chapter 04

January 27, 2005 11:44 L24-ch04 Sheet number 1 Page number 127 black

127

CHAPTER 4

Derivatives of Logarithmic, Exponential, andInverse Trigonometric Functions

EXERCISE SET 4.1

1. y = (2x− 5)1/3; dy/dx =23(2x− 5)−2/3

2. dy/dx =13[2 + tan(x2)

]−2/3sec2(x2)(2x) =

23x sec2(x2)

[2 + tan(x2)

]−2/3

3. dy/dx =23

(x+ 1x− 2

)−1/3x− 2− (x+ 1)

(x− 2)2= − 2

(x+ 1)1/3(x− 2)5/3

4. dy/dx =12

[x2 + 1x2 − 5

]−1/2d

dx

[x2 + 1x2 − 5

]=

12

[x2 + 1x2 − 5

]−1/2 −12x(x2 − 5)2

= − 6x(x2 − 5)3/2

√x2 + 1

5. dy/dx = x3(−23

)(5x2 + 1)−5/3(10x) + 3x2(5x2 + 1)−2/3 =

13x2(5x2 + 1)−5/3(25x2 + 9)

6. dy/dx = −3√2x− 1x2 +

1x

23(2x− 1)2/3

=−4x+ 3

3x2(2x− 1)2/3

7. dy/dx =52[sin(3/x)]3/2[cos(3/x)](−3/x2) = −15[sin(3/x)]3/2 cos(3/x)

2x2

8. dy/dx = −12[cos(x3)

]−3/2 [− sin(x3)](3x2) =

32x2 sin(x3)

[cos(x3)

]−3/2

9. (a) 1 + y + xdy

dx− 6x2 = 0,

dy

dx=

6x2 − y − 1x

(b) y =2 + 2x3 − x

x=

2x+ 2x2 − 1,

dy

dx= − 2

x2 + 4x

(c) From Part (a),dy

dx= 6x− 1

x− 1xy = 6x− 1

x− 1x

(2x+ 2x2 − 1

)= 4x− 2

x2

10. (a)12y−1/2 dy

dx− cosx = 0 or

dy

dx= 2√y cosx

(b) y = (2 + sinx)2 = 4 + 4 sinx+ sin2 x sody

dx= 4 cosx+ 2 sinx cosx

(c) from Part (a),dy

dx= 2√y cosx = 2 cosx(2 + sinx) = 4 cosx+ 2 sinx cosx

11. 2x+ 2ydy

dx= 0 so

dy

dx= −x

y

12. 3x2 + 3y2 dy

dx= 3y2 + 6xy

dy

dx,dy

dx=

3y2 − 3x3y2 − 6xy

=y2 − x2

y2 − 2xy

13. x2 dy

dx+ 2xy + 3x(3y2)

dy

dx+ 3y3 − 1 = 0

(x2 + 9xy2)dy

dx= 1− 2xy − 3y3 so

dy

dx=

1− 2xy − 3y3

x2 + 9xy2


128 Chapter 4

14. x3(2y)dy

dx+ 3x2y2 − 5x2 dy

dx− 10xy + 1 = 0

(2x3y − 5x2)dy

dx= 10xy − 3x2y2 − 1 so

dy

dx=

10xy − 3x2y2 − 12x3y − 5x2

15. − 12x3/2 −

dydx

2y3/2 = 0,dy

dx= −y

3/2

x3/2

16. 2x =(x− y)(1 + dy/dx)− (x+ y)(1− dy/dx)

(x− y)2,

2x(x− y)2 = −2y + 2xdy

dxso

dy

dx=

x(x− y)2 + y

x

17. cos(x2y2)[x2(2y)

dy

dx+ 2xy2

]= 1,

dy

dx=

1− 2xy2 cos(x2y2)2x2y cos(x2y2)

18. − sin(xy2)[y2 + 2xy

dy

dx

]=

dy

dx,dy

dx= − y2 sin(xy2)

2xy sin(xy2) + 1

19. 3 tan2(xy2 + y) sec2(xy2 + y)(2xy

dy

dx+ y2 +

dy

dx

)= 1

sody

dx=

1− 3y2 tan2(xy2 + y) sec2(xy2 + y)3(2xy + 1) tan2(xy2 + y) sec2(xy2 + y)

20.(1 + sec y)[3xy2(dy/dx) + y3]− xy3(sec y tan y)(dy/dx)

(1 + sec y)2= 4y3 dy

dx,

multiply through by (1 + sec y)2 and solve fordy

dxto get

dy

dx=

y(1 + sec y)4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y

21. 4x− 6ydy

dx= 0,

dy

dx=

2x3y

, 4− 6(dy

dx

)2

− 6yd2y

dx2 = 0,

d2y

dx2 = −3(dydx

)2− 2

3y=

2(3y2 − 2x2)9y3 = − 8

9y3

22.dy

dx= −x

2

y2 ,d2y

dx2 = −y2(2x)− x2(2ydy/dx)

y4 = −2xy2 − 2x2y(−x2/y2)y4 = −2x(y3 + x3)

y5 ,

but x3 + y3 = 1 sod2y

dx2 = −2xy5

23.dy

dx= −y

x,d2y

dx2 = −x(dy/dx)− y(1)x2 = −x(−y/x)− y

x2 =2yx2

24. y + xdy

dx+ 2y

dy

dx= 0,

dy

dx= − y

x+ 2y, 2

dy

dx+ x

d2y

dx2 + 2(dy

dx

)2

+ 2yd2y

dx2 = 0,d2y

dx2 =2y(x+ y)(x+ 2y)3

25.dy

dx= (1 + cos y)−1,

d2y

dx2 = −(1 + cos y)−2(− sin y)dy

dx=

sin y(1 + cos y)3


Exercise Set 4.1 129

26.dy

dx=

cos y1 + x sin y

,

d2y

dx2 =(1 + x sin y)(− sin y)(dy/dx)− (cos y)[(x cos y)(dy/dx) + sin y]

(1 + x sin y)2

= −2 sin y cos y + (x cos y)(2 sin2 y + cos2 y)(1 + x sin y)3

,

but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1 so

d2y

dx2 = − sin 2y + y(sin2 y + 1)(1 + x sin y)3

27. By implicit differentiation, 2x + 2y(dy/dx) = 0,dy

dx= −x

y; at (1/2,

√3/2),

dy

dx= −

√3/3; at

(1/2,−√3/2),

dy

dx= +

√3/3. Directly, at the upper point y =

√1− x2,

dy

dx=

−x√1− x2

=

− 1/2√3/4

= −1/√3 and at the lower point y = −

√1− x2,

dy

dx=

x√1− x2

= +1/√3.

28. If y2 − x + 1 = 0, then y =√x− 1 goes through the point (10, 3) so dy/dx = 1/(2

√x− 1). By

implicit differentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = −√x− 1

goes through (10,−3) so dy/dx = −1/(2√x− 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.

29. 4x3 + 4y3 dy

dx= 0, so

dy

dx= −x

3

y3 = − 1153/4 ≈ −0.1312.

30. 3y2 dy

dx+ x2 dy

dx+ 2xy + 2x− 6y

dy

dx= 0, so

dy

dx= −2x y + 1

3y2 + x2 − 6y= 0 at x = 0

31. 4(x2 + y2)(2x+ 2y

dy

dx

)= 25

(2x− 2y

dy

dx

),

dy

dx=

x[25− 4(x2 + y2)]y[25 + 4(x2 + y2)]

; at (3, 1)dy

dx= −9/13

32.23

(x−1/3 + y−1/3 dy

dx

)= 0,

dy

dx= −y

1/3

x1/3 =√3 at (−1, 3

√3)

33. 4a3 da

dt− 4t3 = 6

(a2 + 2at

da

dt

), solve for

da

dtto get

da

dt=

2t3 + 3a2

2a3 − 6at

34.12u−1/2 du

dv+

12v−1/2 = 0 so

du

dv= −√u√v

35. 2a2ωdω

dλ+ 2b2λ = 0 so

dω

dλ= − b2λ

a2ω36. 1 = (cosx)

dx

dyso

dx

dy=

1cosx

= secx


130 Chapter 4

37. (a)

–4 4

–2

2

x

y

(b) Implicit differentiation of the equation of the curve yields (4y3 + 2y)dy

dx= 2x− 1 so

dy

dx= 0

only if x = 1/2 but y4 + y2 ≥ 0, so x = 1/2 is impossible.

(c) x2 − x− (y4 + y2) = 0, so by the Quadratic Formula x =1±

√1 + 4y2 + 4y4

2= 1 + y2,−y2

which gives the parabolas x = 1 + y2, x = −y2.

