Chapter 3: Topics in Differentiation

Transcription

1 Chapter 3: Topics in Differentiation Summary: Having investigated the derivatives of common functions in Chapter (i.e., polynomials, rational functions, trigonometric functions, and their combinations), this chapter begins by asking if the slope of a general curve can be found even if the curve does not have an eplicit function representation. Instead, the assumption here is that the dependent variable y is implicitly described by an equation involving both and y but where y = f() is not known. This process of finding dy/d then allows one to find tangent lines to a curve even if an eplicit description of the curve such as y= f() is not known. Implicit differentiation then leads to an etended version of the power rule, derivatives of eponential functions, and also derivatives of the inverse trigonometric functions. Along the way, derivatives of logarithms are developed based upon a limit given in Chapter and the derivatives of inverse functions are discussed based upon the reflective properties of f and f across the line y =. Finally, the chapter concludes with a discussion of L Hôpital s rule which is useful for evaluating limits that have an appropriate indeterminate form. OBJECTIVES: After reading and working through this chapter you should be able to do the following:. Find the slope of a curve that is defined implicitly ( 3.).. Take derivatives of logarithmic functions ( 3.). 3. Take derivatives of eponential functions and inverse trigonometric functions ( 3.3). 4. Use logarithmic differentiation to find derivatives of complicated functions ( 3.3). 5. Use derivatives to relate the rates of different quantities that depend upon the same parameter such as time ( 3.4). 6. Approimate functions using linearizations or differentials ( 3.5). 7. Use L Hôpital s rule to evaluate limits that have indeterminate forms such as 0 0,, 0,,, 0 0 and 0 ( 3.6). 49

2 50 3. Implicit Differentiation PURPOSE: To give rules for epressing the derivative of a function that is written implicitly. implicit means that y = f() is not known Implicit Differentiation. take derivatives using the chain rule. solve for dy/d This section approaches the problem of being able to find the slope of a curve even if an eplicit formula for the curve is unknown. In other words, most curves are described by a formula of the form y= f(). However, it may not always be practical or possible to find such an epression for y (i.e., consider y+sin(y)=3). In such cases, the chain rule can be used to find the slope dy/d. The basic idea in this section is to assume that y is a function of, that is y= f(), and then to differentiate the relationship describing the curve with respect to by using the chain rule. In the end, the derivative dy/d may be epressed in terms of and y. This brings up two questions.. Can we find dy/d? differentiate with the chain rule. How is dy/d evaluated? ( differentiate implicitly ) ( use the = chain rule ) The first question can be answered using the chain rule. In fact, the phrase differentiate implicity is just another way of saying use the chain rule. For eample, if we have a curve described by 3 + y+y 3 =, then it is assumed that y= f() is a function that corresponds to at least a portion of the graph of this curve. Then the relationship can be rewritten as follows. 3 + f()+[f()] 3 = Now to differentiate, we take the derivative of each term with respect to on both sides of the equation. This leads us to the following. 3 + f()+f ()+3[ f()] f ()=0 Here we have used the product rule to find d [ f()] and the chain rule to find d d ( [ f()] 3 ) as shown. But since y = f() and dy/d = f () we usually write d this statement as 3 + y+ dy dy + 3y d d = 0 Now we only have to solve a linear equation to find dy/d. This is a key feature of implicit differentiation. IDEA: After differentiating an implicit equation, dy/d is found by solving a linear equation.

