Blank B.E., Krantz S.G. Calculus: Single Variable

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12. exp (x

)

13. ln (x 1 1)

14. (x

2 1)(x 1 5)

15. (x 1 1)/(x 2 1)

16. x/(22x)

17. 27(x

1 1)

5/3

18. x sin (x)

19. x cos (2x)

20. tan (x)

21. 2

22. x/3

23. e

sin (x)

24. 2

cos (3x)

25. log

(x)

26. x log

(x)

c In Exercises 2736, calculate the requested derivative. b

27. f

(5)

(x) where f (x) 5 sin (x)

28.

where f (x) 5

x 2 1

29.

€

g(1) where g(t) 5 (3t

2 2)

30. f

000

(π/6) where f (t) 5 sin (t) 1 cos (2t)

31. H

(4)

(x) where H(x) 52x

1 7x

2 9x

1 11

32.

where f ðxÞ5 sec ð5xÞ

33. f

000

(x) where f (x) 5 cos (4x 1 3)

34.

25/3

1 4t

2 8t

7/2

)

35. g

(π/6) where g( x) 5 cos (π sin (x))

36. H

(2)

(π/4) where H(x) 5 x tan (x)

c In Exercises 3740, a velocity υ(t)

is given. Calculate the

acceleration (in m/s

). b

37. υ(t) 5 24

m/s

38. υ(t) 5 t

2 5t m/s

39. υ(t)5216t 1 5 m/s

40. υ(t) 5 8e

2t/2

m/s

c In Exercises 4144, a position p(t)

is given. Calculate the

acceleration (in m/s

). b

41. p(t) 5 15 1 24t m

42. p(t) 5 3

sin (2t)m

43. p(t) 5 3t

1 16t m

44. p(t) 5 10 2 3e

22t

45. The position of a bicycle at time t (measured in minutes)

is p(t) 5 2t

1 t

1 6t meters for 0 , t , 10. What is the

acceleration of the bicycle at time t 5 3s? At time t 5 6s?

46. The position of a moving body is described by p(t) 5

1 bt 1 c.Ifp(0) 5 3, then what is c?Ifp

(0) 5 6, then

what is b?Ifp

(0) 525, then what is a? Where is the

body at time t 5 6?

c Exercises 4750 concern an object that is propelled

straight

up. Its height at time t seconds is given in feet by

H(t) 5216t

1 128t 1 68. b

47. Compute

the velocity and acceleration of the object.

48. For how many seconds does the object rise?

49. What is the maximum height of the object?

50. Supposing that the trajectory is such that the object is

able to fall to the ground (height 0 feet), for how many

seconds does the object fall?

51. An object is thrown straight down. Its height at time t

seconds is given in feet by H(t) 5216t

2 160t 1 84. With

what velocity does it impact the earth (at height 0).

c Exercises 5254 are concerned with a car’s advance p(t)

during

its period of deceleration. Suppose that during the

ﬁrst t seconds of braking, the car moved forward p(t) 5

48t 2 3t

/2 ft. Suppose also that this formula was in effect

until the car came to a stop. b

52. What

was the car’s velocity at the moment brakes were

applied?

53. What was its acceleration a(t) at time t? (The negative

sign of a( t) means that the car was decelerating.)

54. How long was the period of deceleration?

Further Theory and Practice

55. Let

f ðxÞ5

if x . 0

2 x

if x # 0:



a. Prove that f

(0) exists and equals 0.

b. Prove that f

(0) exists and equals 0.

c. Prove that f

000

(0) does not exist.

56. Suppose that f and g are twice differentiable. Calculate a

formula for ( f/g)

57. Suppose that p(x) is a nonzero polynomial. Prove that

p(x) has degree precisely k if and only if

a. p

(k11)

(x) 5 0 for all x; and

b. p

(k)

) 6¼0 for some x

58. Find an explicit formula for the polynomial p(x)of

degree 2 such that p(0) 5 2, p

(0) 5 4, and p

(0)527.

59. Find an explicit formula for the polynomial p(x)of

degree 3 such that p(1) 5 1, p

(2) 522, p

(21)5214, and

000

(5) 5 6.