38. (a) y

–2

0

2

1 2

x

(b) 2ydy

dx= (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If

dy

dx= 0 then

3x2 − 2(a+ b)x+ ab = 0. By the Quadratic Formula

x =2(a+ b)±

√4(a+ b)2 − 4 · 3ab6

=13

[a+ b± (a2 + b2 − ab)1/2

].

(c) y = ±√x(x− a)(x− b). The square root is only defined for nonnegative arguments, so it is

necessary that all three of the factors x, x− a, x− b be nonnegative, or that two of them benonpositive. If, for example, 0 < a < b then the function is defined on the disjoint intervals0 < x < a and b < x < +∞, so there are two parts.

39. (a) y

x

2

–2 2

–2

(b) x ≈ ±1.1547

(c) Implicit differentiation yields 2x− xdy

dx− y + 2y

dy

dx= 0. Solve for

dy

dx=

y − 2x2y − x

. Ifdy

dx= 0

then y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y2 = x2 − 2x2 + 4x2 = 3x2, x = ± 2√3.

40. (a) See Exercise 39 (a)(b) Since the equation is symmetric in x and y, we obtain, as in Exercise 39, x ≈ ±1.1547.



(c) Implicit differentiation yields 2x− xdy

dx− y + 2y

dy

dx= 0. Solve for

dx

dy=

2y − x

y − 2x. If

dx

dy= 0

then 2y − x = 0 or x = 2y. Thus 4 = 4y2 − 2y2 + y2 = 3y2, y = ± 2√3, x = 2y = ± 4√

3.

41. Solve the simultaneous equations y = x, x2−xy+y2 = 4 to get x2−x2+x2 = 4, x = ±2, y = x = ±2,so the points of intersection are (2, 2) and (−2,−2).From Exercise 39 part (c),

dy

dx=

y − 2x2y − x

. When x = y = 2,dy

dx= −1; when x = y = −2, dy

dx= −1;

the slopes are equal.

42. Suppose a2 − 2ab+ b2 = 4. Then (−a)2 − 2(−a)(−b) + (−b)2 = a2 − 2ab+ b2 = 4 so if P (a, b) lieson C then so does Q(−a,−b).

From Exercise 39 part (c),dy

dx=

y − 2x2y − x

. When x = a, y = b thendy

dx=

b− 2a2b− a

, and when

x = −a, y = −b, then dy

dx=

b− 2a2b− a

, so the slopes at P and Q are equal.

43. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use

implicit differentiation to getdy

dx= − 2xy

x2 + 2ayso at (1,1), − 2

1 + 2a= −4

3, 1 + 2a = 3/2, a = 1/4

and hence b = 1 + 1/4 = 5/4.

44. The slope of the line x + 2y − 2 = 0 is m1 = −1/2, so the line perpendicular has slope m = 2(negative reciprocal). The slope of the curve y3 = 2x2 can be obtained by implicit differentiation:

3y2 dy

dx= 4x,

dy

dx=

4x3y2 . Set

dy

dx= 2;

4x3y2 = 2, x = (3/2)y2. Use this in the equation of the curve:

y3 = 2x2 = 2((3/2)y2)2 = (9/2)y4, y = 2/9, x =32

(29

)2

=227

.

45. (a)

–3 –1 2

–3

–1

2

x

y (b) x ≈ 0.84

(c) Use implicit differentiation to get dy/dx = (2y−3x2)/(3y2−2x), so dy/dx = 0 if y = (3/2)x2.Substitute this into x3 − 2xy + y3 = 0 to obtain 27x6 − 16x3 = 0, x3 = 16/27, x = 24/3/3and hence y = 25/3/3.

46. (a)

–3 –1 2

–3

–1

2

x

y (b) Evidently the tangent line at the pointx = 1, y = 1 has slope −1.


132 Chapter 4

(c) Use implicit differentiation to get dy/dx = (2y−3x2)/(3y2−2x), so dy/dx = −1 if 2y−3x2 =−3y2+2x, 2(y−x)+3(y−x)(y+x) = 0. One solution is y = x; this together with x3+y3 = 2xyyields x = y = 1. For these values dy/dx = −1, so that (1, 1) is a solution.

To prove that there is no other solution, suppose y = x. From dy/dx = −1 it followsthat 2(y − x) + 3(y − x)(y + x) = 0. But y = x, so x + y = −2/3. Then x3 + y3 =(x + y)(x2 − xy + y2) = 2xy, so replacing x + y with −2/3 we get x2 + 2xy + y2 = 0, or(x+y)2 = 0, so y = −x. Substitute that into x3 +y3 = 2xy to obtain x3−x3 = −2x2, x = 0.But at x = y = 0 the derivative is not defined.

47. (a) The curve is the circle (x − 2)2 + y2 = 1 about the point (2, 0) of radius 1. One tangentline is tangent at a point P(x,y) in the first quadrant. Let Q(2, 0) be the center of thecircle. Then OPQ is a right angle, with sides |PQ| = r = 1 and |OP | =

√x2 + y2. By

the Pythagorean Theorem x2 + y2 + 12 = 22. Substitute this into (x − 2)2 + y2 = 1 toobtain 3 − 4x + 4 = 1, x = 3/2, y =

√3− x2 =

√3/2. So the required tangent lines are

y = ±(√3/3)x.

(b) Let P (x0, y0) be a point where a line through the origin is tangent to the curvex2 − 4x+ y2 + 3 = 0. Implicit differentiation applied to the equation of the curve givesdy/dx = (2− x)/y. At P the slope of the curve must equal the slope of the line so(2− x0)/y0 = y0/x0, or y2

0 = 2x0 − x20. But x

20 − 4x0 + y2

0 + 3 = 0 because (x0, y0) is on thecurve, and elimination of y2

0 in the latter two equations gives x20 − 4x0 + (2x0 − x2

0) + 3 = 0,x0 = 3/2 which when substituted into y2

0 = 2x0 − x20 yields y2

0 = 3/4, so y0 = ±√3/2. The

slopes of the lines are (±√3/2)/(3/2) = ±

√3/3 and their equations are y = (

√3/3)x and

y = −(√3/3)x.

48. Let P (x0, y0) be a point where a line through the origin is tangent to the curve2x2 − 4x+ y2 + 1 = 0. Implicit differentiation applied to the equation of the curve givesdy/dx = (2− 2x)/y. At P the slope of the curve must equal the slope of the line so(2− 2x0)/y0 = y0/x0, or y2

0 = 2x0(1− x0). But 2x20 − 4x0 + y2

0 + 1 = 0 because (x0, y0) is on thecurve, and elimination of y2

0 in the latter two equations gives 2x0 = 4x0− 1, x0 = 1/2 which whensubstituted into y2

0 = 2x0(1 − x0) yields y20 = 1/2, so y0 = ±

√2/2. The slopes of the lines are

(±√2/2)/(1/2) = ±

√2 and their equations are y =

√2x and y = −

√2x.

49. The linear equation axr−10 x+ byr−1

0 y = c is the equation of a line �. Implicit differentiation of the

equation of the curve yields raxr−1+rbyr−1 dy

dx= 0,

dy

dx= −ax

r−1

byr−1 . At the point (x0, y0) the slope

of the line must be −axr−10

byr−10

, which is the slope of �. Moreover, the equation of � is satisfied by

the point (x0, y0), so this point lies on �. By the point-slope formula, � must be the line tangentto the curve at (x0, y0).

50. Implicit differentiation of the equation of the curve yields rxr−1+ryr−1 dy

dx= 0. At the point (1, 1)

this becomes r + rdy

dx= 0,

dy

dx= −1.

51. By the chain rule,dy

dx=

dy

dt

dt

dx. Use implicit differentiation on 2y3t+ t3y = 1 to get

dy

dt= −2y3 + 3t2y

6ty2 + t3, but

dt

dx=

1cos t

sody

dx= − 2y3 + 3t2y

(6ty2 + t3) cos t.

52. (a) f ′(x) =43x1/3, f ′′(x) =

49x−2/3

(b) f ′(x) =73x4/3, f ′′(x) =

289x1/3, f ′′′(x) =

2827x−2/3

(c) generalize parts (a) and (b) with k = (n− 1) + 1/3 = n− 2/3



53. y′ = rxr−1, y′′ = r(r − 1)xr−2 so 3x2[r(r − 1)xr−2

]+ 4x

(rxr−1

)− 2xr = 0,

3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0,3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3

54. y′ = rxr−1, y′′ = r(r − 1)xr−2 so 16x2[r(r − 1)xr−2

]+ 24x

(rxr−1

)+ xr = 0,

16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0,16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4

55. We shall find when the curves intersect and check that the slopes are negative reciprocals. For theintersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x− k)2 + y2 = k2 to obtain

cy = kx =12(x2 + y2). Thus x2 + y2 = cy + kx, or y2 − cy = −x2 + kx, and

y − c

x= −x− k

y.