3 5 Once the equation of the curve has been differentiated, there will only remain a linear equation for dy/d to be solved. To solve for dy/d we put all of the dy/d terms on one side, factor and divide. In the eample above, this looks like the following: solve a linear equation for dy/d dy dy + 3y d d = 3 y Put dy/d terms on one side. dy( +3y ) = 3 y d Factor out dy/d. dy d = 3 y +3y Solve for dy/d. The second key concern is how to evaluate the derivative dy/d. The answer is that you now need both an and a y value since dy/d will often depend upon both. However, arbitrary values of and y cannot be used. Instead, the values of and y must satisfy the initial relationship that implicitly defined y in terms of. In the eample above, this means that the ordered pair (,y) must satisfy the relationship 3 + y+y 3 = 3. One way to find a valid point is to pick an value (i.e., =0 or some other value) and then solve the resulting equation for y. [ 3 + y+y 3 = 3 ] =0 evaluating dy/d both a value for y and are needed to find dy/d y 3 = 3 y=3 (/3).44 Once dy/d is known at a particular point( 0,y 0 ) then it is a simple matter to find a tangent line using, for eample, the point-slope form of a line. Because of the way that dy/d is solved for, the derivative is often written as some sort of fraction. This information may be used to find where the curve has either horizontal or vertical tangents. To find horizontal tangents, the top of the fraction is set equal to zero. To find vertical tangents, the bottom of the fraction is set equal to zero. It is not quite as simple as this as in some cases, the top and bottom of the fraction may be zero simultaneously. This may be indicative of a vertical tangent and it may not. In the above eample, to find horizontal tangents to the curve 3 + y+y 3 = 3, we set the top of the derivative equal to zero. 3 y=0 Then the curve may have a horizontal tangent where y= 3. This does not give very specific information about and y but it is enough. This relationship may be substituted back into the original curve to find the coordinates of a point with a horizontal tangent.

4 5 Checklist of Key Ideas: y defined eplicitly and implicitly finding the slope of an implicitly defined curve implicit differentiation epressing dy/d in terms of and y finding tangent lines to implicitly defined curves etended power rule for rational powers 3. Derivatives of Logarithmic Functions PURPOSE: To give rules for calculating the derivatives of logarithmic functions. lim h 0 (+h)/h = e Logarithmic Differentiation. take logarithms of the epression. differentiate 3. solve for dy/d logarithm property ln(ab)=lna+lnb This section develops the derivatives of logarithmic functions by using the limit description of e (see margin), the properties of logarithms, and the definition of the derivative. The derivatives of logarithms give rise to a process called logarithmic differentiation. The process can be summarized by taking a logarithm of a given epression and then taking derivatives of the resulting epression. IDEA: The log property of log(ab)=loga+logb can remove the need for the product and/or quotient rules. The idea behind logarithmic differentiation is to take advantage of the property of logarithms to turn multiplication inside a logarithm into a sum of terms. This also simplifies the process of taking derivatives of functions with powers and etends the power rule to cases with arbitrary real eponents. Checklist of Key Ideas: derivatives of logarithmic functions using properties of logarithmic functions before differentiation logarithmic differentiation etended power rule for general powers

5 Derivatives of Eponential and Inverse Trigonometric Functions PURPOSE: To give rules for differentiating eponential functions and inverse trigonometric functions. In this section, implicit differentiation and the properties of inverse functions are both used to find the derivatives of eponential functions and inverse trigonometric functions. The first observation is that if a function is differentiable at a point (a,b) and its slope is not zero then the inverse of the function should be differentiable at the point (b,a). This result can be seen as a geometric description of inverses since f and f are reflections of each other across the line y =. The concepts of increasing and decreasing are briefly mentioned in order to discuss the invertibility of a given function. Essentially, if f () > 0 then f() is said to be increasing and if f ()<0 then f() is decreasing. This is important because when a function is strictly decreasing or increasing then it will be one-to-one (or passes the horizontal line test) and so will have an inverse function. More will be said about whether a function is increasing and decreasing in the net chapter. Using implicit differentiation and the relationship between inverses, the derivatives of eponential functions and inverse trigonometric functions may now be found. For eample, the function y = e may also be written as = lny. Then dy/d may be found using implicit differentiation. This approach also works for the inverse trigonometric functions. differentiability of f increasing and decreasing derivative of a derivative of sin derivative of tan Checklist of Key Ideas: differentiability of inverse functions finding derivatives of inverse functions increasing and decreasing functions and f () derivatives of eponential functions derivatives of functions of the form y= f() g() derivatives of inverse trigonometric functions 3.4 Related Rates PURPOSE: To use derivatives to relate the rates of change of multiple functions. In this section, derivatives are applied to problems that may involve more than one function of an independent variable. Most often, the independent variable that is considered is time, t. The main idea here is that a real situation may eist such that

