Chapter 04

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CHAPTER 4

Derivatives of Logarithmic, Exponential, and Inverse Trigonometric Functions EXERCISE SET 4.1 1. y = (2x − 5)1/3 ; dy/dx =

2 (2x − 5)−2/3 3

2. dy/dx =

−2/3 −2/3 2 1 2 + tan(x2 ) sec2 (x2 )(2x) = x sec2 (x2 ) 2 + tan(x2 ) 3 3

3. dy/dx =

2 3

4. dy/dx =

−1/2 −1/2 1 x2 + 1 d x2 + 1 −12x 6x 1 x2 + 1 √ = =− 2 x2 − 5 dx x2 − 5 2 x2 − 5 (x2 − 5)2 (x2 − 5)3/2 x2 + 1

x+1 x−2

3

5. dy/dx = x

√ 3 6. dy/dx = −

7. dy/dx =

2 − 3

−1/3

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

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

1 2 x (5x2 + 1)−5/3 (25x2 + 9) 3

2 2x − 1 1 −4x + 3 + = 2 2 2/3 x x 3(2x − 1) 3x (2x − 1)2/3

15[sin(3/x)]3/2 cos(3/x) 5 [sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = − 2 2x2

8. dy/dx = −

−3/2 −3/2 1 3 cos(x3 ) − sin(x3 ) (3x2 ) = x2 sin(x3 ) cos(x3 ) 2 2

6x2 − y − 1 dy dy − 6x2 = 0, = dx dx x 2 2 dy 2 + 2x3 − x = + 2x2 − 1, = − 2 + 4x (b) y = x x dx x 1 1 1 1 2 dy 2 2 = 6x − − y = 6x − − + 2x − 1 = 4x − 2 (c) From Part (a), dx x x x x x x

9. (a) 1 + y + x

dy 1 −1/2 dy √ y − cos x = 0 or = 2 y cos x 2 dx dx dy 2 (b) y = (2 + sin x) = 4 + 4 sin x + sin2 x so = 4 cos x + 2 sin x cos x dx dy √ = 2 y cos x = 2 cos x(2 + sin x) = 4 cos x + 2 sin x cos x (c) from Part (a), dx

10. (a)

11. 2x + 2y

dy dy x = 0 so =− dx dx y

12. 3x2 + 3y 2

dy dy dy 3y 2 − 3x y 2 − x2 = 3y 2 + 6xy , = 2 = 2 dx dx dx 3y − 6xy y − 2xy

dy dy + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0 dx dx dy dy 1 − 2xy − 3y 3 (x2 + 9xy 2 ) = 1 − 2xy − 3y 3 so = dx dx x2 + 9xy 2

13. x2

127

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128

Chapter 4

dy dy + 3x2 y 2 − 5x2 − 10xy + 1 = 0 dx dx 10xy − 3x2 y 2 − 1 dy dy = 10xy − 3x2 y 2 − 1 so = (2x3 y − 5x2 ) dx dx 2x3 y − 5x2

14. x3 (2y)

dy

15. −

1 dy y 3/2 dx − = 0, = − dx 2x3/2 2y 3/2 x3/2

(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx) , (x − y)2 dy dy x(x − y)2 + y 2x(x − y)2 = −2y + 2x so = dx dx x

16. 2x =

dy dy 1 − 2xy 2 cos(x2 y 2 ) 17. cos(x2 y 2 ) x2 (2y) + 2xy 2 = 1, = dx dx 2x2 y cos(x2 y 2 ) dy dy dy y 2 sin(xy 2 ) = 18. − sin(xy 2 ) y 2 + 2xy , =− dx dx dx 2xy sin(xy 2 ) + 1 dy dy =1 19. 3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy + y2 + dx dx so

20.

dy 1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y) = dx 3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y)

(1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx) dy = 4y 3 , (1 + sec y)2 dx dy to get multiply through by (1 + sec y)2 and solve for dx dy y(1 + sec y) = 4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y dx

2 dy d2 y dy 2x dy 21. 4x − 6y − 6y 2 = 0, = 0, = , 4−6 dx dx 3y dx dx 2 dy −2 3 dx 2(3y 2 − 2x2 ) d2 y 8 = = − =− 3 3 2 9y 9y dx 3y 22.

23.

dy y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y(−x2 /y 2 ) 2x(y 3 + x3 ) x2 d2 y = − = − = − , = − 2, dx y dx2 y4 y4 y5 d2 y 2x but x3 + y 3 = 1 so =− 5 dx2 y x(dy/dx) − y(1) x(−y/x) − y 2y dy y d2 y =− =− = 2 =− , dx x dx2 x2 x2 x

24. y + x

25.

dy dy dy y dy d2 y + 2y = 0, =− ,2 +x 2 +2 dx dx dx x + 2y dx dx

dy dx

2 + 2y

d2 y d2 y 2y(x + y) = 0, = 2 dx dx2 (x + 2y)3

dy d2 y dy sin y = −(1 + cos y)−2 (− sin y) = (1 + cos y)−1 , = dx dx2 dx (1 + cos y)3

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Exercise Set 4.1

26.

129

cos y dy = , dx 1 + x sin y d2 y (1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y] = dx2 (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 d2 y sin 2y + y(sin2 y + 1) = − dx2 (1 + x sin y)3 √ √ dy dy x 27. By implicit differentiation, 2x + 2y(dy/dx) = 0, = − ; at (1/2, 3/2), = − 3/3; at dx y dx √ √ √ −x dy dy 2 (1/2, − 3/2), = + 3/3. Directly, at the upper point y = 1 − x , = √ = dx dx 1 − x2 √ √ √ x 1/2 dy =√ − = +1/ 3. = −1/ 3 and at the lower point y = − 1 − x2 , 2 dx 1−x 3/4 √ √ 28. If y 2 − 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 + 4y 3

30. 3y 2

dy 1 dy x3 = 0, so = − 3 = − 3/4 ≈ −0.1312. dx dx y 15

dy dy y+1 dy dy + x2 + 2xy + 2x − 6y = 0, so = −2x 2 = 0 at x = 0 dx dx dx dx 3y + x2 − 6y

dy 31. 4(x + y ) 2x + 2y dx 2

2

dy = 25 2x − 2y , dx

dy x[25 − 4(x2 + y 2 )] dy = ; at (3, 1) = −9/13 y[25 + 4(x2 + y 2 )] dx dx

32.

2 3

x−1/3 + y −1/3

dy dx

= 0,

√ √ dy y 1/3 = − 1/3 = 3 at (−1, 3 3) dx x

da da da 2t3 + 3a2 , solve for 33. 4a − 4t = 6 a + 2at to get = 3 dt dt dt dt 2a − 6at 3 da

34.

3

2

√ u du 1 −1/2 du 1 −1/2 = 0 so u + v = −√ 2 dv 2 dv v

35. 2a2 ω

dω dω b2 λ + 2b2 λ = 0 so =− 2 dλ dλ a ω

36. 1 = (cos x)

dx dx 1 = sec x so = dy dy cos x

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130

Chapter 4 y

37. (a) 2

x

–4

4 –2

(b) Implicit differentiation of the equation of the curve yields (4y 3 + 2y) only if x = 1/2 but y 4 + y 2 ≥ 0, so x = 1/2 is impossible. (c) x − x − (y + y ) = 0, so by the Quadratic Formula x = 2

4

2

1±

which gives the parabolas x = 1 + y 2 , x = −y 2 . 38. (a)

dy dy = 2x − 1 so =0 dx dx

1 + 4y 2 + 4y 4 = 1 + y 2 , −y 2 2

y 2 x 0

1

2

–2

dy dy = (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If = 0 then dx dx 3x2 − 2(a + b)x + ab = 0. By the Quadratic Formula 2(a + b) ± 4(a + b)2 − 4 · 3ab 1

a + b ± (a2 + b2 − ab)1/2 . x= = 6 3 (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 be nonpositive. If, for example, 0 < a < b then the function is defined on the disjoint intervals 0 < x < a and b < x < +∞, so there are two parts.