Differentiating the two families yields (black)dy

dx= − x

y − c, and (gray)

dy

dx= −x− k

y. But it was

proven that these quantities are negative reciprocals of each other.

56. Differentiating, we get the equations (black) xdy

dx+ y = 0 and (gray) 2x − 2y

dy

dx= 0. The first

says the (black) slope is = −yx

and the second says the (gray) slope isx

y, and these are negative

reciprocals of each other.

EXERCISE SET 4.2

1.15x

(5) =1x

2.1x/3

13=

1x

3.1

1 + x4.

12 +√x

(1

2√x

)=

12√x(2 +

√x)

5.1

x2 − 1(2x) =

2xx2 − 1

6.3x2 − 14x

x3 − 7x2 − 3

7.1

x/(1 + x2)

[(1 + x2)(1)− x(2x)

(1 + x2)2

]=

1− x2

x(1 + x2)

8.1

(1 + x)/(1− x)1− x+ 1 + x

(1− x)2=

21− x2 9.

d

dx(2 lnx) = 2

d

dxlnx =

2x

10. 3 (lnx)21x

11.12(lnx)−1/2

(1x

)=

12x√lnx

12.1√x

12√x=

12x

13. lnx+ x1x= 1 + lnx

14. x3(1x

)+ (3x2) lnx = x2(1 + 3 lnx) 15. 2x log2(3− 2x)− 2x2

(3− 2x) ln 2

16.[log2(x2 − 2x)

]3 + 3x[log2(x2 − 2x)

]2 2x− 2(x2 − 2x) ln 2


134 Chapter 4

17.2x(1 + log x)− x/(ln 10)

(1 + log x)218. 1/[x(ln 10)(1 + log x)2]

19.1

lnx

(1x

)=

1x lnx

20.1

ln(ln(x))1

lnx1x

21.1

tanx(sec2 x) = secx cscx 22.

1cosx

(− sinx) = − tanx

23. − 1xsin(lnx) 24. 2 sin(lnx) cos(lnx)

1x=

sin(2 lnx)x

=sin(lnx2)

x

25.1

ln 10 sin2 x(2 sinx cosx) = 2

cotxln 10

26.1

(ln 10)(1− sin2 x)(−2 sinx cosx) = − 2 sinx cosx

(ln 10) cos2 x= −2 tanx

ln 10

27.d

dx

[3 ln(x− 1) + 4 ln(x2 + 1)

]=

3x− 1

+8x

x2 + 1=

11x2 − 8x+ 3(x− 1)(x2 + 1)

28.d

dx[2 ln cosx+

12ln(1 + x4)] = −2 tanx+

2x3

1 + x4

29.d

dx

[ln cosx− 1

2ln(4− 3x2)

]= − tanx+

3x4− 3x2

30.d

dx

(12[ln(x− 1)− ln(x+ 1)]

)=

12

(1

x− 1− 1x+ 1

)

31. ln |y| = ln |x|+ 13ln |1 + x2|, dy

dx= x

3√1 + x2

[1x+

2x3(1 + x2)

]

32. ln |y| = 15[ln |x− 1| − ln |x+ 1|], dy

dx=

15

5

√x− 1x+ 1

[1

x− 1− 1x+ 1

]

33. ln |y| = 13ln |x2 − 8|+ 1

2ln |x3 + 1| − ln |x6 − 7x+ 5|

dy

dx=

(x2 − 8)1/3√x3 + 1

x6 − 7x+ 5

[2x

3(x2 − 8)+

3x2

2(x3 + 1)− 6x5 − 7x6 − 7x+ 5

]

34. ln |y| = ln | sinx|+ ln | cosx|+ 3 ln | tanx| − 12ln |x|

dy

dx=

sinx cosx tan3 x√x

[cotx− tanx+

3 sec2 x

tanx− 1

2x

]

35. f ′(x) = exe−1 36. ln y = −√10 lnx,

1y

dy

dx= −√10x

,dy

dx= −

√10

x1+√

10

37. (a) logx e =ln elnx

=1

lnx,d

dx[logx e] = −

1x(lnx)2

(b) logx 2 =ln 2lnx

,d

dx[logx 2] = −

ln 2x(lnx)2



38. (a) From loga b =ln bln a

for a, b > 0 it follows that log(1/x) e =ln e

ln(1/x)= − 1

lnx, hence

d

dx

[log(1/x) e

]=

1x(lnx)2

(b) log(ln x) e =ln e

ln(lnx)=

1ln(lnx)

, sod

dxlog(ln x) e = −

1(ln(lnx))2

1x lnx

= − 1x(lnx)(ln(lnx))2

39. f(x0) = f(e−1) = −1, f ′(x) = 1x, f ′(x0) = e, y − (−1) = e(x− 1/e) = ex− 1, y = ex− 2

40. y0 = log 10 = 1, y = log x = (log e) lnx,dy

dx

∣∣∣∣x=10

= log e110,

y − 1 =log e10

(x− 10), y =log e10

x+ 1− log e

41. f(x0) = f(−e) = 1, f ′(x)∣∣∣∣x=−e

= −1e,

y − 1 = −1e(x+ e), y = −1

ex

42. y − ln 2 = −12(x+ 2), y = −1

2x+ ln 2− 1

43. Let the equation of the tangent line be y = mx and suppose that it meets the curve at (x0, y0).

Then m =1x

∣∣∣∣x=x0

=1x0

and y0 = mx0 = lnx0. So m =1x0

=lnx0

x0and lnx0 = 1, x0 = e,m =

1e

and the equation of the tangent line is y =1ex.

44. Let y = mx + b be a line tangent to the curve at (x0, y0). Then b is the y-intercept and the

slope of the tangent line is m =1x0

. Moreover, at the point of tangency, mx0 + b = lnx0 or

1x0

x0 + b = lnx0, b = lnx0 − 1, as required.

45. The area of the triangle PQR, given by |PQ||QR|/2 isrequired. |PQ| = w, and, by Exercise 44, |QR| = 1, soarea = w/2.

2

–2

1

x

y

P (w, ln w)Q

R

w

46. Since y = 2 lnx, let y = 2z; then z = lnx and we apply the result of Exercise 45 to find that thearea is, in the x-z plane, w/2. In the x-y plane, since y = 2z, the vertical dimension gets doubled,so the area is w.

47. If x = 0 then y = ln e = 1, anddy

dx=

1x+ e

. But ey = x+ e, sody

dx=

1ey

= e−y.

48. When x = 0, y = − ln(e2) = −2. Next,dy

dx=

1e2 − x

. But ey = e− ln(e2−x) = (e2 − x)−1, so

dy

dx= ey.


136 Chapter 4

49. Let y = ln(x+a). Following Exercise 47 we getdy

dx=

1x+ a

= e−y, and when x = 0, y = ln(a) = 0

if a = 1, so let a = 1, then y = ln(x+ 1).

50. Let y = − ln(a− x), thendy

dx=

1a− x

. But ey =1

a− x, so

dy

dx= ey.

If x = 0 then y = − ln(a) = − ln 2 provided a = 2, so y = − ln(2− x).

51. (a) f(x) = lnx; f ′(e2) = lim∆x→0

ln(e2 +∆x)− 2∆x

=d

dx(lnx)

∣∣∣∣x=e2

=1x

∣∣∣∣x=e2

= e−2

(b) f(w) = lnw; f ′(1) = limh→0

ln(1 + h)− ln 1h

= limh→0

ln(1 + h)h

=1w

∣∣∣∣w=1

= 1

52. (a) Let f(x) = ln(cosx), then f(0) = ln(cos 0) = ln 1 = 0, so f ′(0) = limx→0

f(x)− f(0)x

=

limx→0

ln(cosx)x

, and f ′(0) = − tan 0 = 0.

(b) Let f(x) = x√

2, then f(1) = 1, so f ′(1) = limh→0

f(1 + h)− f(1)h

= limh→0

(1 + h)√

2 − 1h

, and

f ′(x) =√2x√

2−1, f ′(1) =√2.

53.d

dx[logb x] = lim

h→0

logb(x+ h)− logb(x)h

= limh→0

1hlogb

(x+ h

x

)Theorem 1.6.2(b)

= limh→0

1hlogb

(1 +

h

x

)

= limv→0

1vx

logb(1 + v) Let v = h/x and note that v → 0 as h→ 0

=1xlimv→0

1vlogb(1 + v) h and v are variable, whereas x is constant

=1xlimv→0

logb(1 + v)1/v Theorem 1.6.2.(c)

=1xlogb lim

v→0(1 + v)1/v Theorem 2.5.5

=1xlogb e Formula 7 of Section 7.1

EXERCISE SET 4.3

1. (a) f ′(x) = 5x4 + 3x2 + 1 ≥ 1 so f is one-to-one on −∞ < x < +∞.