6 54 two or more quantities can be related to each other using algebraic equations (such as area, a, and radius, r of a circle). Then both of these quantities are assumed to be functions of time, t. The most confusing aspect of this is that t does not show up eplicitly in the equations and definitions for the quantities are not known (i.e., a(t)=? and r(t)=? in the previous circle mentioned). As it turns out, all that is needed is the information that these quantities are functions of t. Then the equation that relates them can be differentiated. The result is an equation that relates their rates or relates their derivatives. It may be helpful to write π r(t) as For eample, if a=πr in a circle then we assume that area and radius are funcπ [r(t)] tions of time, t. a(t)=πr(t) Then we differentiate this equation. a (t)=πr(t)r (t) On the right hand side of the last equation, the derivative of r(t) was obtained using the chain rule: let y=u and y=r(t) and then apply the chain rule. dy = du du dy = u r (t)=r(t)r (t) Formulas are still not known for a(t), r(t) or their derivatives. But if numerical information is known about them at a particular time, then frequently we can discover the value of how one of the quantities is changing with respect to time. Checklist of Key Ideas: relating the rates of change of different variables strategy for solving related rates problem identifying equations to relate the variables identifying variables as functions of time, t 3.5 Local Linear Approimations; Differentials PURPOSE: To use a tangent line to approimate a function. This section builds upon the notion at the beginning of this chapter that we can use the derivative of a function to find the equation of a tangent line to the function. The tangent line can then be used to approimate the values of the function nearby the point of tangency. This is the idea of a local linear approimation. The tangent line of a function is this local linear approimation. In some cases, it may be simpler to obtain the tangent line of a function and then use this in place of actual function values. For eample, if the value of a function and its derivative are

7 55 known at a point but a definition of the actual function is not known, then a tangent line approimation can be used. IDEA: The derivative f () is the slope of the tangent line. Differentials take the approimation idea a step further. The notion of a differential for the function y= f() given by dy= f ()d is just another way of using a local linear approimation. The derivative f () is also the slope of the tangent line. But near a point of tangency, the slope should be equal to a small change in y divided by a small change in. Then dy= f ()d says that a small change in y equals the slope of the tangent line times a small change in. This would be eact for a line but this is only an approimation for a general function. In this section and y are used to denote actual (eact) changes in -values and function values. Usually, d is assumed to be equal to a desired. But then dy becomes an approimation to y. Sometimes y is called the error in y and so dy would be an approimation to this error. On the other hand, y gives a percentage y error or relative error in the value of y. Then dy would be approimation of this y relative error. Checklist of Key Ideas: differentiable functions are locally linear general equation of tangent line at = 0 using tangent line as a linear approimation differentials and differential form of derivatives actual change, y, and estimated change, dy error, relative error and percentage error general differential formulas 3.6 L Hôpital s Rule; Indeterminate Forms PURPOSE: To introduce a method for finding limiting values when a limit evaluates as an indeterminate form. The purpose of this section is to develop a method for finding limiting values when a limit evaluates as an indeterminate form. The method in this section is called L Hôpital s rule. The primary usefulness of L Hôpital s rule is when a limit has

8 56 Note: Multiple applications of L Hôpital s rule may be needed L Hôpital s rule cannot be applied directly to indetermine products, quotients or powers 0 is not indeterminate the indeterminate form of 0/0 or /. In either of these cases, the rule says that f() the limit of the ratio of the functions, i.e., lim, will be equivalent to the limit a g() f () of the ratio of their derivatives, i.e., lim a g (). The power of L Hôpital s rule is often abused. If a limit does not have one of the indeterminate forms 0/0 or / then the rule cannot be used. The consequence of this is that the rule cannot be directly applied to cases where a limit has a indeterminate form such as 0,, 0 0, 0 or. In these situations, the limit epressions need to be rewritten before applying the rule. This can often be done by algebraic techniques (using a common denominator, rationalizing an epression with a radical or writing an epression as the reciprocal of its reciprocal, i.e., f() = / f() ) or by using the properties of logarithms (lnab = blna). It should be noted that 0 is not an indeterminate form, nor is a 0 where a < and a 0. Checklist of Key Ideas: indeterminate forms of type 0/0 and / using L Hôpital s rule growth of eponential functions versus polynomials and rational functions indeterminate forms of types 0 and and L Hôpital s rule indeterminate forms of types 0 0, 0 and and L Hôpital s rule

