x = e y is x = e y = x and consider te reciprocal: 2 y = ln x y = x = e y = x. Te derivative of logaritmic function of any base can be obtained converting log a y = log a x = ln x = ln x and using te formula for derivative of ln x. So we ave ln a ln a d log a x = x ln a = x ln a. to ln as Te derivative of ln x is and x te derivative of log a x is. x ln a To summarize, y e x a x ln x log a x y e x a x ln a x x ln a Example 4. Find te derivative of te following functions (a) y = ln(x 2 + 2x) (b) y = log 2 (3x + 4) (c) y = x ln(x 2 + ) (d) y = ln(x + 5e 3x ) 3

4 x 2 +2x Solution. (a) Using te cain rule wit te outer ln u and te inner x 2 + 2x, you ave y = 2x+2 (2x + 2) = = 2(x+). x 2 +2x x(x+2) (b) Using te cain rule wit te outer log 2 u and te inner 3x + 4, you ave y = ln 2(3x+4) 3 = 3 ln 2(3x+4). (c) Use te product rule wit f(x) = x and g(x) = ln(x 2 +) and te cain rule wit derivative of g wit te outer ln u and te inner x 2 +. Obtain tat y = f g + g f = ln(x 2 + ) + x 2 + (2x)(x) = ln(x 2 + ) + 2x2 x 2 +. (d) Use te cain rule wit te outer ln u and te inner u = x + 5e 3x. For derivative of te part e 3x, you will need to use te cain rule again to obtain u = + 5e 3x (3) = + 5e 3x. Tus, y = x+5e 3x ( + 5e 3x ) = +5e3x x+5e 3x. Trigonometric functions and teir derivatives. Te derivative of sin x can be determined using te trigonometric identity sin(x+) = sin x cos + sin cos cos x sin and calculating te limits lim 0 = and lim 0 = 0 using tables. Tus, d sin x = lim 0 sin(x+) sin x = lim 0 sin x cos +cos x sin sin x sin x(cos )+cos x sin = lim 0 cos = sin x lim 0 = sin x (0) + cos x () = cos x + cos x lim 0 sin Example 5. Find derivatives of te following functions. (a) y = sin 3 x (b) y = sin x 3 (c) y = x 3 sin x Solution. (a) In order to see better te inner and outer function in tis composite, note tat te function can be represented also as y = (sin x) 3. In tis representation it is more obvious tat te outer function is u 3 and te inner is sin x. Tus te cain rule produces y = 3(sin x) 2 cos x = 3 sin 2 x cos x. (b) Tis function is a composite of te outer sin u and te inner x 3. Tus te cain rule produces y = cos x 3 (3x 2 ) = 3x 2 cos x 3. (c) Using te product rule wit f(x) = x 2 and g(x) = sin x we obtain y = f g + g f = 3x 2 sin x + cos x(x 3 ) = 3x 2 sin x + x 3 cos x. Te derivative of cos x can be found to be sin x eiter using te trigonometric identity for cosine of a sum and similar arguments as above or te implicit differentiation. In section on Implicit Differentiation we demonstrate tis second metod. Te derivatives of te remaining trigonometric functions can be obtained by expressing tese functions in terms of sine or cosine. In general, you can always express a trigonometric function in terms of sine, cosine or bot and ten use just te following two formulas. Te derivative of sin x is cos x and te derivative of cos x is sin x. Example 6. Find derivatives of tan x and sec x. Simplify your answers. 4