(b) f(1) = 3 so 1 = f−1(3);d

dxf−1(x) =

1f ′(f−1(x))

, (f−1)′(3) =1

f ′(1)=

19

2. (a) f ′(x) = 3x2 + 2ex; for −1 < x < 1, f ′(x) ≥ 2e−1 = 2/e, and for |x| > 1, f ′(x) ≥ 3x2 ≥ 3, sof is increasing and one-to-one

(b) f(0) = 2 so 0 = f−1(2);d

dxf−1(x) =

1f ′(f−1(x))

, (f−1)′(2) =1

f ′(0)=

12



3. f−1(x) =2x− 3, so directly

d

dxf−1(x) = − 2

x2 . Using Formula (1),

f ′(x) =−2

(x+ 3)2, so

1f ′(f−1(x))

= −(1/2)(f−1(x) + 3)2,

d

dxf−1(x) = −(1/2)

(2x

)2

= − 2x2

4. f−1(x) =ex − 1

2, so directly,

d

dxf−1(x) =

ex

2. Next, f ′(x) =

22x+ 1

, and using Formula (1),

d

dxf−1(x) =

2f−1(x) + 12

=ex

2

5. (a) f ′(x) = 2x + 8; f ′ < 0 on (−∞,−4) and f ′ > 0 on (−4,+∞); not enough information. Byinspection, f(1) = 10 = f(−9), so not one-to-one

(b) f ′(x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f ′(x) is positive for all x, so f is one-to-one(c) f ′(x) = 2 + cosx ≥ 1 > 0 for all x, so f is one-to-one(d) f ′(x) = −(ln 2)

( 12

)x< 0 because ln 2 > 0, so f is one-to-one for all x.

6. (a) f ′(x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so not enough information; byobservation (of the graph, and using some guesswork), f(−1 +

√3) = −6 = f(−1−

√3), so

f is not one-to-one.

(b) f ′(x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f ′ is positive for all x, so f is one-to-one

(c) f ′(x) =1

(x+ 1)2; f is one-to-one because:

if x1 < x2 < −1 then f ′ > 0 on [x1, x2], so f(x1) = f(x2)if −1 < x1 < x2 then f ′ > 0 on [x1, x2], so f(x1) = f(x2)if x1 < −1 < x2 then f(x1) > 1 > f(x2) since f(x) > 1 on (−∞,−1) and f(x) < 1 on(−1,+∞)

(d) Note that f(x) is only defined for x > 0.d

dxlogb x =

1x ln b

, which is always negative

(0 < b < 1), so f is one-to-one.

7. y = f−1(x), x = f(y) = 5y3 + y − 7,dx

dy= 15y2 + 1,

dy

dx=

115y2 + 1

;

check: 1 = 15y2 dy

dx+

dy

dx,dy

dx=

115y2 + 1

8. y = f−1(x), x = f(y) = 1/y2,dx

dy= −2y−3,

dy

dx= −y3/2;

check: 1 = −2y−3 dy

dx,dy

dx= −y3/2

9. y = f−1(x), x = f(y) = 2y5 + y3 + 1,dx

dy= 10y4 + 3y2,

dy

dx=

110y4 + 3y2 ;

check: 1 = 10y4 dy

dx+ 3y2 dy

dx,dy

dx=

110y4 + 3y2

10. y = f−1(x), x = f(y) = 5y − sin 2y,dx

dy= 5− 2 cos 2y,

dy

dx=

15− 2 cos 2y

;

check: 1 = (5− 2 cos 2y)dy

dx,dy

dx=

15− 2 cos 2y


138 Chapter 4

11. 7e7x 12. −10xe−5x2

13. x3ex + 3x2ex = x2ex(x+ 3) 14. − 1x2 e

1/x

15.dy

dx=

(ex + e−x)(ex + e−x)− (ex − e−x)(ex − e−x)(ex + e−x)2

=(e2x + 2 + e−2x)− (e2x − 2 + e−2x)

(ex + e−x)2= 4/(ex + e−x)2

16. ex cos(ex)

17. (x sec2 x+ tanx)ex tan x 18.dy

dx=

(lnx)ex − ex(1/x)(lnx)2

=ex(x lnx− 1)

x(lnx)2

19. (1− 3e3x)e(x−e3x) 20.152x2(1 + 5x3)−1/2 exp(

√1 + 5x3)

21.(x− 1)e−x

1− xe−x=

x− 1ex − x

22.1

cos(ex)[− sin(ex)]ex = −ex tan(ex)

23. f ′(x) = 2x ln 2; y = 2x, ln y = x ln 2,1yy′ = ln 2, y′ = y ln 2 = 2x ln 2

24. f ′(x) = −3−x ln 3; y = 3−x, ln y = −x ln 3, 1yy′ = − ln 3, y′ = −y ln 3 = −3−x ln 3

25. f ′(x) = πsin x(lnπ) cosx;

y = πsin x, ln y = (sinx) lnπ,1yy′ = (lnπ) cosx, y′ = πsin x(lnπ) cosx

26. f ′(x) = πx tan x(lnπ)(x sec2 x+ tanx);

y = πx tan x, ln y = (x tanx) lnπ,1yy′ = (lnπ)(x sec2 x+ tanx)

y′ = πx tan x(lnπ)(x sec2 x+ tanx)

27. ln y = (lnx) ln(x3 − 2x),1y

dy

dx=

3x2 − 2x3 − 2x

lnx+1xln(x3 − 2x),

dy

dx= (x3 − 2x)ln x

[3x2 − 2x3 − 2x

lnx+1xln(x3 − 2x)

]

28. ln y = (sinx) lnx,1y

dy

dx=

sinxx

+ (cosx) lnx,dy

dx= xsin x

[sinxx

+ (cosx) lnx]

29. ln y = (tanx) ln(lnx),1y

dy

dx=

1x lnx

tanx+ (sec2 x) ln(lnx),

dy

dx= (lnx)tan x

[tanxx lnx

+ (sec2 x) ln(lnx)]



30. ln y = (lnx) ln(x2 + 3),1y

dy

dx=

2xx2 + 3

lnx+1xln(x2 + 3),

dy

dx= (x2 + 3)ln x

[2x

x2 + 3lnx+

1xln(x2 + 3)

]

31. f ′(x) = exe−1

32. (a) because xx is not of the form ax where a is constant

(b) y = xx, ln y = x lnx,1yy′ = 1 + lnx, y′ = xx(1 + lnx)

33.3√

1− (3x)2=

3√1− 9x2

34. − 1/2√1−

(x+1

2

)2 = − 1√4− (x+ 1)2

35.1√

1− 1/x2(−1/x2) = − 1

|x|√x2 − 1

36.sinx√

1− cos2 x=

sinx| sinx| =

{1, sinx > 0−1, sinx < 0

37.3x2

1 + (x3)2=

3x2

1 + x6 38.5x4

|x5|√(x5)2 − 1

=5

|x|√x10 − 1

39. y = 1/ tanx = cotx, dy/dx = − csc2 x

40. y = (tan−1 x)−1, dy/dx = −(tan−1 x)−2(

11 + x2

)

41.ex

|x|√x2 − 1

+ ex sec−1 x 42. − 1(cos−1 x)

√1− x2

43. 0 44.3x2(sin−1 x)2√

1− x2+ 2x(sin−1 x)3

45. 0 46. −1/√e2x − 1

47. − 11 + x

(12x−1/2

)= − 1

2(1 + x)√x

48. − 1

2√cot−1 x(1 + x2)

49. (a) Let x = f(y) = cot y, 0 < y < π,−∞ < x < +∞. Then f is differentiable and one-to-oneand f ′(f−1(x)) = − csc2(cot−1 x) = −x2 − 1 = 0, andd

dx[cot−1 x]

∣∣∣∣x=0

= limx→0

1f ′(f−1(x))

= − limx→0

1x2 + 1

= −1.

(b) If x = 0 then, from Exercise 50(a) of Section 1.5,d

dxcot−1 x =

d

dxtan−1 1

x= − 1

x2

11 + (1/x)2

= − 1x2 + 1

. For x = 0, Part (a) shows the same;

thus for −∞ < x < +∞,d

dx[cot−1 x] = − 1

x2 + 1.

(c) For −∞ < u < +∞, by the chain rule it follows thatd

dx[cot−1 u] = − 1

u2 + 1du

dx.


