The Derivative. Philippe B. Laval Kennesaw State University

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1 The Derivative Philippe B. Laval Kennesaw State University Abstract This handout is a summary of the material students should know regarding the definition and computation of the derivative 1 Definition of the derivative.7,.8,.10) 1.1 Theory: Definition 1 Derivative of a function at a point) The derivative of the function y = fx) atapointx = a, denoted by f a) or dy is f a) = lim h 0 fa + h) fa) h The function f is said to be differentiable at a whenever the above limit exists. If.f is differentiable at every number in an open interval, then it is said to be differentiable in the interval. Definition Derivative function) The derivative of f, denoted f x) is defined by fx + h) fx) f x) = lim Definition 3 Second derivative) The second derivative of a function f, denoted f is the derivative of f. In other words, f x) =f x)) f a) represents the slope of the tangent line to the curve y = fx) at x = a. So, the equation of the tangent line to the curve y = fx) at x = a becomes y fa) =f a)x a) f a) also represents the instantaneous rate of change of y = fx) at x = a. In particular, if s = ft) is the position function of an object, then f a) represents the velocity of the object at time t = a. 1

2 If s = ft) is the position function of an object, then f a) is the acceleration of the object at time t = a. If y = f x), thenf a) measures how y is changing whenever x changes units of y by 1 unit. In fact, the units of the derivative are units of x. When looking at the graph of a function, it is easy to tell where the function is not differentiable. It will not be differentiable where it is not continuous, where there is a corner point, or where there is a vertical tangent. Places where a graph has a horizontal tangent correspond to places where the derivative is Meaning of the sign of f and f The sign of f is related to a function being increasing or decreasing as follows: If f x) > 0 on an interval then f is increasing on that interval. If f x) < 0 on an interval then f is decreasing on that interval. Definition 4 Concave up, down) The graph of a function is said to be concave up on an interval if it is above its tangents in that interval. It is concave down if it is below its tangents in that interval. Definition 5 Inflection point) An inflection point of a function f is a point where the concavity of the graph of f changes. The sign of f is related to concavity as follows: If f x) > 0 on an interval then f is concave up on that interval. If f x) < 0 on an interval then f is concave down on that interval. 1.3 Things to know: Given the graph of a function, be able to tell where the function is differentiable. The student should know the geometric meaning slope of the tangent) of the derivative. The student should be able to express the rate of change of a quantity as a derivative. The student should know and understand how the derivative is related to a function being increasing, decreasing, concave up, down. Related problems assigned:

3 Pages 155, 156, # 1, 3, 4, 5, 7, 13, 18, 19, 1, 5, 9. Pages , # 1, 3, 19, 1, 3, 9, 31, 3, 45. Pages 178, 180, # 1,, 11, 15. Sample problems: Exercise 6 Find the equation of the tangent line to the curve y = x + when x =0,whenx =1. answer: y =and y =x +1) Exercise 7 When an object is thrown up in the air, its altitude h is given by the following relation: h = 1 gt + v 0 t + h 0 Where g represents the gravity, v 0 is called the initial velocity at time t =0) and h 0 is the initial altitude.. Justify the name initial altitude Answer: Initial means when t =0.Whent =0, we see that h = h 0 this is why h 0 is called the initial altitude.) Compute the velocity of the object at time t, and justify the term initial velocity answer: vt) = gt + v 0.Whent =0, we see that v 0) = v 0 ) Find for which value of t the object will reach its maximum altitude. answer: t = v 0 g ) Exercise 8 Explain why the graph of a quadratic function y = ax + bx + c opens up if a>0 and opens down if a<0. answer: this is related to concavity) Exercise 9 If the population of the US is given as a function of time by the formula P t) =1.44t +180 where t is the number of years after 1940 and P in millions of people, at what rate was the population of the US changing in 1960, give units to your answer. answer: P 1960) = 1.44 million per year) Exercise 10 Study the sign of h t) if h t) represents the height of a person in inches) as a function of the person s age in years). answer: positive, zero, then negative) 3

4 Differentiation rules Ch. 3).1 Polynomial and exponential functions 3.1, 3.).1.1 Constant Rule Let C denote any constant. Then we have C) =0 or using Leibniz notation dc =0 It is easy to see why this rule holds. A constant function does not change. Therefore, its rate of change should be 0. Since the derivative is precisely the rate of change, the result follows. For people who prefer to think geometrically, it is also easy to see why the result is true. The graph of a constant function is a horizontal line, that is a line of slope 0. Since the derivative is the slope, the result follows. For a more mathematical proof, we need to apply the definition. See.1.11 onpage1foraproof.welookatsomeexamples. Example 11 Find d ) Since is a constant, d =0 d sin t) Example 1 Find assuming that x and t are not related. Since x and t are not related, this means that t and therefore sin t do not depend d sin t) on x. Therefore, =0.1. Differentiation of x x) =1 or using Leibniz notation =1 Again, it is easy to see why this rule is true. The slope of the function y = x is 1. Since the derivative is the slope, the result follows. In this formula, it is important not to take the variable x literally. What is important is to understand that we mean "rate of change of a variable with respect to itself". It does not matter what the variable is called. Example 13 Find dt dt dt dt =1. 4