5 Solution. Recall tat tan x = sin x so, using te quotient rule wit f(x) = sin x and g(x) = cos x, cos x obtain d tan x cos x cos x ( sin x) sin x = = cos2 x+sin x x = or cos 2 x cos 2 x cos 2 x sec2 x. Recall tat sec x = = (cos cos x x) so, using te cain rule, obtain tat d sec x = (cos x) 2 ( sin x) = (cos x) 2 sin x or sec 2 x sin x. Sometimes tis function is also written as sec x tan x. Practice problems.. Find te derivative of te given functions. (a) y = e 3x (x 3 + 2x 5) (b) y = 3 2x2 +5 (c) y = x 5 3x (d) y = (2x + e x2 ) 4 (e) y = e2x +e 2x x 2 (f) y = ln(5x e 5x ) (g) y = log 3 (x 2 + 5) () y = log 2 (x 2 + 7x) (i) y = sin(2x 2 + 4) (j) y = x 2 cos x 2 (k) y = sin 3x cos 5x (m) y = cot x 2 (l) y = log 2 x + 3 sin x xe x (n) y = e csc x 2. Find an equation of te line tangent to te curve at te indicated point. (a) f(x) = e2x e 2x + at x = 0. (b) f(x) = ln 2x at x =. 3. Assume tat f(x) is a function differentiable for every value of x. (a) If F (x) = e 3f(x), f(3) = 0 and f (3) = 2, determine F (3). (b) If F (x) = ln(f(x) + ), f() = 0 and f () =, determine F (). (c) If f(x) as te inverse and f(3) = 2 and f (3) = 6, find (f ) (2). 4. A differential equation is an equation in unknown function tat contains one or more derivatives of te unknown function. For example, y 2 +y = sin x and y +sin(xy) = 0 are differential equations. (a) Ceck if y = x 2 and y = 2 + e x3 are solutions of differential equation y + 3x 2 y = 6x 2. (b) Sow tat y = x+c is a solution of te differential equation y = y 2 for every value of te constant c. (c) Sow tat y = ce 2x is a solution of te differential equation y 3y + 2y = 0 for every value of te constant c. (d) Sow tat y = c cos 2x + c 2 sin 2x is a solution of differential equation y + 4y = 0 for every value of te constants c and c 2. (e) Find a value of te constant A for wic te function y = Ae 3x is a solution of te equation y 3y + 2y = 6e 3x. 5

6 5. Te concentration of pollutants (in grams per liter) in a river is approximated by C(x) =.04e 4x were x is te number of miles downstream from a place were te measurements are taken. (a) Determine te initial pollution and te pollution 2 miles downstream. (b) Determine ow muc te concentration canged on average witin te first two miles. (c) Determine ow fast te concentration canges 2 miles downstream. 6. A mass at te end of a vertical spring is stretced 5 cm beyond its natural lengt and is released. If s measures te lengt from te natural lengt, te position at time t (in seconds) can be described by s(t) = 5 cos 2t. (a) Find te velocity and acceleration at time t and grap te position, velocity and acceleration on te same plot for one period of te motion. (b) Mark te intervals on wic te object speeds up and te intervals on wic it speeds down on te plot in part (a). (c) Determine te times wen te mass returns to equilibrium position. (d) Determine te position, velocity, and acceleration 4 second after te object is released. Using tat data, determine te direction in wic te object is moving and if it is speeding up or slowing down. Do te same for t = 5 minutes after te motion started. Solutions.. (a) Use te product rule wit f(x) = e 3x and g(x) = x 3 +2x 5 and te cain for f (x) = e 3x (3) so tat y = 3e 3x (x 3 + 2x 5) + (3x 2 + 2)e 3x. (b) Use te cain rule wit inner 2x