140 Chapter 4

50. (a) By the chain rule,d

dx[csc−1 x] =

d

dxsin−1 1

x= − 1

x2

1√1− (1/x)2

=−1

|x|√x2 − 1

(b) By the chain rule,d

dx[csc−1 u] =

du

dx

d

du[csc−1 u] =

−1|u|√u2 − 1

du

dx

51. x3 + x tan−1 y = ey, 3x2 +x

1 + y2 y′ + tan−1 y = eyy′, y′ =

(3x2 + tan−1 y)(1 + y2)(1 + y2)ey − x

52. sin−1(xy) = cos−1(x− y),1√

1− x2y2(xy′ + y) = − 1√

1− (x− y)2(1− y′),

y′ =y√1− (x− y)2 +

√1− x2y2√

1− x2y2 − x√

1− (x− y)2

53. (a) f(x) = x3 − 3x2 + 2x = x(x− 1)(x− 2) so f(0) = f(1) = f(2) = 0 thus f is not one-to-one.

(b) f ′(x) = 3x2 − 6x + 2, f ′(x) = 0 when x =6±√36− 246

= 1 ±√3/3. f ′(x) > 0 (f is

increasing) if x < 1−√3/3, f ′(x) < 0 (f is decreasing) if 1−

√3/3 < x < 1 +

√3/3, so f(x)

takes on values less than f(1−√3/3) on both sides of 1−

√3/3 thus 1−

√3/3 is the largest

value of k.

54. (a) f(x) = x3(x− 2) so f(0) = f(2) = 0 thus f is not one to one.(b) f ′(x) = 4x3− 6x2 = 4x2(x− 3/2), f ′(x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]

and increasing on [3/2,+∞) so 3/2 is the smallest value of k.

55. (a) f ′(x) = 4x3 + 3x2 = (4x+ 3)x2 = 0 only at x = 0. But on [0, 2], f ′ has no sign change, so fis one-to-one.

(b) F ′(x) = 2f ′(2g(x))g′(x) so F ′(3) = 2f ′(2g(3))g′(3). By inspection f(1) = 3, sog(3) = f−1(3) = 1 and g′(3) = (f−1)′(3) = 1/f ′(f−1(3)) = 1/f ′(1) = 1/7 becausef ′(x) = 4x3 + 3x2. Thus F ′(3) = 2f ′(2)(1/7) = 2(44)(1/7) = 88/7.F (3) = f(2g(3)) = f(2·1) = f(2) = 24, so the line tangent to F (x) at (3, 25) has the equationy − 25 = (88/7)(x− 3), y = (88/7)x− 89/7.

56. (a) f ′(x) = −e4−x2(2 +

1x2

)< 0 for all x > 0, so f is one-to-one.

(b) By inspection, f(2) = 1/2, so 2 = f−1(1/2) = g(1/2). By inspection,

f ′(2) = −(2 +

14

)= −9

4, and

F ′(1/2) = f ′([g(x)]2)d

dx[g(x)2]

∣∣∣∣x=1/2

= f ′([g(x)]2)2g(x)g′(x)∣∣∣∣x=1/2

= f ′(22)2 · 2 1f ′(g(x))

∣∣∣∣x=1/2

= 4f ′(4)f ′(2)

= 4e−12(2 + 1

16 )(2 + 1

4 )=

339e12 =

113e12

57. (a) f ′(x) = kekx, f ′′(x) = k2ekx, f ′′′(x) = k3ekx, . . . , f (n)(x) = knekx

(b) g′(x) = −ke−kx, g′′(x) = k2e−kx, g′′′(x) = −k3e−kx, . . . , g(n)(x) = (−1)nkne−kx

58.dy

dt= e−λt(ωA cosωt− ωB sinωt) + (−λ)e−λt(A sinωt+B cosωt)

= e−λt[(ωA− λB) cosωt− (ωB + λA) sinωt]



59. f ′(x)=1√2πσ

exp

[−12

(x− µ

σ

)2]

d

dx

[−12

(x− µ

σ

)2]

=1√2πσ

exp

[−12

(x− µ

σ

)2] [−(x− µ

σ

)(1σ

)]

= − 1√2πσ3

(x− µ) exp

[−12

(x− µ

σ

)2]

60. y = Aekt, dy/dt = kAekt = k(Aekt) = ky

61. y = Ae2x +Be−4x, y′ = 2Ae2x − 4Be−4x, y′′ = 4Ae2x + 16Be−4x soy′′ + 2y′ − 8y = (4Ae2x + 16Be−4x) + 2(2Ae2x − 4Be−4x)− 8(Ae2x +Be−4x) = 0

62. (a) y′ = −xe−x + e−x = e−x(1− x), xy′ = xe−x(1− x) = y(1− x)

(b) y′ = −x2e−x2/2 + e−x

2/2 = e−x2/2(1− x2), xy′ = xe−x

2/2(1− x2) = y(1− x2)

63.dy

dx= 100(−0.2)e−0.2x = −20y, k = −0.2

64. ln y = (5x+ 1) ln 3− (x/2) ln 4, soy′/y = 5 ln 3− (1/2) ln 4 = 5 ln 3− ln 2, andy′ = (5 ln 3− ln 2)y

65. ln y = ln 60− ln(5 + 7e−t),y′

y=

7e−t

5 + 7e−t=

7e−t + 5− 55 + 7e−t

= 1− 112y, so

dy

dt= r

(1− y

K

)y, with r = 1,K = 12.

66. (a) 12

00 9

(b) P tends to 12 as t gets large; limt→+∞

P (t) = limt→+∞

605 + 7e−t

=60

5 + 7 limt→+∞

e−t=

605

= 12

(c) the rate of population growth tends to zero

3.2

00 9

67. limh→0

10h − 1h

=d

dx10x∣∣∣∣x=0

=d

dxex ln 10

∣∣∣∣x=0

= ln 10


142 Chapter 4

68. limh→0

tan−1(1 + h)− π/4h

=d

dxtan−1 x

∣∣∣∣x=1

=1

1 + x2

∣∣∣∣x=1

=12

69. lim∆x→0

9[sin−1(√

32 +∆x)]2 − π2

∆x=

d

dx(3 sin−1 x)2

∣∣∣∣x=√

3/2= 2(3 sin−1 x)

3√1− x2

∣∣∣∣x=√

3/2

= 2(3π

3)

3√1− (3/4)

= 12π

70. lim∆x→0

(2 + ∆x)(2+∆x) − 4∆x

=d

dxxx∣∣∣∣x=2

=d

dxex ln x

∣∣∣∣x=2

= (1 + lnx)ex ln x∣∣∣∣x=2

= (1 + ln 2)22 = 4(1 + ln 2)

71. limw→2

3 sec−1 w − π

w − 2=

d

dx3 sec−1 x

∣∣∣∣x=2

=3

|2|√22 − 1

=√32

72. limw→1

4(tan−1 w)w − π

w − 1=

d

dx4(tan−1 x)x

∣∣∣∣x=1

=d

dx4ex ln tan−1 x

∣∣∣∣x=1

= 4(tan−1 x)x(ln tan−1 x+ x

1/(1 + x2)tan−1 x

) ∣∣∣∣x=1

= π

(ln(π/4) +

124π

)== 2 + π ln(π/4)

EXERCISE SET 4.4

1. (a) limx→2

x2 − 4x2 + 2x− 8

= limx→2

(x− 2)(x+ 2)(x+ 4)(x− 2)

= limx→2

x+ 2x+ 4

=23

(b) limx→+∞

2x− 53x+ 7

=2− lim

x→+∞

5x

3 + limx→+∞

7x

=23

2. (a)sinxtanx

= sinxcosxsinx

= cosx so limx→0

sinxtanx

= limx→0

cosx = 1

(b)x2 − 1x3 − 1

=(x− 1)(x+ 1)

(x− 1)(x2 + x+ 1)=

x+ 1x2 + x+ 1

so limx→1

x2 − 1x3 − 1

=23

3. Tf (x) = −2(x+ 1), Tg(x) = −3(x+ 1),limit = 2/3

4. Tf (x) = −(x− π

2

), Tg(x) = −

(x− π

2

)limit = 1

5. limx→0

ex

cosx= 1 6. lim

x→3

16x− 13

= 1/5

7. limθ→0

sec2 θ

1= 1 8. lim

t→0

tet + et

−et = −1

9. limx→π+

cosx1

= −1 10. limx→0+

cosx2x

= +∞

11. limx→+∞

1/x1

= 0 12. limx→+∞

3e3x

2x= limx→+∞

9e3x

2= +∞



13. limx→0+

− csc2 x

1/x= limx→0+

−xsin2 x

= limx→0+

−12 sinx cosx

= −∞

14. limx→0+

−1/x(−1/x2)e1/x = lim

x→0+

x

e1/x = 0

15. limx→+∞

100x99

ex= limx→+∞

(100)(99)x98

ex= · · · = lim

x→+∞

(100)(99)(98) · · · (1)ex

= 0

16. limx→0+

cosx/ sinxsec2 x/ tanx

= limx→0+

cos2 x = 1 17. limx→0

2/√1− 4x2

1= 2

18. limx→0

1− 11 + x2

3x2 = limx→0

13(1 + x2)