(b) 2y

(b) x ≈ ±1.1547

y

39. (a) 2

x –2

2 –2

dy dy y − 2x dy dy − y + 2y = 0. Solve for = . If =0 dx dx dx 2y − x dx 2 then y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y 2 = x2 − 2x2 + 4x2 = 3x2 , x = ± √ . 3

(c) Implicit differentiation yields 2x − x

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.

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Exercise Set 4.1

131

dy dx 2y − x dx dy − y + 2y = 0. Solve for = . If =0 dx dx dy y − 2x dy 2 4 then 2y − x = 0 or x = 2y. Thus 4 = 4y 2 − 2y 2 + y 2 = 3y 2 , y = ± √ , x = 2y = ± √ . 3 3

(c) Implicit differentiation yields 2x − x

41. Solve the simultaneous equations y = x, x2 −xy+y 2 = 4 to get x2 −x2 +x2 = 4, x = ±2, y = x = ±2, so the points of intersection are (2, 2) and (−2, −2). y − 2x dy dy dy = . When x = y = 2, = −1; when x = y = −2, = −1; From Exercise 39 part (c), dx 2y − x dx dx 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) lies on C then so does Q(−a, −b). dy y − 2x dy b − 2a From Exercise 39 part (c), = . When x = a, y = b then = , and when dx 2y − x dx 2b − a b − 2a dy = , so the slopes at P and Q are equal. x = −a, y = −b, then dx 2b − a 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 dy 2xy 2 4 implicit differentiation to get =− 2 so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4 dx x + 2ay 1 + 2a 3 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 y 3 = 2x2 can be obtained by implicit differentiation: dy 4x 4x dy dy = 4x, = 2 . Set = 2; 2 = 2, x = (3/2)y 2 . Use this in the equation of the curve: 3y 2 dx dx 3y dx 3y 2 3 2 2 . y 3 = 2x2 = 2((3/2)y 2 )2 = (9/2)y 4 , y = 2/9, x = = 2 9 27 (b) x ≈ 0.84

y

45. (a) 2

x

–3

–1

2 –1

–3

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

46. (a)

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

2

x

–3

–1

2 –1

–3

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132

Chapter 4

(c) Use implicit differentiation to get dy/dx = (2y −3x2 )/(3y 2 −2x), so dy/dx = −1 if 2y −3x2 = −3y 2 +2x, 2(y−x)+3(y−x)(y+x) = 0. One solution is y = x; this together with x3 +y 3 = 2xy yields 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 follows that 2(y − x) + 3(y − x)(y + x) = 0. But y = x, so x + y = −2/3. Then x3 + y 3 = (x + y)(x2 − xy + y 2 ) = 2xy, so replacing x + y with −2/3 we get x2 + 2xy + y 2 = 0, or (x + y)2 = 0, so y = −x. Substitute that into x3 + y 3 = 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 + y 2 = 1 about the point (2, 0) of radius 1. One tangent line is tangent at a point P(x,y) in the first quadrant. Let Q(2, 0) be the center of the circle. Then OP Q is a right angle, with sides |P Q| = r = 1 and |OP | = x2 + y 2 . By this into (x − 2)2 + y 2 = 1 to the Pythagorean Theorem x2 + y 2 + 1√2 = 22 . Substitute √ obtain √ 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 curve x2 − 4x + y 2 + 3 = 0. Implicit differentiation applied to the equation of the curve gives dy/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 y02 = 2x0 − x20 . But x20 − 4x0 + y02 + 3 = 0 because (x0 , y0 ) is on the 2 curve, and elimination of y02 in the latter two equations gives x20 − 4x0 + (2x0 − x√ 0 ) + 3 = 0, 2 2 2 into y = 2x − x yields y = 3/4, so y = ± x0 = 3/2 which when substituted 0 0 0 0 0 √ √ 3/2. The √ 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and slopes of the lines are (± √ y = −( 3/3)x. 48. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve 2x2 − 4x + y 2 + 1 = 0. Implicit differentiation applied to the equation of the curve gives dy/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 y02 = 2x0 (1 − x0 ). But 2x20 − 4x0 + y02 + 1 = 0 because (x0 , y0 ) is on the curve, and elimination of y02 in the latter two equations gives 2x√ 0 = 4x0 − 1, x0 = 1/2 which when 2 2 substituted into y = 2x (1 − x ) yields y = 1/2, so y = ± 2/2. 0 0 0 0 √ 0 √ √ 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−1 x + by0r−1 y = c is the equation of a line . Implicit differentiation of the 0 dy axr−1 dy = 0, = − r−1 . At the point (x0 , y0 ) the slope equation of the curve yields raxr−1 + rby r−1 dx dx by axr−1 0 of the line must be − r−1 , which is the slope of . Moreover, the equation of is satisfied by by0 the point (x0 , y0 ), so this point lies on . By the point-slope formula, must be the line tangent to the curve at (x0 , y0 ). dy = 0. At the point (1, 1) 50. Implicit differentiation of the equation of the curve yields rxr−1 + ry r−1 dx dy dy this becomes r + r = 0, = −1. dx dx 51. By the chain rule,

dy dt dy = . Use implicit differentiation on 2y 3 t + t3 y = 1 to get dx dt dx

2y 3 + 3t2 y dt dy 1 dy 2y 3 + 3t2 y =− , but . = so = − dt 6ty 2 + t3 dx cos t dx (6ty 2 + t3 ) cos t 4 1/3 4 x , f (x) = x−2/3 3 9 7 28 28 −2/3 x1/3 , f (x) = x (b) f (x) = x4/3 , f (x) = 3 9 27 (c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3

52. (a) f (x) =

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Exercise Set 4.2

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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 the intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain y−c 1 x−k cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and =− . 2 x y x dy x−k dy =− , and (gray) =− . But it was Differentiating the two families yields (black) dx y−c dx y proven that these quantities are negative reciprocals of each other. dy dy 56. Differentiating, we get the equations (black) x + y = 0 and (gray) 2x − 2y = 0. The first dx dx x y says the (black) slope is = − and the second says the (gray) slope is , and these are negative x y reciprocals of each other.

EXERCISE SET 4.2 1.

1 1 (5) = 5x x

2.

1 1 1 = x/3 3 x

3.

1 1+x

4.

1 √ 2+ x

5.

2x 1 (2x) = 2 2 x −1 x −1

6.

3x2 − 14x x3 − 7x2 − 3

7.

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

8.

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

9.

d 2 d (2 ln x) = 2 ln x = dx dx x

2

10. 3 (ln x)

12.

1 1 1 √ √ = 2x x2 x

14. x3

16.

1 x

1 + (3x2 ) ln x = x2 (1 + 3 ln x) x

3 2 log2 (x2 − 2x) + 3x log2 (x2 − 2x)

11.