60. Suppose that p(x) is a polynomial, and | p

(x)|#1 for all x.

Prove that p is of degree at most 2, and the coefﬁcient of

its degree 2 term is less than or equal to 1/2 in absolute

value.

c In each of Exercises 6166, use equation (3.7.2) to com-

pute

the requested derivative. b

61. D

( f g)(x) where f (x) 5 x

and g(x) 5 cos (x)

62.

ðx

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

x 1 1

63. ( f g)

(x) where f (x) 5 1/x and g(x) 5 ln (x)

64. D

( f g)(x) where f (x) 5 x

and g(x) 5 cos (x)

3.7 Higher Derivatives 235

65. ( f g )

(2)

(x) where f (x) 5

ﬃﬃﬃ

and g(x) 5 exp (x)

66.



ðx 1 3Þ

ð2x 1 1Þ



67. Suppose that f is twice differentiable at c. Show that the

graphs of y 5 f ( x) and y 5 f (c) 1 f

ðcÞ(x2c) 1 f

(c)

(x2c)

/2 both pass through the point (c, f (c)) and have

the same tangent line there.

c Exercises 6873 involve the factorial numbers n !,

which

were introduced in Section 1.4. They can be deﬁned by 0! 5 1

and, for a positive integer n, n! 5 n(n 2 1) ... 3... 2 1. b

68. Let n be

a positive integer. If k is a positive integer no

greater than n, then the expression





k!ðn 2 kÞ!

nðn 2 1Þðn 2 2Þ  ðn 2 k 1 1Þ

k ðk 2 1Þ   3  2  1

is called a binomial coefﬁcient. For example, if n 5 7, and

k 5 3, then n2k 1 155, and





7  6  5

3  2  1

5 35:

By deﬁnition





5 1. Leibniz’s Rule states that ( f g)

(n)

(c)

is the sum of the n 1 1 expressions that are obtained

by substituting k 5 0, 1, 2, . . . , n in the formula





 f

(n2k)

(k)

(c). Explicitly write out Leibniz’s

Rule for n 5 1, 2, and 3. Observe that the n 5 1 case is the

Product Rule, the n 5 2 case is equation (3.7.2). Verify

Leibniz’s Rule for n 5 2.

69. The Legendre polynomials P

, introduced in 1782 by

A.M. Legendre, can be deﬁned by the following formula,

which was discovered by Benjamin Olinde Rodrigues

(17941850 or 1851):

ðxÞ5

ðx

21Þ

n 5 0; 1; 2; 3;: : : :

Calculate P

(x), P

(x), and P

(x).

70. In general, the Legendre polynomial P

(x) (as deﬁned in

the preceding exercise) satisﬁes the differential equation

ð1 2 x

ÞP

ðxÞ2 2xP

ðxÞ1 nðn 1 1ÞP

ðxÞ5 0:

Verify this equation for P

(x) 5 (63x

2 70x

1 15x)/8.

71. For each nonnegative integer n, the Hermite polynomial

is deﬁned by

ðxÞ5 ð21Þ

Calculate H

(x), H

(x), and H

(x).

72. In general, the Hermite polynomial H

(x) (as deﬁned in

the preceding exercise) satisﬁes the differential equation

ðxÞ2 2xH

ðxÞ1 2nH

ðxÞ5 0:

Verify this equation for H

(x) 5 32x

2 160x

1 120x.

73. Prove that any polynomial p(x) of degree k satisﬁes

pðxÞ5 pð0Þ1 p

ð0Þx 1

ð0Þ

 x

1 1

ðkÞ

ð0Þ

 x



74. Show that there is a polynomial p (x) of degree n 1 1 such

that

tanðxÞ5 pðtanðxÞÞ:

75. Let p(x) be a polynomial. We say that p has a root of

order k at a if there is a polynomial q such that p(x) 5

(x 2 a)

q(x) and q (a) 6¼ 0. Prove that p has a root of

order 2 at x 5 a if and only if p(a) 5 0, p

(a) 5 0, and

(a) 6¼0. (In general, p has a root of order k at a if and

only if p(a) 5 p

(a) 5 p

(k21)

(a) 5 0 and p

(k)

(a) 6¼ 0.)