=13

19. limx→+∞

xe−x = limx→+∞

x

ex= limx→+∞

1ex

= 0

20. limx→π

(x− π) tan(x/2) = limx→π

x− π

cot(x/2)= limx→π

1−(1/2) csc2(x/2)

= −2

21. limx→+∞

x sin(π/x) = limx→+∞

sin(π/x)1/x

= limx→+∞

(−π/x2) cos(π/x)−1/x2 = lim

x→+∞π cos(π/x) = π

22. limx→0+

tanx lnx = limx→0+

lnxcotx

= limx→0+

1/x− csc2 x

= limx→0+

− sin2 x

x= limx→0+

−2 sinx cosx1

= 0

23. limx→(π/2)−

sec 3x cos 5x = limx→(π/2)−

cos 5xcos 3x

= limx→(π/2)−

−5 sin 5x−3 sin 3x =

−5(+1)(−3)(−1) = −5

3

24. limx→π

(x− π) cotx = limx→π

x− π

tanx= limx→π

1sec2 x

= 1

25. y = (1− 3/x)x, limx→+∞

ln y = limx→+∞

ln(1− 3/x)1/x

= limx→+∞

−31− 3/x

= −3, limx→+∞

y = e−3

26. y = (1 + 2x)−3/x, limx→0

ln y = limx→0−3 ln(1 + 2x)

x= limx→0− 61 + 2x

= −6, limx→0

y = e−6

27. y = (ex + x)1/x, limx→0

ln y = limx→0

ln(ex + x)x

= limx→0

ex + 1ex + x

= 2, limx→0

y = e2

28. y = (1 + a/x)bx, limx→+∞

ln y = limx→+∞

b ln(1 + a/x)1/x

= limx→+∞

ab

1 + a/x= ab, lim

x→+∞y = eab

29. y = (2− x)tan(πx/2), limx→1

ln y = limx→1

ln(2− x)cot(πx/2)

= limx→1

2 sin2(πx/2)π(2− x)

= 2/π, limx→1

y = e2/π

30. y= [cos(2/x)]x2, limx→+∞

ln y = limx→+∞

ln cos(2/x)1/x2 = lim

x→+∞

(−2/x2)(− tan(2/x))−2/x3

= limx→+∞

− tan(2/x)1/x

= limx→+∞

(2/x2) sec2(2/x)−1/x2 = −2, lim

x→+∞y = e−2

31. limx→0

(1

sinx− 1x

)= limx→0

x− sinxx sinx

= limx→0

1− cosxx cosx+ sinx

= limx→0

sinx2 cosx− x sinx

= 0


144 Chapter 4

32. limx→0

1− cos 3xx2 = lim

x→0

3 sin 3x2x

= limx→0

92cos 3x =

92

33. limx→+∞

(x2 + x)− x2√x2 + x+ x

= limx→+∞

x√x2 + x+ x

= limx→+∞

1√1 + 1/x+ 1

= 1/2

34. limx→0

ex − 1− x

xex − x= limx→0

ex − 1xex + ex − 1

= limx→0

ex

xex + 2ex= 1/2

35. limx→+∞

[x− ln(x2 + 1)] = limx→+∞

[ln ex − ln(x2 + 1)] = limx→+∞

lnex

x2 + 1,

limx→+∞

ex

x2 + 1= limx→+∞

ex

2x= limx→+∞

ex

2= +∞ so lim

x→+∞[x− ln(x2 + 1)] = +∞

36. limx→+∞

lnx

1 + x= limx→+∞

ln1

1/x+ 1= ln(1) = 0

38. (a) limx→+∞

lnxxn

= limx→+∞

1/xnxn−1 = lim

x→+∞

1nxn

= 0

(b) limx→+∞

xn

lnx= limx→+∞

nxn−1

1/x= limx→+∞

nxn = +∞

39. (a) L’Hopital’s Rule does not apply to the problem limx→1

3x2 − 2x+ 13x2 − 2x

because it is not a00form.

(b) limx→1

3x2 − 2x+ 13x2 − 2x

= 2

40. L’Hopital’s Rule does not apply to the probleme3x2−12x+12

x4 − 16, which is of the form

e0

0, and from

which it follows that limx→2−

and limx→2+

exist, with values −∞ if x approaches 2 from the left and

+∞ if from the right. The general limit limx→2

does not exist.

41. limx→+∞

1/(x lnx)1/(2√x)

= limx→+∞

2√x lnx

= 0 0.15

0100 10000

42. y = xx, limx→0+

ln y = limx→0+

lnx1/x

= limx→0+

−x = 0, limx→0+

y = 1 1

00 0.5



25

190 0.5

43. y = (sinx)3/ ln x,

limx→0+

ln y = limx→0+

3 ln sinxlnx

= limx→0+

(3 cosx)x

sinx= 3,

limx→0+

y = e3

4.1

3.31.4 1.6

44. limx→π/2−

4 sec2 x

secx tanx= limx→π/2−

4sinx

= 4

0

–16

0 345. lnx− ex = lnx− 1

e−x=

e−x lnx− 1e−x

;

limx→+∞

e−x lnx = limx→+∞

lnxex

= limx→+∞

1/xex

= 0 by L’Hopital’s Rule,

so limx→+∞

[lnx− ex] = limx→+∞

e−x lnx− 1e−x

= −∞

–0.6

–1.2

0 1246. lim

x→+∞[ln ex − ln(1 + 2ex)] = lim

x→+∞ln

ex

1 + 2ex

= limx→+∞

ln1

e−x + 2= ln

12;

horizontal asymptote y = − ln 2

1.02

1100 10000

47. y = (lnx)1/x,

limx→+∞

ln y = limx→+∞

ln(lnx)x

= limx→+∞

1x lnx

= 0;

limx→+∞

y = 1, y = 1 is the horizontal asymptote


146 Chapter 4

48. y =(x+ 1x+ 2

)x, limx→+∞

ln y = limx→+∞

lnx+ 1x+ 21/x

= limx→+∞

−x2

(x+ 1)(x+ 2)= −1;

limx→+∞

y = e−1 is the horizontal asymptote

1

00 50

49. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f) −∞

50. (a) Type 00; y = x(ln a)/(1+ln x); limx→0+

ln y = limx→0+

(ln a) lnx1 + lnx

= limx→0+

(ln a)/x1/x

= limx→0+

ln a = ln a,

limx→0+

y = eln a = a

(b) Type ∞0; same calculation as Part (a) with x→ +∞

(c) Type 1∞; y = (x+ 1)(ln a)/x, limx→0

ln y = limx→0

(ln a) ln(x+ 1)x

= limx→0

ln ax+ 1

= ln a,

limx→0

y = eln a = a

51. limx→+∞

1 + 2 cos 2x1

does not exist, nor is it ±∞; limx→+∞

x+ sin 2xx

= limx→+∞

(1 +

sin 2xx

)= 1

52. limx→+∞

2− cosx3 + cosx

does not exist, nor is it ±∞; limx→+∞

2x− sinx3x+ sinx

= limx→+∞

2− (sinx)/x3 + (sinx)/x

=23

53. limx→+∞

(2 + x cos 2x+ sin 2x) does not exist, nor is it ±∞; limx→+∞

x(2 + sin 2x)x+ 1

= limx→+∞

2 + sin 2x1 + 1/x

,

which does not exist because sin 2x oscillates between −1 and 1 as x→ +∞

54. limx→+∞

(1x+

12cosx+

sinx2x

)does not exist, nor is it ±∞;

limx→+∞

x(2 + sinx)x2 + 1

= limx→+∞

2 + sinxx+ 1/x

= 0

55. limR→0+

V tL e−Rt/L

1=

V t

L

56. (a) limx→π/2

(π/2− x) tanx = limx→π/2

π/2− x

cotx= limx→π/2

−1− csc2 x

= limx→π/2

sin2 x = 1

(b) limx→π/2

(1

π/2− x− tanx

)= limx→π/2

(1

π/2− x− sinx

cosx

)= limx→π/2

cosx− (π/2− x) sinx(π/2− x) cosx

= limx→π/2

−(π/2− x) cosx−(π/2− x) sinx− cosx

= limx→π/2

(π/2− x) sinx+ cosx−(π/2− x) cosx+ 2 sinx

= 0

(c) 1/(π/2− 1.57) ≈ 1255.765534, tan 1.57 ≈ 1255.765592;1/(π/2− 1.57)− tan 1.57 ≈ 0.000058


Review Exercises, Chapter 4 147

57. (b) limx→+∞

x(k1/x − 1) = limt→0+

kt − 1t

= limt→0+

(ln k)kt

1= ln k

(c) ln 0.3 = −1.20397, 1024(

1024√0.3− 1

)= −1.20327;

ln 2 = 0.69315, 1024(

1024√2− 1

)= 0.69338

58. If k = −1 then limx→0

(k + cos �x) = k + 1 = 0, so limx→0

k + cos �xx2 = ±∞. Hence k = −1, and by the

rule limx→0

−1 + cos �xx2 = lim

x→0

−� sin �x2x

= limx→0

−�2 cos �x2

= −�2

2= −4 if � = ±2

√2.