1 √

2 x

1 (ln x)−1/2 2

13. ln x + x

1 √ = √ 2 x(2 + x)

1 1 = √ x 2x ln x

1 = 1 + ln x x

15. 2x log2 (3 − 2x) − 2x − 2 (x2 − 2x) ln 2

2x2 (3 − 2x) ln 2

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134

17.

19.

21.

Chapter 4

2x(1 + log x) − x/(ln 10) (1 + log x)2 1 1 1 = ln x x x ln x

18. 1/[x(ln 10)(1 + log x)2 ]

1 (sec2 x) = sec x csc x tan x

1 23. − sin(ln x) x

24.

20.

1 1 1 ln(ln(x)) ln x x

22.

1 (− sin x) = − tan x cos x

2 sin(ln x) cos(ln x)

1 sin(2 ln x) sin(ln x2 ) = = x x x

25.

1 cot x 2 (2 sin x cos x) = 2 ln 10 ln 10 sin x

26.

1 2 sin x cos x 2 tan x (−2 sin x cos x) = − =− 2 2 (ln 10) cos x ln 10 (ln 10)(1 − sin x)

27.

8x 11x2 − 8x + 3 d 3 3 ln(x − 1) + 4 ln(x2 + 1) = + 2 = dx x−1 x +1 (x − 1)(x2 + 1)

28.

29.

30.

1 d 2x3 [2 ln cos x + ln(1 + x4 )] = −2 tan x + dx 2 1 + x4 1 d 3x ln cos x − ln(4 − 3x2 ) = − tan x + dx 2 4 − 3x2 d dx

1 1 1 1 [ln(x − 1) − ln(x + 1)] = − 2 2 x−1 x+1

31. ln |y| = ln |x| +

2x 1 dy 1 3 ln |1 + x2 |, = x 1 + x2 + 3 dx x 3(1 + x2 )

1 dy 1 32. ln |y| = [ln |x − 1| − ln |x + 1|], = 5 dx 5

5

1 x−1 1 − x+1 x−1 x+1

1 1 ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5| 3 2 √ (x2 − 8)1/3 x3 + 1 2x 3x2 6x5 − 7 dy = + − dx x6 − 7x + 5 3(x2 − 8) 2(x3 + 1) x6 − 7x + 5

33. ln |y| =

34. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| −

1 ln |x| 2

3 sec2 x sin x cos x tan3 x 1 dy √ cot x − tan x + = − dx tan x 2x x

e−1

35. f (x) = ex

37. (a) logx e = (b) logx 2 =

36.

√ √ √ 10 dy 10 1 dy √ =− , =− ln y = − 10 ln x, y dx x dx x1+ 10

ln e 1 1 d = , [logx e] = − ln x ln x dx x(ln x)2 ln 2 d ln 2 , [logx 2] = − ln x dx x(ln x)2

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Exercise Set 4.2

135

1 ln b ln e for a, b > 0 it follows that log(1/x) e = =− , hence 38. (a) From loga b = ln a ln(1/x) ln x

d 1 log(1/x) e = dx x(ln x)2 (b) log(ln x) e =

1 ln e 1 1 d 1 = , so log(ln x) e = − =− ln(ln x) ln(ln x) dx (ln(ln x))2 x ln x x(ln x)(ln(ln x))2

1 , f (x0 ) = e, y − (−1) = e(x − 1/e) = ex − 1, y = ex − 2 x dy 1 = log e , 40. y0 = log 10 = 1, y = log x = (log e) ln x, dx x=10 10

39. f (x0 ) = f (e−1 ) = −1, f (x) =

log e log e (x − 10), y = x + 1 − log e 10 10 1 1 1 =− , 41. f (x0 ) = f (−e) = 1, f (x) 42. y − ln 2 = − (x + 2), y = − x + ln 2 − 1 e 2 2 x=−e y−1=

1 1 y − 1 = − (x + e), y = − x e e 43. Let the equation of the tangent line be y = mx and suppose that it meets the curve at (x0 , y0 ). 1 1 1 ln x0 1 Then m = = and y0 = mx0 = ln x0 . So m = = and ln x0 = 1, x0 = e, m = x x=x0 x0 x0 x0 e 1 and the equation of the tangent line is y = x. e 44. Let y = mx + b be a line tangent to the curve at (x0 , y0 ). Then b is the y-intercept and the 1 . Moreover, at the point of tangency, mx0 + b = ln x0 or slope of the tangent line is m = x0 1 x0 + b = ln x0 , b = ln x0 − 1, as required. x0 45. The area of the triangle P QR, given by |P Q||QR|/2 is required. |P Q| = w, and, by Exercise 44, |QR| = 1, so area = w/2.

y

1 P (w, ln w)

x

Q w

2

R –2

46. Since y = 2 ln x, let y = 2z; then z = ln x and we apply the result of Exercise 45 to ďŹ nd that the area 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, and

1 1 dy dy = . But ey = x + e, so = y = e−y . dx x+e dx e

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

dy 1 2 . But ey = e− ln(e −x) = (e2 − x)−1 , so = 2 dx e −x

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Chapter 4

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

1 dy = = e−y , and when x = 0, y = ln(a) = 0 dx x+a

if a = 1, so let a = 1, then y = ln(x + 1). 1 dy dy 1 = . But ey = , so = ey . dx a−x a−x dx If x = 0 then y = − ln(a) = − ln 2 provided a = 2, so y = − ln(2 − x).

50. Let y = − ln(a − x), then

d ln(e2 + ∆x) − 2 1 = (ln x) = = e−2 ∆x→0 ∆x dx x 2 2 x=e x=e ln(1 + h) − ln 1 ln(1 + h) 1 (b) f (w) = ln w; f (1) = lim =1 = lim = h→0 h→0 h h w w=1

51. (a) f (x) = ln x; f (e2 ) = lim

f (x) − f (0) = x→0 x

52. (a) Let f (x) = ln(cos x), then f (0) = ln(cos 0) = ln 1 = 0, so f (0) = lim lim

x→0

ln(cos x) , and f (0) = − tan 0 = 0. x

√ 2

f (1 + h) − f (1) (1 + h) (b) Let f (x) = x , then f (1) = 1, so f (1) = lim = lim h→0 h→0 h h √ √ √ f (x) = 2x 2−1 , f (1) = 2. √ 2

53.

logb (x + h) − logb (x) d [logb x] = lim h→0 dx h x+h 1 = lim logb h→0 h x h 1 = lim logb 1 + h→0 h x = lim

v→0

1 logb (1 + v) vx

−1

, and

Theorem 1.6.2(b)

Let v = h/x and note that v → 0 as h → 0

=

1 1 lim logb (1 + v) x v→0 v

h and v are variable, whereas x is constant

=

1 lim log (1 + v)1/v x v→0 b

Theorem 1.6.2.(c)

=

1 logb lim (1 + v)1/v v→0 x

Theorem 2.5.5

=

1 logb e x

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 −1 1 1 1 , (f −1 ) (3) = = f (x) = −1 dx f (f (x)) f (1) 9

2. (a) f (x) = 3x2 + 2ex ; for −1 < x < 1, f (x) ≼ 2e−1 = 2/e, and for |x| > 1, f (x) ≼ 3x2 ≼ 3, so f is increasing and one-to-one (b) f (0) = 2 so 0 = f −1 (2);

d −1 1 1 1 f (x) = −1 , (f −1 ) (2) = = dx f (f (x)) f (0) 2

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Exercise Set 4.3

137

d −1 2 2 − 3, so directly f (x) = − 2 . Using Formula (1), x dx x −2 1 = −(1/2)(f −1 (x) + 3)2 , f (x) = , so −1 (x + 3)2 f (f (x)) 2 2 2 d −1 f (x) = −(1/2) =− 2 dx x x