5 Example 14 Find dr dr dr dr =1. Example 15 Find assuming that x does not depend on t. dt Thisissimilartoapreviousexample.Ifx does not depend on t, then dt = Power Rule This rule deals with powers of x. Assuming that n is a constant, then we have x n ) = nx n 1 or, using Leibniz notation n = nxn 1 Here again, the variable x should not be taken literally. The rule applies as long as the variable raised to the n th power is the same as the variable of dt n differentiation. In other words, = nt n 1. Similarly, drn dt dr = nrn 1. It is not as obvious to see why this rule should be true. It can be proven using the definition of the derivative. See.1.11 on page 1 for a proof. Example 16 Find By the power rule, Example 17 Find dt10 dt By the power rule, =x 1 =x dt 10 dt =10t 10 1 =10x 9 Example 18 Find 5, assuming that x is not a function of t. dt Since x is not a function of t, x does not change with t and therefore 5 dt =0. ) 1 Example 19 Find x 3 1 At first, it seems that this is not a power of x. However, x 3 = x 3 which is a 5

6 power of x. Therefore, ) 1 = x 3) x 3 = 3x 3 1 = 3x 4 = 3 x 4 You should always try to write your answer in a way similar to the way the problem was given. Example 0 Find x) As before, this does not look like a power of x. However, we know that x = x 1 which is a power of x. Therefore ) ) x = x 1 = 1 x 1 1 = 1 x 1 = 1 x 1 = 1 x Example 1 Find d x)3 Using x) 3) =3x) wouldbewrongbecausethisfunctionisnotapower of x. It is not x being raised to a power, but x, so the power rule does not apply. We need to learn one more rule before we can do it. The point of this problem is that when learning rules, it is important to know the rules. It is equally important to understand their limitations..1.4 Constant Multiple Rule Let C be a constant and f be a differentiable function. Then Or, using Leibniz notation Cf x)) = Cf x) d Cf x)) df x) = C In other words, the constant does not play a role in the differentiation. See.1.11 on page 1 for a proof. 6

7 Example Find x 4) By the constant multiple rule, we have x 4 ) = x 4 ) = 4x 3) we used the power rule) =8x 3 Example 3 Find d x)3 Remember, this is the problem we could not do earlier. Now, we can. First, we write x) 3 = 3 x 3 =8x 3 And therefore d x) 3 = d 8x 3) =8 3 =8 3x ) =4x.1.5 Sum and Difference Rules Suppose that f and g are two differentiable functions. Then, we have: f x)+g x)) = f x)+g x) and f x) g x)) = f x) g x) Or, using Leibniz notation d f x)+g x)) and d f x) g x)) = See.1.11onpage1foraproof. Example 4 Find 5x 3 +3x ) = df x) df x) dg x) + dg x) 5x 3 +3x ) = 5x 3 ) + 3x ) sum rule) =5 x 3) +3 x ) constant multiple rule) =15x +6x power rule) 7

8 .1.6 Differentiation of Exponential Functions Definition 5 Recall that an exponential function is a function of the form f x) =a x where a is a constant such that a>0 and a 1 Example 6 Some examples of exponential functions: 1. f x) = x. f x) =10 x 3. f x) = ) x 1 4. f x) =e x where e more on e below). Remark Sometimes, students are confused between exponential functions a x ) and power functions x n ). The distinction between them is easy, it has to do with where the variable is. In an exponential function, the variable is the exponent. In a power function, the variable is the base. We can find the derivative of an exponential function using the definition. Let f x) =a x.wewanttofindf x). f f x + h) f x) x) = lim a x+h a x = lim a x a h a x = lim rules of exponents) = a x a h 1 lim factored a x ) a h 1 The problem is that we cannot find lim as we do not know a. So far, this is what we have a x ) = a x a h 1 lim Let us investigate this limit further by looking at specific values of a. We will evaluate the limit by making a table of values. In the case a =,wehave h h 1 h h 1 Therefore, it appears that lim.693. The exact value is ln but you are not supposed to know that yet). 8

9 In the case a =3,wehave h h 1 h h 1 Therefore, it appears that lim The exact value is ln 3 but you are not supposed to know that yet). We see that when a =, lim h 0 a h 1 h a h < 1. When a =3, lim > 1. This suggests that the exists a a h 1 value of a between and 3 such that lim =1. This value is the number h 0 we named e. e h 1 Definition 7 The number e isthenumbersuchthat lim =1. It can be proven that it is an irrational number its decimal expansion is infinite and has no pattern, like π). Its approximate value is e Recall, we had a x ) = a x a h 1 lim e h 1 In the special case a = e and since lim =1,wewillhave e x ) = e x For now, this is the only exponential function we will know how to differentiate. Example 8 Find x +5e x) x +5e x) = x ) +5e x ) sum rule) =x +5e x ) power and constant multiple rules) =x +5e x h Example 9 Find der dr de r dr = er Example 30 Find der assuming that r is not a function of x. If r is not a function of x, thene r does not change with x. Therefore, de r =0 9