76. Let f ðxÞ5



150

1 x

100



1 x

1 16



: Compute

(83)

(0). (Hint: Simple observations go farther than brute

force.)

77. Suppose that f is invertible and twice differentiable, that

f (c) 5 γ, and that f

ðf

ðγÞ52

ðcÞ

ðf

ðcÞÞ



78. Verify that

yðtÞ5 h 1 t

ﬃﬃﬃ



1 1 expð2t

ﬃﬃﬃﬃﬃﬃ

gκ



satisﬁes the differential equation

ðtÞ52g 1 kðy

ðtÞÞ

Calculator/Computer Exercises

c In Exercises 7982, you are given a function f, a value c,

and a viewing rectangle R containing the point P 5 (c, f (c)).

In R, graph the four functions f, g, h, and k, where g(x) 5

f (c) 1 f

(c)(x 2 c), h(x) 5 g(x) 1

2 f

(c)(x 2 c)

, and k(x) 5

h(x) 1

6 f w(c)(x 2 c)

. The graphs of g, h, and k are,

respectively, linear, parabolic, and cubic approximations of

the graph of f near c. The method of constructing these

approximating functions, which are called Taylor polynomials

of f with base point c, is studied in Chapter 8. b

79. f ðxÞ5

ﬃﬃﬃ

ﬃ

c 5 0:5; R 5 ½0; 1 3 ½0; 1:1

80. f (x) 5 x

3/2

, c 5 1, R 5 [0, 4] 3 [21/2, 9]

81. f (x) 5 sin

(x), c 5 π/3, R 5 [0, 2] 3 [20.2, 1.2]

82. f (x) 5 xe

, c 5 1, R 5 [0, 2] 3 [22.7, 14.8]

c In Exercises 8386, use second central difference quo-

tients

to approximate f

83. f ðxÞ5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ﬃ

x 1 1=x

; c 5 e

85. f ðxÞ5 lnð

ﬃﬃﬃ

1 1=

ﬃﬃﬃ

Þ; c 5 3

86. f (x) 5 tan



sin

(x)



, c 5 π/4

236 Chapter 3 The Derivative

3.8 Implicit Differentiation

Examine the locus of the circle x

5 4 in Figure 1. We know that, at each point

of the graph, there is a tangent line to the curve. However, we cannot calculate the

equation of the tangent line using the techniques of calculus that we have learned

so far because the circle is not the graph of a function. One way to remedy this

problem is to solve for y in terms of x. We obtain two curves: the upper half of the

circle, which is the graph of y 51

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

, and the lower half of the circle, which is

the graph of y 52

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

. Thus we have realized the circle as the union of the

graphs of two functions. Refer to Figure 2.

Finding the slope of the tangent line to the circle at the point P 5 (1,

ﬃﬃﬃ

) is just

the same as ﬁnding the slope of the tangent line to the graph of f ( x ) 51

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

P. Similarly, ﬁnding the tangent line to the graph of x

1 y

5 4atQ 5 (

ﬃﬃﬃ

; 2

ﬃﬃﬃ

)

is just the same as ﬁnding the slope of the tangent line to the graph of the function

g(x) 52

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

. Differentiating using the Chain RulePower Rule, we ﬁnd that

ðxÞ 5

ð4 2 x

1=2

ð4 2 x

21=2

ð2 2xÞ5

2 x

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

and

ðxÞ 52f

ðxÞ5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x



Because f

(1) 521/

ﬃﬃﬃ

and g

(

ﬃﬃﬃ

) 5 1, the slopes of the tangent lines to the circle

at (1,

ﬃﬃﬃ

) and (

ﬃﬃﬃ

, 2

ﬃﬃﬃ

) are 21/

ﬃﬃﬃ

and 1, respectively.

It turns out that there is a simpler way to handle the graph of an equation that

does not express y explicitly as a function of x. Called the method of implicit

differentiation, this technique eliminates the need to solve for y in terms of x.