3. f −1 (x) =

4. f −1 (x) =

d −1 ex − 1 ex 2 , so directly, f (x) = . Next, f (x) = , and using Formula (1), 2 dx 2 2x + 1

ex d −1 2f −1 (x) + 1 f (x) = = dx 2 2 5. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not enough information. By inspection, 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 + cos x ≥ 1 > 0 for all x, so f is one-to-one

x (d) f (x) = −(ln 2) 12 < 0 because ln 2 > 0, so f is one-to-one for all x. not enough information; 6. (a) f (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so √ √ by observation (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 1 (c) f (x) = ; f is one-to-one because: (x + 1)2 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 1 logb x = (d) Note that f (x) is only defined for x > 0. , which is always negative dx x ln b (0 < b < 1), so f is one-to-one. 7. y = f −1 (x), x = f (y) = 5y 3 + y − 7, check: 1 = 15y 2

dx dy 1 ; = 15y 2 + 1, = dy dx 15y 2 + 1

dy dy 1 dy + , = 2 dx dx dx 15y + 1

8. y = f −1 (x), x = f (y) = 1/y 2 , check: 1 = −2y −3

dx dy = −2y −3 , = −y 3 /2; dy dx

dy dy , = −y 3 /2 dx dx

1 dx dy = 10y 4 + 3y 2 , = ; 4 dy dx 10y + 3y 2 1 dy dy dy + 3y 2 , = check: 1 = 10y 4 dx dx dx 10y 4 + 3y 2

9. y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1,

dx dy 1 = 5 − 2 cos 2y, = ; dy dx 5 − 2 cos 2y 1 dy dy = check: 1 = (5 − 2 cos 2y) , dx dx 5 − 2 cos 2y

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

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138

Chapter 4

11. 7e7x

12. −10xe−5x

13. x3 ex + 3x2 ex = x2 ex (x + 3)

14. −

15.

2

1 1/x e x2

(ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x ) dy = dx (ex + e−x )2 =

(e2x + 2 + e−2x ) − (e2x − 2 + e−2x ) = 4/(ex + e−x )2 (ex + e−x )2

16. ex cos(ex ) 17. (x sec2 x + tan x)ex tan x

19. (1 − 3e3x )e(x−e

21.

3x

)

x−1 (x − 1)e−x = x 1 − xe−x e −x

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

18.

dy ex (x ln x − 1) (ln x)ex − ex (1/x) = = dx (ln x)2 x(ln x)2

20.

15 2 x (1 + 5x3 )−1/2 exp( 1 + 5x3 ) 2

22.

1 [− sin(ex )]ex = −ex tan(ex ) cos(ex )

1 y = ln 2, y = y ln 2 = 2x ln 2 y

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

1 y = − ln 3, y = −y ln 3 = −3−x ln 3 y

25. f (x) = π sin x (ln π) cos x; y = π sin x , ln y = (sin x) ln π,

1 y = (ln π) cos x, y = π sin x (ln π) cos x y

26. f (x) = π x tan x (ln π)(x sec2 x + tan x); y = π x tan x , ln y = (x tan x) ln π,

1 y = (ln π)(x sec2 x + tan x) y

y = π x tan x (ln π)(x sec2 x + tan x) 1 dy 3x2 − 2 1 = 3 ln x + ln(x3 − 2x), y dx x − 2x x 2 dy 1 3 ln x 3x − 2 3 = (x − 2x) ln x + ln(x − 2x) dx x3 − 2x x

27. ln y = (ln x) ln(x3 − 2x),

sin x dy 1 dy sin x sin x = + (cos x) ln x, =x + (cos x) ln x 28. ln y = (sin x) ln x, y dx x dx x 1 dy 1 = tan x + (sec2 x) ln(ln x), y dx x ln x dy tan x + (sec2 x) ln(ln x) = (ln x)tan x dx x ln x

29. ln y = (tan x) ln(ln x),

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Exercise Set 4.3

139

2x 1 1 dy = 2 ln x + ln(x2 + 3), y dx x +3 x 2x dy 1 = (x2 + 3)ln x 2 ln x + ln(x2 + 3) x +3 x dx

30. ln y = (ln x) ln(x2 + 3),

31. f (x) = exe−1 32. (a) because xx is not of the form ax where a is constant 1 (b) y = xx , ln y = x ln x, y = 1 + ln x, y = xx (1 + ln x) y 3

35.

37.

3x2 3x2 = 3 2 1 + (x ) 1 + x6

1−

(3x)2

1 1 − 1/x2

=√

3 1 − 9x2

33.

(−1/x2 ) = −

34. − 1 √ |x| x2 − 1

1/2 1

x+1 2 = − 4 − (x + 1)2 1− 2

sin x sin x = = | sin x| 1 − cos2 x

1, sin x > 0 −1, sin x < 0

36.

√

38.

5x4 5 √ = |x| x10 − 1 |x5 | (x5 )2 − 1

39. y = 1/ tan x = cot x, dy/dx = − csc2 x −1

40. y = (tan

41.

−1

x)

−1

, dy/dx = −(tan

ex √ + ex sec−1 x |x| x2 − 1

−2

x)

1 1 + x2

42. −

(cos−1

1 √ x) 1 − x2

3x2 (sin−1 x)2 √ + 2x(sin−1 x)3 1 − x2

43. 0

44.

45. 0

√ 46. −1/ e2x − 1

1 47. − 1+x

1 −1/2 x 2

=−

1 √ 2(1 + x) x

1 48. − √ −1 2 cot x(1 + x2 )

49. (a) Let x = f (y) = cot y, 0 < y < Ď€, −∞ < x < +∞. Then f is dierentiable and one-to-one and f (f −1 (x)) = − csc2 (cot−1 x) = −x2 − 1 = 0, and d 1 1 = lim −1 = − lim 2 = −1. [cot−1 x] x→0 f (f x→0 x + 1 (x)) dx x=0

(b) If x = 0 then, from Exercise 50(a) of Section 1.5, 1 d 1 1 d 1 cot−1 x = tan−1 = − 2 . For x = 0, Part (a) shows the same; =− 2 dx dx x x 1 + (1/x)2 x +1 d 1 thus for −∞ < x < +∞, [cot−1 x] = − 2 . x +1 dx (c) For −∞ < u < +∞, by the chain rule it follows that

1 du d [cot−1 u] = − 2 . dx u + 1 dx

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140

Chapter 4

50. (a) By the chain rule, (b) By the chain rule,

1 1 1 d d −1 √ [csc−1 x] = sin−1 = − 2 = 2 dx dx x x |x| x2 − 1 1 − (1/x) du d −1 du d √ [csc−1 u] = [csc−1 u] = dx dx du |u| u2 − 1 dx

51. x3 + x tan−1 y = ey , 3x2 +

x (3x2 + tan−1 y)(1 + y 2 ) y + tan−1 y = ey y , y = 2 1+y (1 + y 2 )ey − x 1

52. sin−1 (xy) = cos−1 (x − y),

1−

x2 y 2

y 1 − (x − y)2 + 1 − x2 y 2 y = 1 − x2 y 2 − x 1 − (x − y)2

1

(xy + y) = −

1 − (x − y)2

(1 − y ),

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. √ √ 6 Âą 36 − 24 2 (b) f (x) = 3x − 6x + 2, f (x) = 0 when x = = 1 Âą 3/3. f (x) > 0 (f is 6 √ √ √ 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 f is one-to-one. (b) F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f (1) = 3, so g(3) = f −1 (3) = 1 and g (3) = (f −1 ) (3) = 1/f (f −1 (3)) = 1/f (1) = 1/7 because f (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 equation y − 25 = (88/7)(x − 3), y = (88/7)x − 89/7.