10 .1.7 Product rule Suppose that f and g are two differentiable functions. Then See.1.11onpage1foraproof. Example 31 Find x e x).1.8 Quotient rule f x) g x)) = f x) g x)+f x) g x) x e x) = x ) e x + x e x ) product rule) =xe x + x e x = xe x + x) Suppose that f and g are two differentiable functions. Then ) f x) = f x) g x) f x) g x) g x) g x)) See.1.11onpage1foraproof. Example 3 Find y for y = x + ex x. Since this function is a quotient, we first apply the quotient rule. ) x + e y x =.1.9 More Examples x = x + ex ) x x + e x ) x ) x ) = 1 + ex ) x x + e x )x) x 4 = x + x e x x xe x x 4 = x + xe x x ) x 4 = x + ex x ) x 3 All the applications we studied earlier which used the derivative can now be done more quickly by using the rules of differentiation. We look at some examples. 10

11 Example 33 Find xe x ) and determine where xe x is increasing. This function is a product of two functions, so we first apply the product rule. xe x ) =x) e x + x e x ) =1)e x + xe x = e x 1 + x) A function is increasing where its derivative is positive. Since e x is always positive, it follows that 1 + x) must also be positive, that is 1 + x > 0 or x> 1. So,xe x is increasing on the interval 1, ). Example 34 Find y for y = x + ex x. Since this function is a quotient, we first apply the quotient rule. ) x + e y x = x = x + ex ) x x + e x ) x ) x ) = 1 + ex ) x x + e x )x) x 4 = x + x e x x xe x x 4 = x + xe x x ) x 4 = x + ex x ) x 3 Example 35 Find f x) for f x) =x 3 By definition, f x) =f x)). So, we must first find f x). Therefore f x) = 3x ) f x) =3x power rule) =3 x ) constant multiple rule) =6x Example 36 The position of an object is given by h t) = 16t +64t+6 where h is in feet and t in seconds. Find the instantaneous velocity of the object for t =1, t =. The instantaneous velocity of the object is h t). We compute it using the rules of differentiation. h t) = 16t +64t +6 ) = 3t

12 Now, the instantaneous velocity at t =1is h 1) = 3 ft/s. The instantaneous velocity at t =is h ) = 0 ft/s Summary of the Rules Learned in this Subsection Let c and n denote any constants. Suppose that f and g are two differentiable functions. We have the following rules:: 1. c) =0. x) =1 3. cf x)) = c f x)) 4. x n ) = nx n 1 5. f x) ± g x)) = f x) ± g x) sum and difference rules) 6. f x) g x)) = f x) g x)+f x) g x) product rule) 7. ) f x) = g x) f x) f x) g x) g x) g x)) quotient rule) 8. e x ) = e x Remark When computing derivatives, you should always try to simplify the function first. You should also try to simplify your answer as much as possible. Chances are that after having computed a derivative, you will have to use it. You may need to find where it is 0, positive, negative. If it has been simplified, these tasks will be easier to perform Proofs of the Rules Learned in this Subsection 1. Proof of the constant rule. Let f x) =C. Wewanttoprovethatf x) =0. By definition, we have f f x + h) f x) x) = lim C C =lim =lim0) h 0 =0 1

13 . Proof of =1 Let f x) =x. Then, by definition f f x + h) f x) x) = lim x + h x =lim =lim h =1 h 0 h 3. Proof of the power rule. Note: though the rule is true for any constant n, we only prove it in the case n is a positive integer. Let f x) =x n,wheren is a constant. Then, by definition f f x + h) f x) x) = lim x + h) n x n =lim This computation is a little bit more difficult than the previous ones, we do it in several steps. First, x + h) n = x n + nx n 1 n n 1) h + x n h nxh n 1 + h n Therefore x + h) n x n = x n + nx n 1 n n 1) h + x n h nxh n 1 + h n x n = nx n 1 n n 1) h + x n h nxh n 1 + h n ) = h nx n 1 n n 1) + x n h nxh n + h n 1 It follows that f x + h) f x) h Therefore = h nx n 1 + = nx n 1 + f x + h) f x) lim =lim h 0 = nx n 1 ) n n 1) x n h nxh n + h n 1 h n n 1) x n h nxh n + h n 1 nx n 1 + Since all the other terms contain h as a factor. ) n n 1) x n h nxh n + h n 1 13

14 4. Proof of the constant multiple rule. Let C be a constant and suppose that f is differentiable, prove that Cf x)) = Cf x). Since we are assuming that f is differentiable, it means that f x) exists and f f x + h) f x) x) = lim. Let us define F x) =Cf x). We need to prove that F x) =Cf x). By definition, F F x + h) F x) x) = lim Cf x + h) Cf x) = lim since F x) =Cf x)) C x + h) f x)) = lim factor C) f x + h) f x) = C lim since C is a constant) = Cf x) definition of the derivative) 5. Proof of the sum rule Suppose that f and g are two differentiable functions. Let F x) =f x)+ g x). WewanttoprovethatF x) =f x)+g x). By definition F F x + h) F x) x) = lim fx + h)+gx + h) fx) gx) = lim since F x) =f x)+g x) ) fx + h) fx)+gx + h) gx) = lim [ ] fx + h) fx) gx + h) gx) = lim + h fx + h) fx) gx + h) gx) = lim + lim property of limits) = f x)+g x) definition of the derivative) 6. Proof of the difference rule The proof is very similar to the proof of the sum rule. We leave it as an exercise. 7. Proof of the product rule Suppose that f and g are two differentiable functions. Let F x) = f x) g x). WewanttoprovethatF x) =f x) g x)+f x) g x). By definition F F x + h) F x) x) = lim f x + h) g x + h) f x) g x) =lim definition of F ) 14