The Main Idea of

Implicit Differentiation

The example x

1 y

5 4 that we have been considering suggests a framework for

the general problem we wish to investigate. Because the expression x

1 y

depends

on the variables x and y, we may write it as F(x,y), using the notation for functions

of two variables (Section 1.4). The right side of x

1 y

5 4 is a constant, which we

may represent by C. Thus x

1 y

5 4 is an example of an equation that has the form

F(x, y) 5 C.

It should be understood that, when we write F(x, y), the two variables x and y

are independent variables. Assigning a value to one of them has no bearing on the

22

2

 y

 4

m Figure 1

22

f(x)   4  x

(1, 3)



m Figure 2

22

2

g(x)   4  x

( 2,  2)





3.8 Implicit Differentiation 237

value of the other. Thus if we set x 5 1inF (x, y) 5 x

1 y

, the equation F (1, y) 5

1 1 y

imposes no restriction on the value of y, which clearly is not a funct ion of x.

The matter is quite different when the equation F(x, y) 5 C holds. For example,

setting x 5 1 in the equation x

1 y

5 4 forces y to be

ﬃﬃﬃ

if (x, y) lies on the upper

semicircle and 2

ﬃﬃﬃ

if (x, y) lies on the lower semicircle. This suggests that, on each

of the two semicircles, y is a function of x. Indeed, as we have already noticed,

we can solve for y and obtain explicit formulas, y 5 f (x) 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

and y 5

g(x) 52

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

, for y in terms of x.

In general, we cannot obtain an explicit expression for y from an equation of

the form F(x, y) 5 C. Instead, we understand the equation F(x, y) 5 C to implicitly

deﬁne y as a function of x. It may seem impossible to obtain an expression for dy/dx

if we do not have a formula for y, but we can ﬁnd dy/dx, if it exists, by differ-

entiating each side of the equation F(x, y) 5 C with respect to x. When we do this,

we treat y as though it were a differentiable function of x on an open interval

centered at x

. This is the method of implicit differentiation.

Some Examples In illustrating the method of implicit differentiation, we will use Leibniz notation

because it contains a built-in reminder of the variable with respect to which we

differentiate.

⁄ EX

AMPLE 1 Use the method of implicit differentiation to ﬁnd the slopes

of the tangent lines to the curve x

1 y

5 4 at the points (1,

ﬃﬃﬃ

) and (

ﬃﬃﬃ

, 2

ﬃﬃﬃ

Solution We

apply d/ dx to both sides of the equation. Remembering to treat y as

though it were a function of x, we have

ðx

1 y

Þ5

ð4Þ

2x 

1 2y 

5 0

Take particular notic e that we have applied the Chain RulePower Rule to the

ﬁrst term, even though this was not strictly necessary. There was no need to write

, but in doing so, we have made our treatment of the term x

just the same as our

treatment of the term y

. As we work our ﬁrst examples, this will lead to less

confusion. Because dx/dx 5 1, our equation becomes

2x 1 2y 

5 0

ð2yÞ

522x:

We may solve for dy/dx by dividing both sides by 2y if y 6¼0. The result is

if y 6¼ 0:

238 Chapter 3 The Derivative

Now it is a sim ple matter to ﬁnd the slope of the tangent line to the curve at

(1,

ﬃﬃﬃ

), for at this point we have

ﬃﬃﬃ

This is the same answer that we obtained in the introduction, but the technique is

easier.

Similarly, we ﬁnd the slope of the tangent line to the curve at (

ﬃﬃﬃﬃ

ﬃﬃﬃ

)tobe

ﬃﬃﬃ

5 1: ¥

INSIGHT

Compare Example 1 with the discussion at the beginning of this section.

Notice that the method of implicit differentiation does not require separate consideration

of the two functions f (x) 5

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

and g(x) 52

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4 2 x

whose graphs make up the

curve x

1 y

5 4.