4−x2

56. (a) f (x) = −e

1 2+ 2 x

< 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, 9 1 = − , and f (2) = − 2 + 4 4 2 d 2 2 = f ([g(x)] )2g(x)g (x) F (1/2) = f ([g(x)] ) [g(x) ] dx x=1/2 x=1/2 1 −12 e (2 + 16 ) 1 f (4) 11 33 =4 = f (22 )2 ¡ 2 =4 = 12 = 12 1 f (g(x)) x=1/2 f (2) 9e 3e (2 + 4 ) 57. (a) f (x) = kekx , f (x) = k 2 ekx , f (x) = k 3 ekx , . . . , f (n) (x) = k n ekx (b) g (x) = −ke−kx , g (x) = k 2 e−kx , g (x) = −k 3 e−kx , . . . , g (n) (x) = (−1)n k n e−kx 58.

dy = eâˆ’Îťt (ωA cos ωt − ωB sin ωt) + (âˆ’Îť)eâˆ’Îťt (A sin ωt + B cos ωt) dt = eâˆ’Îťt [(ωA − ÎťB) cos ωt − (ωB + ÎťA) sin ωt]

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Exercise Set 4.3

141

2 2 1 xâˆ’Âľ d 1 1 xâˆ’Âľ − 59. f (x) = √ exp − 2 Ďƒ dx 2 Ďƒ 2Ď€Ďƒ 2 1 1 xâˆ’Âľ 1 xâˆ’Âľ =√ − exp − 2 Ďƒ Ďƒ Ďƒ 2Ď€Ďƒ 2 1 1 xâˆ’Âľ = −√ (x − Âľ) exp − 2 Ďƒ 2Ď€Ďƒ 3

60. y = Aekt , dy/dt = kAekt = k(Aekt ) = ky 61. y = Ae2x + Be−4x , y = 2Ae2x − 4Be−4x , y = 4Ae2x + 16Be−4x so y + 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 = −x2 e−x

2

63.

/2

+ e−x

2

/2

= e−x

2

/2

(1 − x2 ), xy = xe−x

2

/2

(1 − x2 ) = y(1 − x2 )

dy = 100(−0.2)e−0.2x = −20y, k = −0.2 dx

64. ln y = (5x + 1) ln 3 − (x/2) ln 4, so y /y = 5 ln 3 − (1/2) ln 4 = 5 ln 3 − ln 2, and y = (5 ln 3 − ln 2)y 7e−t y 7e−t + 5 − 5 1 = = = 1 − y, so 65. ln y = ln 60 − ln(5 + 7e−t ), y 5 + 7e−t 5 + 7e−t 12 dy y y, with r = 1, K = 12. =r 1− K dt 12

66. (a)

0

9 0

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

t→+∞

60 60 60 = = = 12 5 + 7e−t 5 5 + 7 lim e−t t→+∞

(c) the rate of population growth tends to zero 3.2

0

9 0

67.

d x 10h − 1 d x ln 10 = 10 e = = ln 10 h→0 h dx dx x=0 x=0 lim

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142

68.

69.

Chapter 4

tan−1 (1 + h) − π/4 1 1 d = = = tan−1 x 2 h→0 h dx 1 + x 2 x=1 x=1 lim

lim

9[sin−1 (

√ 3 2

∆x→0

+ ∆x)]2 − π 2 d 3 = (3 sin−1 x)2 √ = 2(3 sin−1 x) √ ∆x dx 1 − x2 x=√3/2 x= 3/2

π 3 = 2(3 ) = 12π 3 1 − (3/4) d x (2 + ∆x)(2+∆x) − 4 d x ln x = x e = 70. lim ∆x→0 ∆x dx x=2 dx x=2 = (1 + ln 2)22 = 4(1 + ln 2) = (1 + ln x)ex ln x x=2

√ 3 sec−1 w − π 3 3 d −1 71. lim = √ = = 3 sec x 2 w→2 w−2 dx 2 |2| 2 − 1 x=2 72.

d 4(tan−1 w)w − π d x ln tan−1 x = 4(tan−1 x)x 4e = w→1 w−1 dx dx x=1 x=1 2 1/(1 + x ) 14 == 2 + π ln(π/4) = 4(tan−1 x)x ln tan−1 x + x = π ln(π/4) + 2π tan−1 x x=1 lim

EXERCISE SET 4.4 1. (a)

lim

x→2 x2

2 x2 − 4 (x − 2)(x + 2) x+2 = lim = lim = + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3

5 2 − lim 2x − 5 2 x→+∞ x = = (b) lim 7 x→+∞ 3x + 7 3 3 + lim x→+∞ x 2. (a) (b)

cos x sin x sin x = sin x = cos x so lim = lim cos x = 1 x→0 tan x x→0 tan x sin x x2 − 1 x2 − 1 (x − 1)(x + 1) x+1 2 = = 2 so lim 3 = 3 2 x→1 x − 1 x −1 (x − 1)(x + x + 1) x +x+1 3

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

π π , Tg (x) = − x − 4. Tf (x) = − x − 2 2 limit = 1

5.

lim

ex =1 x→0 cos x

6.

7.

sec2 θ =1 θ→0 1

8. lim

9.

lim

lim+

x→π

11.

cos x = −1 1

1/x =0 x→+∞ 1 lim

lim

x→3

1 = 1/5 6x − 13

tet + et = −1 t→0 −et

10.

12.

lim

x→0+

cos x = +∞ 2x

3e3x 9e3x = lim = +∞ x→+∞ 2x x→+∞ 2 lim

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Exercise Set 4.4

13.

lim+

− csc2 x −x −1 = lim = −∞ = lim+ 2 + 1/x x→0 sin x x→0 2 sin x cos x

lim

−1/x x = lim 1/x = 0 x→0+ e (−1/x2 )e1/x

x→0

14.

15.

143

x→0+

100x99 (100)(99)x98 (100)(99)(98) · · · (1) = lim = · · · = lim =0 x x x→+∞ x→+∞ x→+∞ e e ex lim

√ 2/ 1 − 4x2 17. lim =2 x→0 1

cos x/ sin x 16. lim = lim+ cos2 x = 1 x→0+ sec2 x/ tan x x→0 1− 18.

20.

21.

22.

23.