15 It helps when we know the result to prove. We are trying to obtain f x) g x)+f x) g x). If we think in terms of the definition of the derivative, that means that in the numerator we must have g x)f x + h) f x))+ f x)g x + h) g x)) that is, if we expand it, we realize that we must have in the numerator g x) f x + h) g x) f x)+f x) g x + h) g x) g x). This is more terms than what we currently have in the numerator. This means that we need to insert new terms. Of course, we cannot simply insert what is missing, we would change the problem. We use the standard trick of adding something and subtracting it immediately. This way we haven t changed anything. We then rearrange the terms. We decide to insert the following term: g x + h) f x)+g x + h) f x) g x + h) f x). You can see that this term is simply 0, so we have not changed anything. Here is what we obtain after inserting this term: F f x + h) g x + h) g x + h) f x)+g x + h) f x) f x) g x) x) = lim g x + h)fx + h) fx)) + f x)gx + h) gx)) = lim h 0 = lim h 0 [ h g x + h)fx + h) fx)) + h f x)gx + h) gx)) h g x + h)fx + h) fx)) f x)gx + h) gx)) = lim + lim property of limits) Let us evaluate each limit separately. g x + h)fx + h) fx)) lim = h 0 h lim g x + h) h 0 ) lim h 0 ] ) f x + h) f x) h Since g is differentiable, by a theorem studied in class, we also know that it is continuous. Therefore, lim g x + h) =g x). The second limit is, by h 0 definition, f x). So,we see that g x + h)fx + h) fx)) lim = f x) g x) h 0 h Similarly, we have f x)gx + h) gx)) lim = f x) g x) and therefore F x) =f x) g x)+f x) g x) 8. Proof of the quotient rule Suppose that f and g are two differentiable functions. Let F x) = f x) g x). 15

16 We want to prove that F x) = f x) g x) f x) g x) g x)). By definition F F x + h) F x) x) = lim = lim h 0 f x + h) g x + h) f x) g x) h f x + h) g x) f x) g x + h) = lim reduced to the same denominator g x + h) g x) and combined fractions) As in the proof of the product rule, we realize that we are missing terms to obtain what we need. So, we insert some extra terms. This time, we insert f x) g x)+f x) g x). We obtain F f x + h) g x) f x) g x)+f x) g x) f x) g x + h) x) = lim g x + h) g x) g x)fx + h) fx)) f x)gx + h) gx)) =lim g x + h) g x) g x)fx + h) fx)) f x)gx + h) gx)) =lim lim g x + h) g x) g x + h) g x) We evaluate each limit separately. As in the proof of the product rule, g x + h) =g x). So,wehave lim h 0 g x)fx + h) fx)) lim g x + h) g x) fx + h) fx) = g x) lim g x + h) g x) = g x) [ fx + h) fx) lim lim h 0 = g x) f x) g x)) = f x) g x) g x)) Similarly, we obtain that f x)gx + h) gx)) gx + h) gx) lim = f x) lim g x + h) g x) g x + h) g x) gx + h) gx) = f x) lim = f x) g 1 x) g x)) = f x) g x) g x)) lim h 0 ] 1 g x + h) g x) 1 g x + h) g x) 16

17 And therefore, combining the two limits don t forget the minus sign in between), we obtain the desired result. F x) = f x) g x) f x) g x) g x)).1.1 Problems Assigned for this Subsection Beabletodoproblemssuchas#1,3,5,7,9,11,13,15,17,19,31,37, 41, 43, 45, 47, 49, 54 on pages 196, 198. Be able to do problems such as # 1, 3, 5, 7, 11, 13, 17, 1, 7, 35, 36, 37, 4, 44 on pages Findarulefor ) 1 x n. Findarulefor n x) 3. Prove the difference rule. 17

18 . Trigonometric functions 3.4) In this section, we study differentiation of the trigonometric functions sin x, cos x, tan x, cot x, sec x, csc x. Before we derive the formulae for the derivative of these functions, we must first revisit limits...1 Fundamental Trigonometric Limits When deriving the formula for the derivative of sin x and cos x, two limits play sin x cos x 1 an important role. They are lim and lim. These are not limits x 0 x x 0 x we can evaluate by using the rules we have developed earlier. To first get an idea of what the value of these limits might be, we look at a table of value. x sin x x sin x It therefore appears that lim =1. We are going to prove this result by x 0 x using the squeeze theorem. The goal, when using the squeeze theorem, is to find Figure 1: Trigonometric Function in a Unit Circle two functions f and g such that f x) sin x x g x), such that lim x 0 f x) = 18