Knowing how to use the method of implicit differentiation includes knowing when

the method cannot be applied. There are a number of reasons why



may fail to exist at a point P

5 (x

, y

) on the graph of F(x, y) 5 C. In using the method

of implicit differentiation, we assume that y is a function of x on an open interval

centered at x

. This may not actually be the case. In the case of the circle x

1 y

5 4,

when P

5 (22, 0) or P

5 (2, 0), neither x

522 nor x

5 2 is the center of an x2interval

on which y is deﬁned. Even if y is a function of x on an interval centered at x

, it may

happen that y is not differentiable with respect to x at P

. This occurs if there is no

tangent line at P

or if the tangent line exists but is vertical.

Implicit differentiation always expresses the derivative dy/dx as a fraction involving x

or y or both. The points on the curve where the numerator of this fraction is nonzero but

where the denominator is zero correspond to places where the tangent line is vertical or

where it fails to exist. (No simple rule is available for points at which both the numerator

and denominator are zero.)

⁄ EXAMPLE 2 At what points on the curve (y 2 1)

1 y(x 2 2) 5 does dy/dx

not exist?

Solution If

we differentiate the equ ation (y 2 1)

1 y(x 2 2) 5 0 implicitly with

respect to x, using the Chain RulePower Rule for the ﬁrst summand and the

Product Rule for the second, then we obtain

3ðy 2 1Þ



ðx 2 2Þ1 y ð1 2 0Þ5 0 ;



3ðy 2 1Þ

1 ðx 2 2Þ



52y:

Thus if 3(y 2 1)

1 (x 2 2) 6¼ 0, the derivative dy/dx exists and

3.8 Implicit Differentiation 239

3ðy21Þ

1 ðx 2 2Þ

: ð3:8:1Þ

We must investigate the points (x, y) that satisfy the equations

3ðy 2 1Þ

1 ðx 2 2 Þ5 0 ð3:8:2Þ

(because equation (3.8.1) is not valid at such points) and

ðy 2 1Þ

1 yðx 2 2Þ5 0 ð3:8:3Þ

(because we consi der only points on the given curve). Equation (3.8.2) gives us

x 2 2 523 ðy21Þ

: ð3:8:4Þ

Substituting this formula for x 2 2 into equation (3.8.3) results in (y 2 1)

2 3y 

(y 2 1)

5 0. Factoring, we obtain (y 2 1)

(y 2 1 2 3y) 5 0, or (y 2 1)

(2y 1 1) 5 0.

The solutions to this last equation are y 5 1, and y 521/2. Substi tuting these values

back into (3.8.4) allows us to determine that x 5 2 when y 5 1, and x 5219/4 when

y 521/2. We have graphed portions of the curve (y 2 1)

1 y(x 2 2) 5 0 in Figure 3.

We see that, although y is a function of x near the point (2, 1), dy/dx fails to exist

because the tangent line is vertical. On the other hand, we cannot even deﬁne y as a

continuous function of x on an open interval containing 219/4 with the graph of y

passing through the point (219/4, 21/2). Therefore dy/dx has no meaning at this

point.

⁄ EXAMPLE 3 Find the slope of the tangent line to the curve 8 xy

9/2

5 9 2 x

27/5

at the point (1, 1).

Solution Using

implicit differentiation, we have

ð8xy

9=2

Þ5

ð9 2 x

27=5

Þ;

or, by using the Product Rule and the Chain Rule Power Rule on the left side,

9=2

1 8x 

 y

7=2



5 0 1

 x

212=5

To make the algebra simpler, we will substitut e in the point (1,1) into this equation

before we solve for dy=dx (but notice that we have ﬁnished differentiating). We

obtain 8 1 36 dy/dx 5 7/5, or dy/dx 5211/60.

INSIGHT

You might have noticed that the equation considered in Example 3 is not

of the form F(x, y) 5 C, as speciﬁed in the method of implicit differentiation. Although it

would have been easy to rearrange the equation as 8xy

9/2

1 x

27/5

5 9, which does have

the standard form F(x, y) 5 C, it is not necessary to bring all variables to one side of the

equation in order to use the method of implicit differentiation. We may differentiate

the equation as it is presented.

⁄ EXAMPLE 4 Find the equation of the tangent line to the curve

1 x 5 17 ð3:8:5Þ

at each point where x 5 1.