24.

lim

x→0

1 1 1 1 + x2 = lim = 2 2 x→0 3x 3(1 + x ) 3

lim (x − π) tan(x/2) = lim

x→π

x→π

19.

lim xe−x = lim

x→+∞

x→+∞

x 1 = lim x = 0 x x→+∞ e e

x−π 1 = lim = −2 cot(x/2) x→π −(1/2) csc2 (x/2)

sin(π/x) (−π/x2 ) cos(π/x) = lim = lim π cos(π/x) = π x→+∞ x→+∞ x→+∞ 1/x −1/x2

lim x sin(π/x) = lim

x→+∞

lim tan x ln x = lim

x→0+

x→0+

lim

x→(π/2)−

ln x 1/x − sin2 x −2 sin x cos x = lim+ = lim = lim+ =0 cot x x→0 − csc2 x x→0+ x 1 x→0

sec 3x cos 5x =

lim (x − π) cot x = lim

x→π

x→π

lim

x→(π/2)−

−5(+1) 5 cos 5x −5 sin 5x = lim = =− cos 3x x→(π/2)− −3 sin 3x (−3)(−1) 3

x−π 1 = lim =1 x→π sec2 x tan x

25. y = (1 − 3/x)x , lim ln y = lim x→+∞

x→+∞

26. y = (1 + 2x)−3/x , lim ln y = lim − x→0

x→0

ln(1 − 3/x) −3 = lim = −3, lim y = e−3 x→+∞ 1 − 3/x x→+∞ 1/x

3 ln(1 + 2x) 6 = lim − = −6, lim y = e−6 x→0 x→0 x 1 + 2x

ln(ex + x) ex + 1 = lim x = 2, lim y = e2 x→0 x→0 e + x x→0 x

27. y = (ex + x)1/x , lim ln y = lim x→0

28. y = (1 + a/x)bx , lim ln y = lim x→+∞

x→+∞

b ln(1 + a/x) ab = lim = ab, lim y = eab x→+∞ 1 + a/x x→+∞ 1/x

ln(2 − x) 2 sin2 (πx/2) = lim = 2/π, lim y = e2/π x→1 cot(πx/2) x→1 x→1 π(2 − x)

29. y = (2 − x)tan(πx/2) , lim ln y = lim x→1

2

ln cos(2/x) (−2/x2 )(− tan(2/x)) = lim 2 x→+∞ x→+∞ 1/x −2/x3

30. y = [cos(2/x)]x , lim ln y = lim x→+∞

− tan(2/x) (2/x2 ) sec2 (2/x) = lim = −2, x→+∞ x→+∞ 1/x −1/x2

= lim 31.

lim

x→0

1 1 − sin x x

= lim

x→0

lim y = e−2

x→+∞

x − sin x 1 − cos x sin x = lim = lim =0 x→0 x cos x + sin x x→0 2 cos x − x sin x x sin x

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144

32.

Chapter 4

lim

x→0

9 1 − cos 3x 3 sin 3x 9 = lim cos 3x = = lim 2 x→0 x→0 x 2x 2 2

33.

(x2 + x) − x2 x 1 √ = 1/2 = lim √ = lim 2 2 x→+∞ x→+∞ x→+∞ x +x+x x +x+x 1 + 1/x + 1

34.

ex − 1 − x ex − 1 ex = lim = lim = 1/2 x→0 xex − x x→0 xex + ex − 1 x→0 xex + 2ex

35.

lim

lim

lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln

x→+∞

x→+∞ x

x

lim

x→+∞

36.

x→+∞

ex , x2 + 1

x

e e e = lim = lim = +∞ so lim [x − ln(x2 + 1)] = +∞ x→+∞ x2 + 1 x→+∞ 2x x→+∞ 2

lim ln

x→+∞

x 1 = lim ln = ln(1) = 0 1 + x x→+∞ 1/x + 1

ln x 1/x 1 = lim = lim =0 x→+∞ nxn−1 x→+∞ nxn xn xn nxn−1 = lim = lim nxn = +∞ (b) lim x→+∞ ln x x→+∞ 1/x x→+∞

38. (a)

lim

x→+∞

0 3x2 − 2x + 1 because it is not a form. x→1 3x2 − 2x 0

39. (a) L’Hˆ opital’s Rule does not apply to the problem lim 3x2 − 2x + 1 =2 x→1 3x2 − 2x

(b) lim

e3x −12x+12 e0 40. L’Hˆ opital’s Rule does not apply to the problem , which is of the form , and from x4 − 16 0 which it follows that lim− and lim+ exist, with values −∞ if x approaches 2 from the left and 2

x→2

x→2

+∞ if from the right. The general limit lim does not exist. x→2

41.

lim

x→+∞

1/(x ln x) 2 √ = lim √ =0 x→+∞ 1/(2 x) x ln x

0.15

100

10000 0

42. y = xx , lim ln y = lim x→0+

x→0+

ln x = lim −x = 0, lim y = 1 1/x x→0+ x→0+

1

0

0.5 0

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Exercise Set 4.4

145

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

25

3 ln sin x x = lim (3 cos x) = 3, ln x sin x x→0+

lim ln y = lim

x→0+

x→0+

lim y = e3

x→0+

0

0.5 19

44.

lim −

x→π/2

4 sec2 x 4 = lim =4 sec x tan x x→π/2− sin x

4.1

1.4

1.6 3.3

45. ln x − ex = ln x −

1 e−x

lim e−x ln x = lim

x→+∞

=

x→+∞

e−x ln x − 1 ; e−x

0 0

3

ln x 1/x = lim = 0 by L’Hˆ opital’s Rule, x→+∞ ex ex

e−x ln x − 1 = −∞ x→+∞ e−x

so lim [ln x − ex ] = lim x→+∞

–16

46.

lim [ln ex − ln(1 + 2ex )] = lim ln

x→+∞

= lim ln x→+∞

x→+∞

ex 1 + 2ex

–0.6 0

12

1 1 = ln ; +2 2

e−x

horizontal asymptote y = − ln 2 –1.2

47. y = (ln x)1/x ,

1.02

ln(ln x) 1 = lim = 0; x→+∞ x→+∞ x ln x x

lim ln y = lim

x→+∞

lim y = 1, y = 1 is the horizontal asymptote

x→+∞

100

10000 1

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146

Chapter 4

48. y =

x+1 x+2

x

x+1 x +2 , lim ln y = lim x→+∞ x→+∞ 1/x −x2 = lim = −1; x→+∞ (x + 1)(x + 2) ln

1

lim y = e−1 is the horizontal asymptote

x→+∞

0

50 0

49. (a) 0

(b) +∞

(d) −∞

(c) 0

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

lim y = eln a = a

x→0+

(e) +∞

(f ) −∞

(ln a) ln x (ln a)/x = lim+ = lim+ ln a = ln a, 1 + ln x 1/x x→0 x→0

x→0+

(b) Type ∞0 ; same calculation as Part (a) with x → +∞ (c) Type 1∞ ; y = (x + 1)(ln a)/x , lim ln y = lim x→0

x→0

lim y = eln a = a

(ln a) ln(x + 1) ln a = lim = ln a, x→0 x + 1 x

x→0

sin 2x 1 + 2 cos 2x x + sin 2x does not exist, nor is it ±∞; lim = lim =1 1+ 51. lim x→+∞ x→+∞ x→+∞ 1 x x 52.