9 57 Chapter 3 Sample Tests Section 3.. Answer true or false. If y= 5 4+ then dy d = Answer true or false. If y 3 = then dy d = 3y. 3. Find dy/d if 3 y cos =. dy/d= 3y /3 sin dy/d=3y /3 sin dy/d= 6y /3 sin dy/d=6y /3 sin 4. Find dy/d if + y = 5. 5 y y y 5 y 5. Answer true or false. If y + y= 5 then dy d = 5 y+. 6. Suppose that + 3y = 9. Find d y/d. d y d = 6 + 4y 36y 3 d y d = 6 + 4y 6y 3 d y d = 6 4y 36y 3 d y d = 6y 4 9y 3 7. Find the slope of the tangent line to + y = 5 at the point (,) Find the slope of the tangent line to y = 4 at the point(,) Find dy/d if y 4 = y 4 3 y 4 4y 3 3 4y y 4 4y 3 0. Find dy/d if =cos(y). sin(y) sin(y) +ysin(y) sin(y) +ysin(y) sin(y). Answer true or false. If cos=siny then dy/d=tan.. Answer true or false. If tan(y)=4 then dy d = ysec (y). 3. y = 3 has a tangent line parallel to the -ais at (,) (0,0) (,) (,) 4. + y = 6 has tangent lines parallel to the y-ais at (0, 6) and (0,6) (0, 4) and(0, 4) ( 4, 0) and(4, 0) ( 6,0) and (6,0) 5. If y= 3, find the formula for y. y= y= 3 + y = 3 3 (+ ) 3 3 y = 3 3 (+ )

10 58 6. If y= /3 + then dy d = 6 7/6 + / 6 7/6 + / ( ) 3 /3 + ( ) ) 3 /3 + + /( / Find the equation of the tangent line to the function y=( + 4) /3 at =. y= ( )+ y= ( )+8 y= 3 ( )+ y= 3 ( )+8 Section 3.. If y = ln 6 then find dy/d ln6. If y = ln(cos ) then find dy/d. tan tan cos cos 3. If y= 3+ln then dy/d= 3+ln 3+ln 3+ln ln 3+ln 4. Answer true or false. If y=ln( 3 ) then dy d = 3. [ ] 5. Find dy/d if y=ln. sin cos sin cot tan cot + dy 6. If y= 8 then find by logarithmic differentiation. +3 d ( ) + 7/ (+3) ( 8 + ) If y=ln(k) then dn y d n = k n n ( ) n k n n ( ) n n (n )! 8. Answer true or false. If the position of a particle along a line is given by s(t)=lnt for t > 0, then the limiting velocity of the particle as t is zero. 9. If y=ln() then y has a horizontal tangent line at which of the following -values? = =0 = y has no horizontal tangents. 0. Suppose that f()=(cos()) π. What is dy/d? π(cos()) π (cos()) π π( sin()) π π sin()(cos()) π. If y + ln(y)=y 3 then dy/d=

11 59 y+ y 3y (y+ y 3y ) +y y ( y 3y ) none of the above. If y=sec(ln) then y = sec(ln)tan(ln) sec(ln) tan(ln) sec ( ) tan sec (ln) ( 3. If y= dy then ln d = (ln) (ln) (ln) then which of the following epres- 4. If y = ( + ) 3 sin cos(3) sions is y y? ) cot+3tan(3) csc sec(3) 5. If y= ln 3sin(3) 4 ( + )cos tan( ) ln 3 then dy/d= Section 3.3. Answer true or false. If y= 8 e 7 then dy/d=56 7 e 7.. If y=(ln)e then dy/d= (ln)e + e (ln)e e (ln)e e 3. Answer true or false. If +e y = then dy d = yey e y. 4. Let f()=5. Find d d [ f()]. 5 ln5 5 ln5 5 ln 5. Answer true or false. If f()=π sin+cos then dy/d=(sin+cos)π sin+cos. 6. Answer true or false. If y= cos then dy ( cos ) d = sinln cos. 7. The equation y = y is satisfied by y=e y=cos y=sin y=e 3 h 8. Evaluate the limit. lim = h 0 h 0 ln3 9. If y=sin () then find dy/d. 4 4