19 sin x lim g x) =L. By the squeeze theorem, it will follow that lim x 0 x 0 x find these functions using geometry. From figure 1, one can see that = L. We Area of triangle OAP Area of sector OAP Area of triangle OAQ 1) But, remembering that the area of a triangle is 1 base height, andthatwe have a unit circle, that is OA =1, we see that and Area of triangle OAP = 1 1 sin x ) = sin x Area of triangle OAQ = 1 1 tan x 3) = tan x = sin x cosx Also remembering see your calculus book, on the first page) that the area of the sector of angle θ in a circle of radius r is 1 r θ, since we are in a unit circle, we see that Area of sector OAP = 1 x 4) Using equations, 3 and 4 in equation 1, we have sin x Multiplying each side by gives 1 x sin x cosx sin x x sin x cos x Dividing each side by sin x gives 1 x sin x 1 cos x Since lim cos x = 1, it follows that lim x 0 cos x = 1. theorem, x lim x 0 sin x =1 x 0 1 Hence, by the squeeze 19

20 This is not the limit we wanted. We can now get the limit we wanted using our limit rules. sin x lim x 0 x = lim 1 x 0 x sin x = lim x 0 1 x sin x = 1 1 =1 cos x 1 We are now ready to find lim. x 0 x cos x 1 cos x 1) cos x +1) lim = lim x 0 x x 0 x cos x +1) cos x 1 = lim x 0 x cos x +1) sin x = lim x 0 x cos x +1) sin x = lim sin x x 0 x cos x +1 ) ) sin x sin x = lim lim x 0 x x 0cos x +1 ) 0 = 1) 1+1 =0 Therefore, we have proven the following theorem: Theorem 37 Fundamental Trigonometric Limits) lim x 0.. Differentiation of Sine sin x lim x 0 x cos x 1 x =1 5) We are now ready to find the derivative of sin x, using the definition. Let us first remind the reader of some trigonometric identities which will be useful sin a + b) = sin a cos b +sinb cos a 6) sin a b) = sin a cos b sin b cos a cos a + b) =cosa cos b sin a sin b cos a b) =cosa cos b +sina sin b =0 0

21 sin x) sin x + h) sin x) =lim sin x) cos h)+sinh) cos x) sin x) =lim using equation 6) =lim h 0 sin x)cosh) 1) + sin h) cos x) h sin x)cosh) 1) sin h)cosx) =lim + lim Since we are taking the limit as h 0, both sin x and cos x are constants with respect to this limit. Therefore, sin x) cos h) 1 sin h) =sinx lim +cosx lim =sinx 0) + cos x 1) by equation 5) =cosx..3 Differentiation of Cosine The derivation of the formula for cos x is almost identical to that of sin x and is left as an exercise. We have: You will notice the minus sign. cos x) = sin x..4 Differentiation of Tangent BY definition, tan x = sin x cos x. Therefore, tan x) = sin x cos x ) = sin x) cos x sin x cos x) cos x = cos x sin x ) cos x = 1 cos x =sec x quotient rule)..5 Differentiation of the Other Trigonometric Functions Since the remaining trigonometric functions can be expressed as ratios of sin x and cos x,wecanusetheruleswehavetoderivetheformulafortheirderivatives. 1

22 The derivation of these formulae is left as an exercise. You will recall that cot x = cos x sin x = 1 tan x, sec x = 1 cos x, csc x = 1 sin x.wehave: cot x) = csc x sec x) =secx tan x csc x) = csc x cot x..6 Summary for this Subsection Theorem 38 We have proven and you must know the following rules: 1. sin x) =cosx or d sin x = cos x. cos x) = sin x or d cos x = sin x 3. tan x) =sec x or d tan x =sec x 4. cot x) = csc x or d cot x = csc x 5. sec x) =secx tan x or d sec x =secx tan x 6. csc x) = csc x cot x or d csc x = csc x cot x Remark As in the formulas studied before, it is important to understand that x should not be taken literally. What matters is that the variable of differentiation be the same as the variable in the function. In other words, we have d sin x = cos x butwealsohave d sin u =cosu or d sin θ =cosθ, du dθ d sin u and so on. On the other hand, cannot be done using this rule because the variable of the function is not the variable of differentiation. If u is some function of x, thenwehavetowaituntilthenextsection before we can do it. This means that we cannot find derivatives such as d sin x ) or d cos ex ) d tan x) or even. On the other hand, as noticed several times above, if u isnotafunctionofx, then d sin u =0...7 Examples Example 39 Find f x) for f x) =x sin x. f x) =x sin x) =x) sin x + x sin x) =sinx + x cos x product rule)

23 Example 40 Find f x) for f x) = sin x cos x f x) =sinx cos x) = sin x) cos x +sinx cos x) product rule) = cos x sin x Example 41 Find f x) for f x) = 1 sec x tan x f x) = 1 sec x) tan x 1 sec x) tan x) tan x = 0 sec x tan x)tanx 1 sec x) sec x ) tan x = sec x tan x sec x + sec 3 x tan x = sec x sec x tan x ) sec x = tan x sec x 1 sec x) tan x recall that 1+tan x =sec x) Example 4 Find the slope of the tangent to f x) = cos x at x = π 3. This is a classical problem. We must find a point on the tangent and the slope of the tangent. When x = π π ) π ) 3, y = f =cos = 1 π 3 3. Therefore, 3, 1 ) is a point on the tangent. The slope of the tangent to f x) at x = π π ) 3 is f.sincef x) = cos x, 3 π ) f x) = sin x and therefore f = sin π =. It follows that the equation of the tangent is y 1 = 3 x π ) 3..8 Problems Assigned for this Subsection 1. Be able to do problems such as # 1, 3, 5, 7, 9, 17, 5, 7, 36 on pages 3, 4 sin 5x. Find lim Answer: 1) x 0 5x sin 5x 3. Find lim Answer: 5) x 0 x 3