59. (a) No; sin(1/x) oscillates as x→ 0. (b) 0.05

–0.05

–0.35 0.35

(c) For the limit as x→ 0+ use the Squeezing Theorem together with the inequalities−x2 ≤ x2 sin(1/x) ≤ x2. For x→ 0− do the same; thus lim

x→0f(x) = 0.

60. (a) Apply the rule to get limx→0

− cos(1/x) + 2x sin(1/x)cosx

which does not exist (nor is it ±∞).

(b) Rewrite as limx→0

[ x

sinx

][x sin(1/x)], but lim

x→0

x

sinx= lim

x→0

1cosx

= 1 and limx→0

x sin(1/x) = 0,

thus limx→0

[ x

sinx

][x sin(1/x)] = (1)(0) = 0

61. limx→0+

sin(1/x)(sinx)/x

, limx→0+

sinxx

= 1 but limx→0+

sin(1/x) does not exist because sin(1/x) oscillates between

−1 and 1 as x→ +∞, so limx→0+

x sin(1/x)sinx

does not exist.

62. Since f(0) = g(0) = 0, then for x = a,f(x)g(x)

=(f(x)− f(0)/(x− a)(g(x)− g(0))/(x− a)

. Now take the limit:

limx→a

f(x)g(x)

= limx→a

(f(x)− f(0)/(x− a)(g(x)− g(0))/(x− a)

=f ′(a)g′(a)

REVIEW EXERCISES, CHAPTER 4

1.1

4(6x− 5)3/4(6) =

32(6x− 5)3/4

2.1

3(x2 + x)2/3(2x+ 1) =

2x+ 13(x2 + x)2/3

3. dy/dx =32

[x− 1x+ 2

]1/2d

dx

[x− 1x+ 2

]=

92(x+ 2)2

[x− 1x+ 2

]1/2

4. dy/dx =x2 4

3(3− 2x)1/3(−2)− (3− 2x)4/3(2x)

x4 =2(3− 2x)1/3(2x− 9)

3x3


148 Chapter 4

5. (a) 3x2 + xdy

dx+ y − 2 = 0,

dy

dx=

2− y − 3x2

x

(b) y = (1 + 2x− x3)/x = 1/x+ 2− x2, dy/dx = −1/x2 − 2x

(c)dy

dx=

2− (1/x+ 2− x2)− 3x2

x= −1/x2 − 2x

6. (a) xy = x− y, xdy

dx+ y = 1− dy

dx,dy

dx=

1− y

x+ 1

(b) y(x+ 1) = x, y =x

x+ 1, y′ =

1(x+ 1)2

(c)dy

dx=

1− y

x+ 1=

1− xx+1

1 + x=

1x2 + 1

7. − 1y2

dy

dx− 1x2 = 0 so

dy

dx= −y

2

x2

8. 3x2 − 3y2 dy

dx= 6(x

dy

dx+ y), −(3y2 + 6x)

dy

dx= 6y − 3x2 so

dy

dx=

x2 − 2yy2 + 2x

9.(xdy

dx+ y

)sec(xy) tan(xy) =

dy

dx,dy

dx=

y sec(xy) tan(xy)1− x sec(xy) tan(xy)

10. 2x =(1 + csc y)(− csc2 y)(dy/dx)− (cot y)(− csc y cot y)(dy/dx)

(1 + csc y)2,

2x(1 + csc y)2 = − csc y(csc y + csc2 y − cot2 y)dy

dx,

but csc2 y − cot2 y = 1, sody

dx= −2x(1 + csc y)

csc y

11.dy

dx=

3x4y

,d2y

dx2 =(4y)(3)− (3x)(4dy/dx)

16y2 =12y − 12x(3x/(4y))

16y2 =12y2 − 9x2

16y3 =−3(3x2 − 4y2)

16y3 ,

but 3x2 − 4y2 = 7 sod2y

dx2 =−3(7)16y3 = − 21

16y3

12.dy

dx=

y

y − x,

d2y

dx2 =(y − x)(dy/dx)− y(dy/dx− 1)

(y − x)2=

(y − x)(

y

y − x

)− y

(y

y − x− 1)

(y − x)2

=y2 − 2xy(y − x)3

but y2 − 2xy = −3, sod2y

dx2 = − 3(y − x)3

13.dy

dx= tan(πy/2) + x(π/2)

dy

dxsec2(πy/2),

dy

dx= 1 + (π/4)

dy

dx(2),

dy

dx=

2π − 2

14. Let P (x0, y0) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope ofthe tangent to y2 = 2x3 at P must be −3/4. By implicit differentiation dy/dx = 3x2/y, so at P ,3x2

0/y0 = −3/4, or y0 = −4x20. But y2

0 = 2x30 because P is on the curve y2 = 2x3. Elimination of

y0 gives 16x40 = 2x3

0, x30(8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x2

0 it follows that y0 = 0when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solutionbecause dy/dx = 3x2/y (the slope of the curve as determined by implicit differentiation) is validonly if y = 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point(1/8,−1/16) is the only solution.



15. Substitute y = mx into x2 +xy+ y2 = 4 to get x2 +mx2 +m2x2 = 4, which has distinct solutionsx = ±2/

√m2 +m+ 1. They are distinct because m2 + m + 1 = (m + 1/2)2 + 3/4 ≥ 3/4, so

m2 +m+ 1 is never zero.Note that the points of intersection occur in pairs (x0, y0) and (−x0, y0). By implicit differentiation,the slope of the tangent line to the ellipse is given by dy/dx = −(2x+ y)/(x+2y). Since the slopeis unchanged if we replace (x, y) with (−x,−y), it follows that the slopes are equal at the two pointof intersection.Finally we must examine the special case x = 0 which cannot be written in the form y = mx. Ifx = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal.

16. Use implicit differentiation to get dy/dx = (y−3x2)/(3y2−x), so dy/dx = 0 if y = 3x2. Substitutethis into x3 − xy + y3 = 0 to obtain 27x6 − 2x3 = 0, x3 = 2/27, x = 3

√2/3 and hence y = 3

√4/3.

17. By implicit differentiation, 3x2 − y− xy′ + 3y2y′ = 0, so y′ = (3x2 − y)/(x− 3y2). This derivativeexists except when x = 3y2. Substituting this into the original equation x3 − xy+ y3 = 0, one has27y6 − 3y3 + y3 = 0, y3(27y3 − 2) = 0. The unique solution in the first quadrant isy = 21/3/3, x = 3y2 = 22/3/3

18. By implicit differentiation, dy/dx = k/(2y) so the slope of the tangent to y2 = kx at (x0, y0) is

k/(2y0) if y0 = 0. The tangent line in this case is y− y0 =k

2y0(x− x0), or 2y0y− 2y2

0 = kx− kx0.

But y20 = kx0 because (x0, y0) is on the curve y2 = kx, so the equation of the tangent line becomes

2y0y − 2kx0 = kx − kx0 which gives y0y = k(x + x0)/2. If y0 = 0, then x0 = 0; the graph ofy2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0y = k(x + x0)/2 gives thesame result when x0 = y0 = 0.

19. y = ln(x+ 1) + 2 ln(x+ 2)− 3 ln(x+ 3)− 4 ln(x+ 4), dy/dx =1

x+ 1+

2x+ 2

− 3x+ 3

− 4x+ 4

20. y =12lnx+

13ln(x+ 1)− ln sinx+ ln cosx, so

dy

dx=

12x

+1

3(x+ 1)− cosx

sinx− sinx

cosx=

5x+ 36x(x+ 1)

− cotx− tanx

21.12x

(2) = 1/x 22. 2(lnx)(1x

)=

2 lnxx

23.1

3x(lnx+ 1)2/324. y = 1

3 ln(x+ 1), y′ =1

3(x+ 1)

25. log10 lnx =ln lnxln 10

, y′ =1

(ln 10)(x lnx)

26. y =1 + lnx/ ln 101− lnx/ ln 10

=ln 10 + lnxln 10− lnx

, y′ =(ln 10− lnx)/x+ (ln 10 + lnx)/x

(ln 10− lnx)2=

2 ln 10x(ln 10− lnx)2

27. y =32lnx+

12ln(1 + x4), y′ =

32x

+2x3

(1 + x4)

28. y =12lnx+ ln cosx− ln(1 + x2), y′ =

12x− sinx

cosx− 2x

1 + x2 =1− 3x2

2x(1 + x2)− tanx

29. y = x2 + 1 so y′ = 2x.