12 60 0. If y=tan then find dy/d. (+) + (+ ) +. If y=e sin then dy/d= e sin (sin )e sin (cos)e sin (cos )e sin. Find dy/d if y=ln(cos ). cos +(cos ) ( )cos ( )cos 3. Let y= sin. Then dy/d= y= 4 y= ( sin )( ) y= 4 y= ( sin )( 4 ) 4. If sin y=ln then find dy/d. ( ) y ( ) + y sin y sin y sin y+ sin y Section 3.4. The volume of a sphere is given by V = 4 3 πr3. Find dv terms of dr. dv dv dv dv = 4πr dr = 4 dr πr3 3 = 4 dr πr 3 = 3r dr. A cylinder of length m and radius m is epanding such that dl/ = 0.0 m/sec and dr/ = 0.0 m/sec. Find dv/ m/s m/s m/s m/s 3 3. A 0-ft ladder rests against a wall. If it were to slip so that when the bottom of the ladder is 6 feet from the wall it will be moving at 0.0 ft/s, how fast would the ladder be moving down the wall? 0.0 ft/s ft/s 0.05 ft/s 0. ft/s 4. A plane is approaching an observer with a horizontal speed of 00 ft/s and is currently 0,000 ft from being directly overhead at an altitude of 0,000 ft. Find the rate at which the angle of elevation, θ, is changing with respect to time, dθ/ rad/s 0.00 rad/s 0.08 rad/s rad/s 5. Answer true or false. Suppose that z=y 5 3. Then dz/ =(dy/) 5 +(d/) Suppose that z= + y. Find dz/. d d + dy + y dy in

13 6 ( ) d ( ) dy + d + y dy + y 4. Answer true or false. If V = 3 y then dy = dv d Answer true or false. If V = 0 3 then d = dv The power in watts for a certain circuit is given by P=I R. How fast is the power changing if the resistance, R, of the circuit is,000 Ω, the current, I is A, and the current is decreasing with respect to time at a rate of 0.03 A/s w/s 60 w/s 0 w/s.8 w/s 8. Gravitational force is inversely proportional to the distance between two objects squared. If F = 5 N at a distance d d = 3 m, how fast is the force diminishing if the objects are moving away from each other at m/s? N/s 0.74 N/s 6.7 N/s. N/s 9. A point P is moving along a curve whose equation is given by y= When P=(,5), y is increasing at a rate of units/s. How fast is changing? units/s 5 8 units/s 64 units/s 5 6 units/s 0. Water is running out of an inverted conical tank at a rate of 3 ft 3 /s. How fast is the height of the water in the tank changing if the height is currently 5 ft and the radius is 5 ft? ft/s ft/s.3 ft/s 9.45 ft/s. Answer true or false. If z=lny then dz =. Answer true or false. If z=e lny then dz = e dy y + e lny d. 3. Answer true or false. If sinθ = 3y then dθ ( d )( ) dy. dy d = 3 +3y. Section 3.5. If y= 4, find the formula for y. y=(+ ) 4 y=4 3 y=4( ) 3 y=(+ ) 4 4. If y=cos, find the formula for y. y=cos(+ ) cos y=cos(+ ) y= sin y= +cos 3. Answer true or false. The formula for dy is obtained from the formula for y by replacing with d. 4. Let y= 5. Find the formula for dy. dy=(+d) 5 dy=(+d) 5 5 dy= 5 +(d) 5 dy=5 4 d 5. Let y=tan. Find the formula for dy. dy=(sec )d dy=(sectan)d dy=tan(+d) tan dy=tan(+d) 6. Let y= 3 sin. Find the formula for dy. dy=(3 sin+ 3 cos)d dy=3(+d) 3 sin(+d) dy=(3 cos)d dy=(+d) 3 sin(+d) 3 sin 7. Let y=. Find dy at = if d= Let y= 5. Find dy at = if d= 0.0.