53.

lim

x→+∞

2 2 − cos x 2x − sin x 2 − (sin x)/x does not exist, nor is it ±∞; lim = lim = x→+∞ x→+∞ 3 + cos x 3x + sin x 3 + (sin x)/x 3

x(2 + sin 2x) 2 + sin 2x = lim , x→+∞ 1 + 1/x x+1 which does not exist because sin 2x oscillates between −1 and 1 as x → +∞ lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim

x→+∞

x→+∞

54.

lim

x→+∞

lim

x→+∞

55.

lim+

R→0

sin x 1 1 + cos x + x 2 2x

does not exist, nor is it ±∞;

x(2 + sin x) 2 + sin x = lim =0 x→+∞ x + 1/x x2 + 1 V t −Rt/L L e

1

=

Vt L

π/2 − x −1 = lim = lim sin2 x = 1 cot x x→π/2 − csc2 x x→π/2 1 1 cos x − (π/2 − x) sin x sin x = lim (b) lim − tan x = lim − cos x (π/2 − x) cos x x→π/2 π/2 − x x→π/2 π/2 − x x→π/2

56. (a)

lim (π/2 − x) tan x = lim

x→π/2

x→π/2

= lim x→π/2

−(π/2 − x) cos x −(π/2 − x) sin x − cos x

(π/2 − x) sin x + cos x =0 x→π/2 −(π/2 − x) cos x + 2 sin x

= lim

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

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Review Exercises, Chapter 4

147

kt − 1 (ln k)k t = lim+ = ln k x→+∞ t 1 t→0

√ (c) ln 0.3 = −1.20397, 1024 1024 0.3 − 1 = −1.20327;

√ ln 2 = 0.69315, 1024 1024 2 − 1 = 0.69338

57. (b)

lim x(k 1/x − 1) = lim

t→0+

k + cos x = ±∞. Hence k = −1, and by the x→0 x→0 x2 √ −1 + cos x − sin x − 2 cos x 2 = lim rule lim = lim = − = −4 if = ±2 2. x→0 x→0 x→0 x2 2x 2 2

58. If k = −1 then lim (k + cos x) = k + 1 = 0, so lim

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

0.05

(b)

–0.35

0.35

–0.05

(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 f (x) = 0. +

x→0

− cos(1/x) + 2x sin(1/x) which does not exist (nor is it ±∞). x→0 cos x

x x 1 [x sin(1/x)], but lim = lim = 1 and lim x sin(1/x) = 0, (b) Rewrite as lim x→0 sin x x→0 sin x x→0 cos x x→0

x [x sin(1/x)] = (1)(0) = 0 thus lim x→0 sin x

60. (a) Apply the rule to get lim

61.

lim

x→0+

sin(1/x) sin x , lim+ = 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between (sin x)/x x→0 x x→0

−1 and 1 as x → +∞, so lim+ x→0

x sin(1/x) does not exist. sin x

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

(f (x) − f (0)/(x − a) f (x) = . Now take the limit: g(x) (g(x) − g(0))/(x − a)

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

REVIEW EXERCISES, CHAPTER 4 1.

1 3 (6) = 4(6x − 5)3/4 2(6x − 5)3/4

3. dy/dx =

2.

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

1/2 1/2 9 x−1 d x−1 3 x−1 = 2 x+2 dx x + 2 2(x + 2)2 x + 2

4 x2 (3 − 2x)1/3 (−2) − (3 − 2x)4/3 (2x) 2(3 − 2x)1/3 (2x − 9) 3 4. dy/dx = = x4 3x3

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148

Chapter 4

dy 2 − y − 3x2 dy + y − 2 = 0, = dx dx x (b) y = (1 + 2x − x3 )/x = 1/x + 2 − x2 , dy/dx = −1/x2 − 2x

5. (a) 3x2 + x

2 − (1/x + 2 − x2 ) − 3x2 dy = = −1/x2 − 2x dx x

(c)

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

6. (a) xy = x − y, x

x 1 − x+1 dy 1−y 1 = = = 2 dx x+1 1+x x +1

(c) 7. −

1 y2 1 dy dy − = − = 0 so y 2 dx x2 dx x2

dy x2 − 2y dy dy dy = 6(x + y), −(3y 2 + 6x) = 6y − 3x2 so = 2 dx dx dx dx y + 2x dy dy dy y sec(xy) tan(xy) 9. x + y sec(xy) tan(xy) = , = dx dx dx 1 − x sec(xy) tan(xy)

8. 3x2 − 3y 2

(1 + csc y)(− csc2 y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx) , (1 + csc y)2 dy 2x(1 + csc y)2 = − csc y(csc y + csc2 y − cot2 y) , dx 2x(1 + csc y) dy =− but csc2 y − cot2 y = 1, so dx csc y

10. 2x =

11.

dy (4y)(3) − (3x)(4dy/dx) 12y − 12x(3x/(4y)) 12y 2 − 9x2 −3(3x2 − 4y 2 ) 3x d2 y = = = = , = , 2 2 2 3 dx 4y dx 16y 16y 16y 16y 3 but 3x2 − 4y 2 = 7 so

12.

d2 y −3(7) 21 = =− dx2 16y 3 16y 3

y dy = , dx y−x d2 y (y − x)(dy/dx) − y(dy/dx − 1) = = 2 dx (y − x)2 =

13.

(y − x)

y y −y −1 y−x y−x (y − x)2

y 2 − 2xy d2 y 3 2 but y − 2xy = −3, so =− (y − x)3 dx2 (y − x)3

dy dy dy 2 dy dy = tan(πy/2) + x(π/2) sec2 (πy/2), = 1 + (π/4) (2), = dx dx dx dx dx π−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 of the tangent to y 2 = 2x3 at P must be −3/4. By implicit differentiation dy/dx = 3x2 /y, so at P , 3x20 /y0 = −3/4, or y0 = −4x20 . But y02 = 2x30 because P is on the curve y 2 = 2x3 . Elimination of y0 gives 16x40 = 2x30 , x30 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x20 it follows that y0 = 0 when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution because dy/dx = 3x2 /y (the slope of the curve as determined by implicit differentiation) is valid only 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.

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Review Exercises, Chapter 4

149

15. Substitute√y = mx into x2 + xy + y 2 = 4 to get x2 + mx2 + m2 x2 = 4, which has distinct solutions x = ±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 slope is unchanged if we replace (x, y) with (−x, −y), it follows that the slopes are equal at the two point of intersection. Finally we must examine the special case x = 0 which cannot be written in the form y = mx. If x = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal. = 0 if y = 3x2 . Substitute 16. Use implicit differentiation to get dy/dx = (y − 3x2 )/(3y 2 − x), so dy/dx √ √ 3 3 3 6 3 3 this into x − xy + y = 0 to obtain 27x − 2x = 0, x = 2/27, x = 2/3 and hence y = 3 4/3. 17. By implicit differentiation, 3x2 − y − xy + 3y 2 y = 0, so y = (3x2 − y)/(x − 3y 2 ). This derivative exists except when x = 3y 2 . Substituting this into the original equation x3 − xy + y 3 = 0, one has 27y 6 − 3y 3 + y 3 = 0, y 3 (27y 3 − 2) = 0. The unique solution in the first quadrant is y = 21/3 /3, x = 3y 2 = 22/3 /3 18. By implicit differentiation, dy/dx = k/(2y) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is k (x − x0 ), or 2y0 y − 2y02 = kx − kx0 . k/(2y0 ) if y0 = 0. The tangent line in this case is y − y0 = 2y0 But y02 = kx0 because (x0 , y0 ) is on the curve y 2 = kx, so the equation of the tangent line becomes 2y0 y − 2kx0 = kx − kx0 which gives y0 y = k(x + x0 )/2. If y0 = 0, then x0 = 0; the graph of y 2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k(x + x0 )/2 gives the same result when x0 = y0 = 0. 19. y = ln(x + 1) + 2 ln(x + 2) − 3 ln(x + 3) − 4 ln(x + 4), dy/dx = 20. y =

2 3 4 1 + − − x+1 x+2 x+3 x+4

1 1 ln x + ln(x + 1) − ln sin x + ln cos x, so 2 3

dy 1 1 cos x sin x 5x + 3 = + − − cot x − tan x − = dx 2x 3(x + 1) sin x cos x 6x(x + 1) 21.

1 (2) = 1/x 2x

2 ln x 1 = 22. 2(ln x) x x

23.