so tat y = 3 2x2 +5 ln 3 4x = 4x ln 3 3 2x2 +5. (c) Use te product rule wit f(x) = x and g(x) = 5 3x and te cain for g (x) = 5 3x ln 3(3) so tat y = 5 3x + 3x ln 5 5 3x. (d) Te cain rule wit inner 2x + e x2 and anoter cain rule for derivative of e x2 produces y = 4(2x + e x2 ) 3 (2 + e x2 2x) = 8( + xe x2 )(2x + e x2 ) 3. (e) Te quotient rule wit f(x) = e 2x +e 2x and g(x) = x 2 and te cain for f (x) = 2e 2x 2e 2x produces y = (2e2x 2e 2x )x 2 2x(e 2x +e 2x ) = 2x((x )e2x (x+)e 2x ) = 2((x )e2x (x+)e 2x ). x 4 x 4 x 3 (f) Use te cain rule wit inner 5x e 5x and anoter cain rule for derivative of e 5x so tat y = (5 5e 5x ) = 5 5e5x. 5x e 5x 5x e 5x (g) Te cain rule wit te outer log 3 u and te inner x 2 +5 produces y = ln 3(x 2 +5) 2x = 2x. ln 3(x 2 +5) () Te cain rule wit te outer log 2 u and te inner x 2 + 7x produces y = ln 2(x 2 +7x) (2x+7) = 2x+7 ln 2(x 2 +7x). (i) Te cain rule wit te outer cos u and te inner 2x produces y = cos(2x 2 + 4) (4x) = 4x cos(2x 2 + 4). (j) Te product rule wit f(x) = x 2 and g(x) = cos x 2 and te cain for g (x) = sin x 2 (2x) so tat y = 2x cos x 2 sin x 2 (2x)(x 2 ) = 2x cos x 2 2x 3 sin x 2. (k) Te product rule wit f(x) = sin 3x and g(x) = cos 5x and te cain for f (x) = cos 3x(3) and g (x) = sin 5x(5) so tat y = 3 cos 3x cos 5x 5 sin 5x sin 3x. 6

7 (l) Since te function is a sum of tree terms, you can differentiate term by term. Use te product rule for te last term. Obtain y = x ln cos x ex xe x. (m) Representing te function as y = cot x 2 cos x2 = and using te quotient rule wit f(x) = sin x 2 cos x 2 and g(x) = sin x 2 and te cain for f (x) = sin x 2 (2x) and g (x) = cos x 2 (2x) obtain tat y = sin x2 (2x) sin x 2 cos x 2 (2x) cos x 2 = 2x(sin2 x 2 +cos 2 x 2 ) = 2x. sin 2 x 2 sin 2 x 2 sin 2 x 2 (n) Use te cain rule wit te outer e u and te inner u = csc x = = (sin sin x x) so tat u = (sin x) 2 cos x and obtain tat y = e csc x (sin x) 2 cos x or y = e csc x csc 2 x cos x. 2. (a) Use te quotient rule to find f (x) = 2e2x (e 2x +) 2e 2x (e 2x ) and evaluate it at x = 0 to get (e 2x +) 2 te slope f (0) = 2(+) 2( ) = 4 0 = so te slope is. Since f(0) = = 0, te tangent (+) is y 0 = (x 0) y = x. (b) Eiter represent te function as f(x) = ln(2x ) /2 = ln(2x ) and use a single cain 2 rule to get f (x) = (2) = or keep it as f(x) = ln(2x 2 2x 2x )/2 and use two cain rules to obtain f (x) = (2x (2x ) /2 2 ) /2 (2) = =. In eiter case (2x ) /2 (2x ) /2 2x f () = = so te slope is. Since f() = ln 2 = ln = 0, te tangent line is 2 y 0 = (x ) y = x. 3. (a) Use te cain rule wit te outer e u and te inner 3f(x) to find te derivative of F (x) = e 3f(x) to be F (x) = e 3f(x) 3f (x). Since f(3) = 0 and f (3) = 2, F (3) = e 0 3 (2) = 6. (b) Use te cain rule wit te outer ln u and te inner f(x) + to find te derivative of F (x) = ln(f(x) + ) to be F (x) = f (x). Since f() = 0 and f () =, F () = =. f(x)+ 0+ (c) Since f(3) = 2, (f ) (2) = f (3). Ten since f (3) = 6, (f ) (2) = f (3) = (a) y = x 2 y = 2x. Plug te function and its derivative into te equation y + 3x 2 y = 6x 2 2x + 3x 2 (x 2 ) = 6x 2 2x + 3x 4 = 6x 2. Tis equation does not old for every value of x (for