24 sin x 4. lim x 0 5x Answer: 1 5 ) 5. Prove that cos x) = sin x 6. Prove that cot x) = csc x 7. Prove that sec x) =secx tan x 8. Prove that csc x) = csc x cot x 4

25 .3 The chain rule 3.5).3.1 Introduction Before we continue, let us look at what we know so far. To help understand better, it helps classifying the rules we know in two categories. There are rules which apply to specific functions, like the rules which give the derivative of x n, e x, sin x, cos x, tan x, cot x, sec x, csc x. However, most functions are a combination of these elementary functions. So, there are rules for these combinations as well. There are different ways in which functions can be combined. We can add or subtract functions. For this, we use the sum or difference rule. We can multiply function. For this, we use the product rule. We can divide functions. For this, we use the quotient rule. There is another way to combine functions we haven t looked at yet. We can compose functions. If f and g are two functions, we can form f g x)). Ifg x) =x and f x) = sin x then f g x)) = sin x ). If f x) =x n and g x) =tanx then f g x)) = tan x) n. And so on. The chain rule will show us how to find the derivative of this way of combining functions. In other words, we will have to use the chain rule in examples similar to the examples below: d sin x 6) because we do not have sin x but sin x 6). de tan x d sin x) 3 because we do not have e x but e tan x. because we do not have x 3 but sin x) 3 As mentioned above, the chain rule is applied when we are differentiating functions of the form f g x)). Wewillrefertof as the outside function, and g as the inside function. For example, sin x ) is of the form f g x)) with g x) =x and f x) = sin x. Another example, sin 3 x is of the form f g x)) with g x) = sin x and f x) =x 3. You should know start to have an idea of when the chain rule should be applied. Now, let us see how it is applied. Proposition 43 If f gx)) and g x) both exist then, fgx))) = f gx))g x). If we let u = g x), then the formula becomes f u)) = f u) u. We can also write the chain rule using Leibniznotation. If u = g x) and y = f u) then dy = dy du du. The formula fu)) = f u)u suggests that the derivative of f u) is the derivative of the outside function, f, evaluated at the inside function, u, times the derivative of the inside function. Let us look at some examples. 5

26 Example 44 Find y for y =sin x ) The outside function is sin x, the inside function is x. In other words, it is of the form sin u) with u = x.therefore, y = sin x )) =cos x ) x ) since sin x) =cosx) =x cos x ) Example 45 Find y for y = e sin x. This is of the form e u with u =sinx. The differentiation formula for e u gives us: y = e sin x) = sin x) e sin x since e x ) = e x ) =cosxe sin x Example 46 Find y for y =sin 4 x). This is of the form u 4 with u =sinx. Hence, we use the power rule. y = sin 4 x ) = 4 sin 3 x sin x) = 4 sin 3 x cos x Example 47 Find y for y = tan u) where u is a differentiable function. y = tan u)) = u sec u Example 48 Find y for y = f sin x) where f is a differentiable function. y =f sin x)) = cos xf sin x) Sometimes, the chain rule will have to be applied more than once. Example 49 Find y for y =tan sin x )). Thus is of the form tan u with u =sin x ). Therefore, y = tan sin x ))) =sec sin x )) sin x )) We need the chain rule again to find sin x )) sin x )) =cos x ) x ) And therefore, =x cos x ) y =x sec sin x )) cos x ) 6

27 .3. Differentiation of Exponential Functions Revisited) We are now ready to derive a formula for the derivative of every exponential function. Recall, an exponential function is a function of the form a x. To find the derivative of a x, we first need to rewrite it. We do it as follows. Since ln x and e x are inverses of each other, we have: And therefore We are now ready to find a x ). a x ) = e x ln a) Example 50 Find y for y = x. a = e ln a a x = e ln a) x = e x ln a = e x ln a x ln a) chain rule) =lnae x ln a ln a is a constant) Example 51 Find y for y = sin x. y = x ) =ln) x y = sin x) =ln) sin x sin x) = ln cos x sin x chain rule).3.3 Summary We can now combine the chain rule to all the previous rules we learned to obtain more general rules. These are the rules we will remember. For example, we know that sin x) =cosx. If instead of sin x, wehavesin u where u is some function of x, then its derivative is sin u) = cos u) u = u cos u So, we see that the rule is almost the same as the original rule, with one extra term, the derivative of the inner function. If we do the same to all the rules we already know, assuming that u denotes some function of x, we obtain: 7