1 3x(ln x + 1)2/3

24. y =

25. log10 ln x =

1 3

ln(x + 1), y =

1 3(x + 1)

ln ln x 1 ,y = ln 10 (ln 10)(x ln x)

26. y =

(ln 10 − ln x)/x + (ln 10 + ln x)/x 2 ln 10 1 + ln x/ ln 10 ln 10 + ln x = = ,y = 1 − ln x/ ln 10 ln 10 − ln x (ln 10 − ln x)2 x(ln 10 − ln x)2

27. y =

3 3 1 2x3 ln x + ln(1 + x4 ), y = + 2 2 2x (1 + x4 )

28. y =

2x 1 1 − 3x2 1 sin x − − tan x = ln x + ln cos x − ln(1 + x2 ), y = − 2 2x cos x 1 + x2 2x(1 + x2 )

29. y = x2 + 1 so y = 2x.

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150

Chapter 4

30. y = ln

ex (1 + ex + e2x ) dy = − ln(1 − ex ), = x x 2x (1 − e )(1 + e + e ) dx 1 − ex √ x

31. y = 2e 32. y =

√ x

+ 2xe

√ √ √ d √ x = 2e x + xe x dx

abe−x (1 + be−x )2

33. y =

2 π(1 + 4x2 )

−1 ln 2 2sin x 2 1−x

x x x dy 1 1 y = ex + ln x , = xe ex + ln x = ex xe −1 + xe ln x 35. ln y = ex ln x, y x dx x

34. y = e(sin

36. ln y =

−1

x) ln 2

, y = √

1 ln(1 + x) y x/(1 + x) − ln(1 + x) ln(1 + x) = , , = − x y x2 x(1 + x) x2

dy (1 + x)(1/x) 1 ln(1 + x) = (1 + x)(1/x)−1 − x2 dx x 37. y =

2 |2x + 1| (2x + 1)2 − 1

d x 1 1 √ cos−1 x2 = − √ 38. y = √ cos−1 x2 1 − x4 2 cos−1 x2 dx x 3x2 1 3 x4 ln(x2 + 1), y /y = − 2 ,y=√ − 2 2 x x +1 x2 + 1 (x + 1)3/2 2 2x 4x 1 y 4x 1 2x 3 x − 1 2 2 = 40. ln y = (ln(x −1)−ln(x +1)), = − 2 so y = 2 4 4 3 y 3 x −1 x +1 3(x − 1) 3(x − 1) x2 + 1 39. ln y = 3 ln x −

y

41. (b)

(c)

6

1 1 dy dy dy = − so < 0 at x = 1 and > 0 at x = e dx 2 x dx dx

4 2 x 1

2

3

4

(d) The slope is a continuous function which goes from a negative value at x = 1 to a positive value at x = e; therefore it must take the value zero between, by the Intermediate Value Theorem. dy (e) = 0 when x = 2 dx 10 dβ = 42. β = 10 log I − 10 log I0 , dI I ln 10 dβ 1 (a) = db/W/m2 dI I=10I0 I0 ln 10 dβ 1 (c) = db/W/m2 dI I=100I0 100I0 ln 10

(b)

dβ dI

= I=100I0

1 db/W/m2 10I0 ln 10

dx dy dy dx dx dy =3 given y = x ln x. Then = = (1 + ln x) , so 1 + ln x = 3, ln x = 2, dt dt dt dx dt dt x = e2 .

43. Solve

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151

44. x = 2, y = 0; y = −2x/(5 − x2 ) = −4 at x = 2, so y − 0 = −4(x − 2) or y = −4x + 8 1 45. Set y = logb x and solve y = 1: y = = 1 so x ln b 1 x = . The curves intersect when (x, x) lies on the ln b graph of y = logb x, so x = logb x. From Formula (8), ln x Section 1.6, logb x = from which ln x = 1, x = e, ln b ln b = 1/e, b = e1/e ≈ 1.4447. √ 46. (a) Find the point of intersection: f (x) = x + k = ln x. The 1 1 √ slopes are equal, so m1 = = m2 = √ , x = 2, x = 4. x 2 x √ Then ln 4 = 4 + k, k = ln 4 − 2. k 1 √ = m2 = , so x 2 x √ √ k x = 2. At the point of intersection k x = ln x, 2 = ln x,

y

2 x 2

y 2 x 2

y

(b) Since the slopes are equal m1 =

x = e2 , k = 2/e.

2 x 0

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 −1 3 1 (f −1 (x) + 1)2 (3/x)2 3 f (x) = − 2 ; and −1 =− =− = − 2. dx x f (f (x)) 3 3 x d −1 2 f (x) = ; (b) f (x) = ex/2 , f (x) = 12 ex/2 . If x = f (y) = ey/2 then y = f −1 (x) = 2 ln x, so dx x −1 1 2 and −1 = 2e−f (x)/2 = 2e− ln x = 2x−1 = f (f (x)) x 49. Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an equation of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line passes through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives x0 = 1/3 and thus y0 = e. The point is (1/3, e). 50. ln y = ln 5000 + 1.07x;

dy/dx dy = 1.07, or = 1.07y y dx

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

dy dy/dx = 2 ln 3 + 7 ln 5, or = (2 ln 3 + 7 ln 5)y y dx

52.

dk qk0 q(T − T0 ) q q(T − T0 ) − 2 = − 2 exp − = k0 exp − dT 2T0 T 2T 2T 2T0 T

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.

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Chapter 4

√ √ 54. sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y = y + 2 sin y cos3 y =

1 −2x x + 2√ = 0. 2 )3/2 2 (1 + x2 )2 (1 + x 1+x

1 −2x , y = , hence 1 + x2 (1 + x2 )2

100

55. (a)

0

8 20

(b) as t tends to +∞, the population tends to 19 95 95 95 = = lim P (t) = lim = 19 t→+∞ t→+∞ 5 − 4e−t/4 5 5 − 4 lim e−t/4 t→+∞

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

0

8

–80

d (1 + h)Ď€ − 1 Ď€ Ď€âˆ’1 = (1 + x) = Ď€(1 + x) =Ď€ 56. (a) y = (1 + x) , lim h→0 h dx x=0 x=0 1 − ln x 1 − ln x 1/x 1 dy (b) Let y = =− =− . Then y(e) = 0, and lim = x→e (x − e) ln x ln x dx x=e (ln x)2 e Ď€

57. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, because large 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 f (x) = 0. Consider g(x) = x; lim g(x) = 0. For f (x) choose x→a

cos x, x , and |x| 2

59.

1/2

x→0

cos x x2 |x|1/2 . Then: lim = +∞. does not exist, lim = 0, and lim x→0 x x→0 x x→0 x2

ex ex ex = lim = lim = +∞ x→+∞ x→+∞ x→+∞ x2 x→+∞ 2x x→+∞ 2 so lim (ex /x2 − 1) = +∞ and thus lim x2 (ex /x2 − 1) = +∞ lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim x→+∞

ln x 1/x 1 60. lim 4 = ; lim = lim 3 x→1 x − 1 x→1 4x 4 x→1

x→+∞

ln x = 4 x −1

lim

ln x 1 = −1 2

x→1 x4

(x2 + 2x)ex (x2 + 2x)ex (x2 + 4x + 2)ex 1 = lim = lim = x→0 6 sin 3x cos 3x x→0 x→0 3 sin 6x 18 cos 6x 9

61. = lim 62.

lim ax ln a = ln a

x→0

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