150 Chapter 4

30. y = ln(1 + ex + e2x)

(1− ex)(1 + ex + e2x)= − ln(1− ex),

dy

dx=

ex

1− ex

31. y′ = 2e√x + 2xe

√x d

dx

√x = 2e

√x +√xe√x

32. y′ =abe−x

(1 + be−x)233. y′ =

2π(1 + 4x2)

34. y = e(sin−1 x) ln 2, y′ =ln 2√1− x2

2sin−1 x

35. ln y = ex lnx,y′

y= ex

(1x+ lnx

),dy

dx= xe

x

ex(1x+ lnx

)= ex

[xe

x−1 + xex

lnx]

36. ln y =ln(1 + x)

x,y′

y=

x/(1 + x)− ln(1 + x)x2 =

1x(1 + x)

− ln(1 + x)x2 ,

dy

dx=

1x(1 + x)(1/x)−1 − (1 + x)(1/x)

x2 ln(1 + x)

37. y′ =2

|2x+ 1|√(2x+ 1)2 − 1

38. y′ =1

2√cos−1 x2

d

dxcos−1 x2 = − 1√

cos−1 x2

x√1− x4

39. ln y = 3 lnx− 12ln(x2 + 1), y′/y =

3x− x

x2 + 1, y =

3x2√x2 + 1

− x4

(x2 + 1)3/2

40. ln y =13(ln(x2−1)−ln(x2+1)),

y′

y=

13

(2x

x2 − 1− 2xx2 + 1

)=

4x3(x4 − 1)

so y′ =4x

3(x4 − 1)3

√x2 − 1x2 + 1

41. (b) y

x

2

4

6

1 2 3 4

(c)dy

dx=

12− 1x

sody

dx< 0 at x = 1 and

dy

dx> 0 at x = e

(d) The slope is a continuous function which goes from a negative value at x = 1 to a positivevalue at x = e; therefore it must take the value zero between, by the Intermediate ValueTheorem.

(e)dy

dx= 0 when x = 2

42. β = 10 log I − 10 log I0,dβ

dI=

10I ln 10

(a)dβ

dI

]I=10I0

=1

I0 ln 10db/W/m2 (b)

dβ

dI

]I=100I0

=1

10I0 ln 10db/W/m2

(c)dβ

dI

]I=100I0

=1

100I0 ln 10db/W/m2

43. Solvedy

dt= 3

dx

dtgiven y = x lnx. Then

dy

dt=

dy

dx

dx

dt= (1 + lnx)

dx

dt, so 1 + lnx = 3, lnx = 2,

x = e2.



44. x = 2, y = 0; y′ = −2x/(5− x2) = −4 at x = 2, so y − 0 = −4(x− 2) or y = −4x+ 8

y

x

2

2

45. Set y = logb x and solve y′ = 1: y′ =1

x ln b= 1 so

x =1ln b

. The curves intersect when (x, x) lies on the

graph of y = logb x, so x = logb x. From Formula (8),

Section 1.6, logb x =lnxln b

from which lnx = 1, x = e,

ln b = 1/e, b = e1/e ≈ 1.4447.

46. (a) Find the point of intersection: f(x) =√x+ k = lnx. The

slopes are equal, so m1 =1x= m2 =

12√x,√x = 2, x = 4.

Then ln 4 =√4 + k, k = ln 4− 2.

y

x

2

2

(b) Since the slopes are equal m1 =k

2√x

= m2 =1x, so

k√x = 2. At the point of intersection k

√x = lnx, 2 = lnx,

x = e2, k = 2/e.

y

x

0

2

5

47. Where f is differentiable and f ′ = 0, g must be differentiable; this can be inferred from the graphs.In general, however, g need not be differentiable: consider f(x) = x3, g(x) = x1/3.

48. (a) f ′(x) = −3/(x+ 1)2. If x = f(y) = 3/(y + 1) then y = f−1(x) = (3/x)− 1, so

d

dxf−1(x) = − 3

x2 ; and1

f ′(f−1(x))= − (f−1(x) + 1)2

3= − (3/x)2

3= − 3

x2 .

(b) f(x) = ex/2, f ′(x) = 12ex/2. If x = f(y) = ey/2 then y = f−1(x) = 2 lnx, so

d

dxf−1(x) =

2x;

and1

f ′(f−1(x))= 2e−f

−1(x)/2 = 2e− ln x = 2x−1 =2x

49. Let P (x0, y0) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and anequation of the tangent line at P is y − y0 = 3e3x0(x − x0), y − e3x0 = 3e3x0(x − x0). If the linepasses through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0e

3x0 which givesx0 = 1/3 and thus y0 = e. The point is (1/3, e).

50. ln y = ln 5000 + 1.07x;dy/dx

y= 1.07, or

dy

dx= 1.07y

51. ln y = 2x ln 3 + 7x ln 5;dy/dx

y= 2 ln 3 + 7 ln 5, or

dy

dx= (2 ln 3 + 7 ln 5)y

52.dk

dT= k0 exp

[−q(T − T0)

2T0T

](− q

2T 2

)= − qk0

2T 2 exp[−q(T − T0)

2T0T

]

53. y′= aeax sin bx+ beax cos bx and y′′ = (a2 − b2)eax sin bx+ 2abeax cos bx, so y′′ − 2ay′ + (a2 + b2)y= (a2 − b2)eax sin bx+ 2abeax cos bx− 2a(aeax sin bx+ beax cos bx) + (a2 + b2)eax sin bx = 0.


152 Chapter 4

54. sin(tan−1 x) = x/√1 + x2 and cos(tan−1 x) = 1/

√1 + x2, and y′ =

11 + x2 , y

′′ =−2x

(1 + x2)2, hence

y′′ + 2 sin y cos3 y =−2x

(1 + x2)2+ 2

x√1 + x2

1(1 + x2)3/2

= 0.

55. (a) 100

200 8

(b) as t tends to +∞, the population tends to 19

limt→+∞

P (t) = limt→+∞

955− 4e−t/4

=95

5− 4 limt→+∞

e−t/4=

955

= 19

(c) the rate of population growth tends to zero 0

–80

0 8

56. (a) y = (1 + x)π, limh→0

(1 + h)π − 1h

=d

dx(1 + x)π

∣∣∣∣x=0

= π(1 + x)π−1∣∣∣∣x=0

= π

(b) Let y =1− lnxlnx

. Then y(e) = 0, and limx→e

1− lnx(x− e) lnx

=dy

dx

∣∣∣∣x=e

= − 1/x(lnx)2

= −1e

57. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, becauselarge positive (negative) quantities are added to large positive (negative) quantities. The cases+∞− (+∞) and −∞− (−∞) are indeterminate; large numbers of opposite sign are subtracted,and more information about the sizes is needed.

58. (a) when the limit takes the form 0/0 or ∞/∞(b) Not necessarily; only if lim

x→af(x) = 0. Consider g(x) = x; lim

x→0g(x) = 0. For f(x) choose

cosx, x2, and |x|1/2. Then: limx→0

cosxx

does not exist, limx→0

x2

x= 0, and lim

x→0

|x|1/2x2 = +∞.

59. limx→+∞

(ex − x2) = limx→+∞

x2(ex/x2 − 1), but limx→+∞

ex

x2 = limx→+∞

ex

2x= limx→+∞

ex

2= +∞

so limx→+∞

(ex/x2 − 1) = +∞ and thus limx→+∞

x2(ex/x2 − 1) = +∞

60. limx→1

lnxx4 − 1

= limx→1

1/x4x3 =

14; limx→1

√lnx

x4 − 1=

√limx→1

lnxx4 − 1

=12

61. = limx→0

(x2 + 2x)ex

6 sin 3x cos 3x= limx→0

(x2 + 2x)ex

3 sin 6x= limx→0

(x2 + 4x+ 2)ex

18 cos 6x=

19

62. limx→0

ax ln a = ln a

chapter 04

Education

y dyx2 d2 y y

dx x x x x xx2

sin y cos y

c x2 x y

cos2 y

ysin2 y

y xdx2

x sin yd2 y