28 1. u n ) = nu n 1 u. e u ) = u e u 3. a u ) = u ln aa u 4. sin u) = u cos u 5. cos u) = u sin u 6. tan u) = u sec u 7. cot u) = u csc u 8. sec u) = u sec u tan u 9. csc u) = u csc u cot u.3.4 Problems Assigned for this Subsection 1. Be able to do problems such as # 1, 3, 5, 7, 9, 11, 15, 1, 9, 57, 71 on pages

29 .4 Implicit Differentiation 3.6).4.1 General Idea Until now, all the rules we have studied applied to functions of the form y = f x). In other words, we were given a formula which expressed y explicitly in terms of x. There may be instances in which y is not expressed explicitly in terms of x. For example, the equation of the circle of radius r centered at the origin is given by x + y = r. This relation indicates that x and y are related. However, y is not given explicitly in terms of x. In this case, we say that y is given implicitly in terms of x. Suppose that we need to find thew slope of the tangent at a point on the circle. To do this, we would need to compute y or dy. Howcanwedoitifwedonothaveaformulaoftheformy = f x)? You might argue that in this particular case, one could solve for y and then take the derivative of the obtained formula. What about if the formula is too complicated to allow solving for y? This is when implicit differentiation comes into play. It is a technique which allows to find y even if we do not know y explicitly as a function of x. The technique works as follows: To find y when y is not given explicitly as a function of x, take the derivative on each side of the equal sign of a relation. For example, if you have to find dy in x + y =1,differentiate each side with respect to x to obtain: x + y ) =1) x ) + y ) =0 x +yy =0 y = x y Remark 5 During this process, remember that you are differentiating with respect to x. y must be treated as a function of x. So, for example, if y appears in the relation, its derivative will be y ) =yy by the chain rule. Similarly, if sin y appears, its derivative will be y cos y. Example 53 Find y if sin x +siny = x. Since y is given implicitly in terms of x, we use implicit differentiation. sin x +siny) =x) sin x) + sin y) =1 cos x +cosy) y =1 cos y) y =1 cos x y = 1 cos x if cos y 0 cos y Example 54 Consider the curve given by x 3 + y 3 =6xy. This curve is known as the folium of Descartes. 9

30 1. Find y. Find the equation of the tangent at the point 3, 3) 3. Find where the tangent is horizontal, vertical. We first show the graph of this curve.x 3 + y 3 =6xy Answer 1. To find y, we use implicit differentiation. x 3 + y 3) =6xy) x 3 ) + y 3 ) =6xy) 3x +3y y =6[x y + xy ] 3x +3y y =6y +6xy We now need to solve for y. We do this by grouping the terms containing y. 3y y 6xy =6y 3x 3y y x ) =3 y x ) y = y x y x Answer. Since we already have a point on the tangent, we simply need its slope. The slope of the tangent at 3, 3) is y evaluated at 3, 3). Ifwecallm the 30

31 slope, we have 3) 3 m = 3 3) = 1 Therefore, the equation of the tangent is y 3= x 3) y 3= x +3 y = x +6 Answer 3. Looking at the picture, we guess that there are two horizontal tangents and two vertical tangents. Horizontal tangents. The tangent is horizontal when y =0.Since y is a fraction, this will happen when the numerator of this fraction is 0 in other words when y x =0 y = x Substituting in the original equation gives ) x x =6x x x 3 + x6 8 =3x3 x 6 8 =x3 x 6 =16x 3 x 6 16x 3 =0 x 3 x 3 16 ) =0 This happens when either x =0or x 3 =16or x = 3 16 = 4 3. The corresponding y values are 0 or 8 3 =5 3. Thus, the folium of ) Descartes. has a horizontal tangent at 0, 0) and 4 3, 5 3. Vertical tangents. There will be a vertical tangent where the denominator of y is 0 that is where x = y. We can perform the same computation as we did above. We can also notice that x and y play a similar role. Since the condition we obtain here is similar to the one above, x and y being interchanged. It means that the answer will be similar, x and y being) interchanged. Thus, there is a vertical tangent at 0, 0) and 5 3,

32 .4. Assigned Problems 1. Be able to do problems such as # 3, 11, 13, 4, 45, 47, 49 on pages 43, 45 3

33 .5 Derivative of Inverse Trigonometric functions and Logarithm functions.3.6, 3.7).5.1 General Idea Remember that the inverse of a function f is the function denoted f 1 satisfying the following: y = f 1 x) x = f y) Thegoalistofindy if y = f 1 x). To do this, we start by using the above relation to get x = f y). Then, we use implicit differentiation x = f y) x) =f y)) 1=y f y) y = 1 f y) How we finish depends of the function f. We illustrates this below for three different functions: sin x, tan x, and ln x..5. Derivative of sin 1 x If y = sin 1 x then x = siny when π differentiation y π. Now, we use implicit Since x =siny 1=y cos y y = 1 cos y sin y + cos y =1 cos y = 1 sin y It follows that y = 1 1 sin y = 1 1 x Thus, if we combine this formula with the chain rule, we get: sin 1 u ) u = 1 u 33