14 Let y=. Find y at =3 if = Use dy to approimate 3.96 starting at = Answer true or false. A circular spill is spreading so that when its radius r is m, dr = 0.05 m. The corresponding change in the area covered by the spill, A, is 0.63 m (to the nearest hundreh).. A small suspended droplet of radius 0 microns is evaporating. If dr = 0.00 micron, find the change in the volume, dv, to the nearest thousanh of a cubic micron cubic microns.57 cubic microns 0.49 cubic microns 4.89 cubic microns 3. Answer true or false. A cube is epanding as temperature increases. If the length of the cube is changing at a rate of = mm when is m, the volume is eperiencing a corresponding change of mm 3 (to the nearest thousanh). 4. Answer true or false. The radius of the base of a cylinder is mm with a possible error of ±0.0 mm. The height of the cylinder is eactly 4 m. Using differentials to estimate the maimum error of the volume, it is found to be mm 3 (to the nearest hundreh). Section 3.6. lim 0 sin7 sin = lim 9 3 = 0 tan 3. lim = lim 0 + ln(+) e = 0 e 5. lim 4 = 0 6. lim lne = 0 7. lim 0 (+3) / = 3 ln 3 e 3 sin 8. lim 0 + ln(+) = 0

15 63 ( cos ) 9. Answer true or false. lim 0 cos ( ) =. ( 0. lim e ) = 0. lim 0 +( ln ) = 0 sin3. Answer true or false. lim 0 cos = Answer true or false. lim =. ( ) 4. Answer true or false. lim = 0. ( sin 5. lim ) = Chapter 3 Test. If y=ln(5 ) then find dy/d. 5. Answer true or false. If y=3ln e 3 then dy d = 3e3 + 9ln e If f()=8 then find d d [ f()]. 8 ln ln8 8 8 ln8 4. Find dy/d if y=tan 4. 3/4 (+ ) + (+) Answer true or false. If y= cos + then dy d = ( cos +)( 4 ). 6. lim 0 sin9 sin8 = 9 8 sin 7. lim = Answer true or false. lim 0 ( 8+ ) = e If y= 9 then find the formula for y. y= y= y= y= Answer true or false. If e y = y then y = y yey e y. (. Evaluate the limit: lim 0 + ) sin 0 Cannot be determined.

16 64. Answer true or false. If + y = 4 then y = 4y If y=ln( ) then dy d = ln() ln() 4. Answer true or false. If y=sin(arctan(e )) then dy d = e (+e ) / e 3 (+e ) 3/. 5. Evaluate the limit: lim 0 (cos) sin 0 Cannot be determined. 6. Evaluate the limit: lim (ln(3 ) ln) 0 ln() ln(3) 7. Answer true or false. lim = e 0 8. Evaluate the limit: lim If cos ( y) ( ) = ln y then dy d = y y y y ysin(y/) 0. Answer true or false. The function y=e has no horizontal tangent lines.. Evaluate the limit: lim 0 + ln 0. If y= + then dy d = (+) + (+ + ln) ln() + Cannot be determined. 3. If y= tan, find the formula for dy. dy=(tan)d dy=( sec )d dy=(tan+ sec )d dy=(tan sec )d 4. Answer true or false. If y= 5 and = then dy= 5 6 d. 5. Answer true or false. A spherical balloon is deflating. The rate at which the volume is changing when r = m is given by dv = 6πdr m 3 /s.

17 65 Chapter 3: Answers to Sample Tests Section 3.. false. true 3. a 4. c 5. false 6. d 7. b 8. d 9. b 0. d. false. false 3. a 4. c 5. a 6. a 7. c Section 3.. c. b 3. d 4. true 5. d 6. c 7. c 8. true 9. c 0. d. b. a 3. c 4. a 5. c Section 3.3. false. a 3. true 4. a 5. false 6. true 7. d 8. d 9. d 0. a. a. c 3. b 4. b Section 3.4. a. b 3. c 4. c 5. false 6. d 7. c 8. b 9. b 0. a. false. true 3. false 4. false 5. true Section 3.5. d. a 3. false 4. d 5. a 6. a 7. b 8. d 9. b 0. b. true. b 3. true 4. true Section 3.6. a. a 3. d 4. b 5. c 6. d 7. d 8. b 9. false 0. d. b. false 3. true 4. false 5. d Chapter 3 Test. a. false 3. d 4. a 5. false 6. d 7. b 8. false 9. b 0. true. b. true 3. c 4. false 5. b 6. c 7. false 8. c 9. d 0. false. a. b 3. c 4. false 5. true

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