example if x = te equation false identity = 6) so y = x 2 is not a solution of te given equation. y = 2+e x3 y = 3x 2 e x3. Plug te function and its derivative into te equation y +3x 2 y = 6x 2 3x 2 e x3 + 3x 2 (2 + e x3 ) = 6x 2 3x 2 e x3 + 6x 2 + 3x 2 e x3 = 6x 2 6x 2 = 6x 2. Tis identity olds for every x so te given function is a solution of te equation. (b) Find te derivative of y = to be x+c y = and plug te function and its derivative (x+c) 2 into te equation y = y 2 = ( ) (x+c) 2 x+c =. Tis identity olds for every (x+c) 2 (x+c) 2 x so te given function is a solution of te equation. (c) Find te derivatives of y = ce 2x to be y = 2ce 2x and y = 4ce 2x and plug into te equation y 3y + 2y = 0 4ce 2x 6ce 2x + 2ce 2x = 0 ( )ce 2x = 0 0 = 0. Te given function is a solution of te equation. (d) y = c cos 2x + c 2 sin 2x y = 2c sin 2x + 2c 2 cos 2x y = 4c cos 2x 4c 2 sin 2x. Plugging into te differential equation y +4y = 0 gives you 4c cos 2x 4c 2 sin 2x+4c cos 2x+ 4c 2 sin 2x = 0 0 = 0. Te given function is a solution of te equation. (e) Find te derivatives of y = Ae 3x to be y = 3Ae 3x and y = 9Ae 3x and substitute tem into te equation y 3y + 2y = 6e 3x to get 9Ae 3x 9Ae 3x + 2Ae 3x = 6e 3x 2Ae 3x = 6e 3x 2A = 6 A = 3. 7

8 Tus, y = 3e 3x is a solution of differential equation. 5. (a) Te initial pollution is C(0) = 0.04 and te pollution 2 miles downstream is C(2) grams per liter. (b) Te average rate of cange witin first two miles is C(2) C(0) Tus 2 0 te concentration is decreasing on average by.02 grams per liter per mile during te first two miles. (c) Te derivative is C (x) =.04e 4x ( 4) =.6e 4x so tat C (2) = , tus te concentration is decreasing by grams per liter per mile 2 miles downstream. 6. (a) Differentiate s(t) = 5 cos 2t to find te velocity v(t) = s (t) = 0 sin 2t. Differentiate te velocity to find te acceleration a(t) = v (t) = s (t) = 20 cos 2t.Tese periodic functions ave period 2π = π so you 2 can grap tem on te interval [0, π]. Since te amplitudes of s, v and a are 5, 0 and 20, respectively, you can coose te range [-20, 20] so tat all te functions are displayed. (b) Te object speeds up on intervals were v and a ave te same sign and slows down on intervals were v and a ave te opposite sign. Tese intervals are marked on te grap. (c) Te te mass returns to equilibrium position wen s = 0. 5 cos 2t = 0 cos 2t = 0 2t = π, 3π, 5π... t = π, 3π, 5π (d) s(4) = 5 cos cm meaning tat te object is 0.75 above te equilibrium position. Te velocity is v(4) = 0 sin cm per second and, since it is negative, te object is moving upwards. a(4) = 20 cos cm per second squared. Since te velocity and acceleration ave different signs, te object is slowing down. Wen t = 5 min = 300 sec, s(300) = 5 cos cm meaning tat te object is almost at te igest point above te equilibrium position. Te velocity is v(300) = 0 sin cm per second. Te small value of velocity agrees wit te fact tat te object is almost at te igest point. Te negative sign indicates it is still moving upwards. a(300) = 20 cos cm per second squared. Since te velocity and acceleration ave different signs, te object is slowing down. 8