(y  1)



y(x  2)  0

4 2

(2,1)

2



()

m Figure 3

240 Chapter

3 The Derivative

Solution If we substitute x 5 1 into the equation, then we obtain y

1 1 5 17. This

simpliﬁes to y

5 16 or y 562. Thus there are two points, (1, 2) and (1, 22), with

x 5 1. Now we differentiate equation (3.8.5) to obtain

ðxy

1 xÞ5

ð17Þ:

Performing the different iations, we have

1 x  4y



1 1 5 0;

2 1 2 y

4xy



Thus at the point (1, 2), the tangent line has slope 217/32; whereas, at the point

(1,22), the tangent line has slope 17/32. The equations of these tangent lines are

y 5 (217/32) (x 2 1) 1 2 and y 5 (17/32) (x 2 1) 1 (22), respectively. The curve

and its tangents are illustrated in Figure 4.

INSIGHT

When we apply the method of implicit differentiation to ﬁnd the slope of

the tangent line to F(x, y) 5 C at (x

, y

), the expression for



ðx

will, in general, involve y

as well as x

. As we see in Example 4, there may be more

than one y for a given x

such that (x

, y) is on the curve. The value of dy/dx will depend

on which value of y is used. Geometrically, this corresponds to the vertical line x 5 x

intersecting the curve F(x, y) 5 C in more than one point. The tangents to the curve at

these points need not be parallel, which is why the value of y sometimes must be speciﬁed

when we calculate dy/dx by the method of implicit differentiation. This is unlike the

situation for the explicitly deﬁned functions that we considered prior to this section.

When y is deﬁned as an explicit function of x with x

in its domain, the vertical line x 5 x

intersects the curve in only one point, and dy/dx is determined by x

alone.

Calculating Higher

Derivatives

We can use the metho d of implicit differentiation to calculate higher derivatives.

⁄ EX

AMPLE 5 Consider the equation sin (y) 1 x 5 1/

ﬃﬃﬃ

. Calculate d

y/dx

at the point (0, π/4).

Solution Diffe

rentiating both sides with respect to x yields

cosðyÞ

1 1 5 0: ð3:8:6Þ

We do not yet substitute in the coordinates of the point we wish to study. We must

differentiate another time. Using the Product Rule to differentiate the ﬁrst sum-

mand on the left side of equation (3.8.6), we obtain



2 sinðyÞ:





1 cosðyÞ





1 0 5 0: ð3:8:7Þ

2

 x  17

y  (1732)(x  1)  2

y  (1732)(x  1)  (2)

(1, 2)

(1, 2)

m Figure 4

3.8 Implicit Differentiation 241

Now that we have ﬁnished differentiating, it is correct to put in our numerical

values: Setting y 5 π/4 in equati on (3.8.6) yields

ﬃﬃﬃ



ð0; π=4Þ

1 1 5 0 ;



ð0; π=4Þ

ﬃﬃﬃﬃ

Now we substitute this value for

; together with x 5 0 and y 5 π/4, into equation

(3.8.7) to obtain



ﬃﬃﬃ



ð2

ﬃﬃﬃ



ﬃﬃﬃ





ð0; π=4Þ



5 0;



ð0; π=4Þ

5 2: ¥

INSIGHT

There is a subtlety in going from line (3.8.6) to line (3.8.7). When we

differentiated the term cos (y)  dy/dx, we had to apply the Product Rule to this term.

The differentiation of the factor cos (y) then led to another factor of dy/dx while the

differentiation of the factor dy/dx led to a factor of d

y/dx

. In this step, we must be

careful to keep track of both types of derivative that arise.

Parametric Curves Our focus in this section has been the method of implicit differentiation. Placed in a

broader context, our disc ussion has concerned the calculation of tangent lines

to curves that are not presented in the form y 5 f (x). It is therefore appropriate to

extend that discussion by studying the tangent line problem for parametric curves.

Recall from Section 1.5 that these are curves that have the form x 5 ϕ

(t), y 5 ϕ

(t).