34 .5.3 Derivative of tan 1 x If y =tan 1 x then x =tany. Using implicit differentiation: x =tany 1=y sec y y = 1 sec y = 1 1+tan y = 1 1+x Thus, if we combine this formula with the chain rule, we get:.5.4 Derivative of ln x tan 1 u ) = u 1+u If y =lnx then x = e y. Using implicit differentiation, x = e y 1=y e y y = 1 e y = 1 x Thus, if we combine this formula with the chain rule, we get: ln u) = u u.5.5 Logarithmic Differentiation When we have a complicated expression or one for which we do not have a rule) made of products, quotients, or powers, it might be easier to first take the logarithm of the expression, then use implicit differentiation. This process is called logarithmic differentiation. Here is an illustration. Before we look at a specific example, let us recall the properties of the logarithmic functions we will use: ln ab) =lna +lnb a ) ln =lna ln b b ln a r )=r ln a 34

35 Example 55 Find y for y = x ) x 3 x We first look at ln y which we simplify using the properties of the logarithmic functions. ) x ) x 3 ln y =ln x =lnx ) +lnx 3 ln x =lnx ) + 3 ln x 1 ln x Now, we use implicit differentiation. ln y) = lnx ) + 3 ln x 1 ) ln x Finally, we solve for y y y = x + 3 x 1 x = x + 5 x y = y x + 5 ) x Example 56 Find y for y = x tan x Using implicit differentiation, = x ) x 3 x x + 5 x ln y =lnx tan x =tanx ln x ln y) = tan x ln x) y y = sec x ln x + tan x x ) Thus y = y sec x ln x + tan x ) x = x tan x sec x ln x + tan x ) x 35

36 .5.6 Assigned Problems 1. Be able to do problems such as # 7, 8, 9, 33 on page 44. Be able to do problems such as # 3, 7, 9, 11, 13, 17, 19, 3, 5, 31, 33 on pages 50, 51 36

37 3 Summary of Things to Know in the Entire Document Know the sum, difference, product and quotient rules. Know how to differentiate polynomial and exponential functions. Know the fundamental trig limits. Using the fundamental trig limits, be able to derive the formulas for the derivatives of the trig functions. Know the derivatives of the trig functions. Know the chain rule, and be able to combine it with the other rules you know. Know how to use implicit differentiation to find y or dy between x and y. given a relation Using implicit differentiation, be able to find the derivative of inverse functions. Know the rules to differentiate the inverse of the trigonometric functions. Know the rule to differentiate logarithmic functions. Beabletodoproblemssuchas#1,3,5,7,9,11,13,15,17,19,31,37, 41, 43, 45, 47, 49, 54 on pages 196, 198. Be able to do problems such as # 1, 3, 5, 7, 11, 13, 17, 1, 7, 35, 36, 37, 4, 44 on pages Be able to do problems such as # 1, 3, 5, 7, 9, 17, 5, 7, 36 on pages 3, 4 Be able to do problems such as # 1, 3, 5, 7, 9, 11, 15, 1, 9, 57, 71 on pages Be able to do problems such as # 3, 11, 13, 4, 7, 8, 9, 33, 45, 47, 49 on pages 43, 45 Be able to do problems such as # 3, 7, 9, 11, 13, 17, 19, 3, 5, 31, 33 on pages 50, 51 37

38 4 Additional Practice Problems 1. Find a formula for f n) x) if f x) =xe x. Using the formula, find where f is increasing, decreasing, concave up, concave down. Also, indicate the values of x at which f has a horizontal tangent. answer: f n) x) =e x x + n). increasing on 1, ), decreasing otherwise. Concave up on, ), concave down otherwise. horizontal tangent at x = 1). Find y in each case below a) y =sin 1 e x ) answer: y e x = ) 1 e x b) y = sin x ) 1+x answer: y = x 1+x ) cos x ) sin x )) 1 + x ) ) c) xe y = y 1 answer: y = ey 1 xe y ) d) y =ln x +1 x 1 answer: y = x x x +1)x 1) ) e) y = x ex answer: y =lnx) e x x ex + e x x ex 1 ) 3. Find an equation of the tangent to the curve x + y =3at the point 4, 1). answer: y = x +3) 4. A particle is moving along a vertical line and its position is given by y = t 3 1t +3, t 0 a) Find the velocity and the acceleration functions. answer: v t) =3t 1, a t) =6t) b) When is the particle moving upward? downward? answer: upward when t>) 38

39 5 Differentiation Rules Let C and n denote any constant, let u denote a function of x, leta denote a real number such that a>0 and a 1. We have proven and you must know the following rules: 1. C) =0. Cf) = Cf 3. f ± g) = f ± g sum and difference rules) 4. fg) = f g + fg product rule) 5. ) f = gf fg g g quotient rule) 6. fgx))) = f gx))g x) chain rule) 7. u n ) = nu n 1 u power rule) 8. e u ) = u e u 9. a u ) = u ln aa u 10. sin u) = u cos u 11. cos u) = u sin u 1. tan u) = u sec u 13. cot u) = u csc u 14. sec u) = u sec u tan u 15. csc u) = u csc u cot u 16. sin 1 u ) u = 1 u 17. tan 1 u ) = u 18. ln u) = u u 1+u 39

Derivatives: rules and applications (Stewart Ch. 3/4) The derivative f (x) of the function f(x):

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