THEOREM 1

Suppose that ϕ

and ϕ

are differentiable functions on an interval

I 5 (a, b) and that C 5 {(ϕ

(t), ϕ

(t)) : a , t , b}. Let P

5 (x

, y

) 5 (ϕ

), ϕ

))

for some value t

in I. If the parametric equations x 5 ϕ

(t), y 5 ϕ

(t) determine

y as a differentiable function f of x on an arc of C that extends beyond P

each side and if ϕ

) 6¼0, then



ðx

; y

ðt

: ð3:8:8Þ

Proof. By

hypothesis, ϕ

(t) 5 y 5 f (x) 5 f (ϕ

(t)) for some differentiable function f

deﬁned on an open interval containing t

. Differentiating each side of the equation

(t) 5 f (ϕ

(t)) with respect to t, we obtain ϕ

(t) 5 f

(ϕ

(t)) ϕ

(t). Evaluating

at t 5 t

, we have ϕ

)  f

(ϕ

)) 5 ϕ

), or ϕ

)  f

) 5 ϕ

). If ϕ

)6¼0,

then



ðx

; y

5 f

ðx

Þ5

ðt

: ’

242 Chapter 3 The Derivative

⁄ EXAMPLE 6 At the beginning of this section, we determined that the

slope of the tangent line to the graph of x

1 y

5 4atP 5 (1,

ﬃﬃﬃ

Þ is 21/

ﬃﬃﬃ

. Derive

this result using the parametric equations x 5 2 cos (t), y 5 2 sin (t).

Solution We

may verify that ϕ

(t) 5 2 cos (t) and ϕ

(t) 5 2 sin (t), then

(t) 1 ϕ

(t) 5 4 (cos

(t) 1 sin

(t)) 5 4. Because ϕ

(π/3) 5 2 cos (π/3) 5 1 and

(π/3) 5 2 sin (π/3) 5

ﬃﬃﬃ

, the point P 5 (1,

ﬃﬃﬃ

) corresponds to t

5 π/3. Formula

(3.8.8) tells us that the required slope is



ð1;

ﬃﬃ

ðt

2 cos ðtÞ

22 sin ðtÞ



t5t

cos ðtÞ

sin ðtÞ



t5π=3

cos ðπ=3Þ

sin ðπ=3Þ

ﬃﬃﬃ

: ¥

INSIGHT

In practice, we often do not give names such as ϕ

and ϕ

to the functions

that deﬁne x and y parametrically. When we forgo the extra notation, formula (3.8.8)

simpliﬁes to the more intuitive (and memorable) equation

dy=dt

dx=dt

: ð3:8:9Þ

Thus in Example 6, the parametric equations x 5 2 cos (t) and y 5 2 sin (t) lead to dx/dt 5

22 sin (t), dy/dt 5 2 cos (t), and therefore, by (3.8.9),

2 cosðtÞ

2 2 sinðtÞ

cosðtÞ

sinðtÞ

The result is, of course, the same.

QUICK QUIZ

1. The equation ln (y) 1 4y 2 x 5 1 implicitly deﬁnes y 5 f ( x ). What is f (3)?

2. Calculate dy/dx at (2, 1) if exp (y 2 1) 1 y 5 x.

3. Suppose that y

1 y 5 10x. Calculate dy/dx and d

y/dx

when x 5 1.

4. If x 5 t

1 1, and y 5 t

, calculate dy/dx at (5, 28).

Answers

1. 1 2. 1/2 3. dy/dx 5 10/13

and d

y/dx

52120/13

4. 23

EXERCISES

Problems for Practice

c In each of Exercises 16, y is a function of x. Calculate the

derivative of the given expression with respect to x. (Your

answer should contain the term dy/dx.) b

1. 4y

3/2

2. cos (y)

3. 2x/

ﬃﬃﬃ

4. (y

2 1) /y

5. xe

6. ln (y)/x

c In each of Exercises 714, use the method of implicit

differentiation

to calculate dy/dx at the point P

7. xy

1 yx

5 6 P

5 (1, 2)

8. sin (πxy) 2 xy

1 2y 5 1 P

5 (1, 1)

9. x

3/5

1 4y

3/5

5 12 P

5 (32, 1)

10. x

2 y

5215 P

5 (1, 2)

11. xy

1 3y

1 x

5 21 P

5 (1, 2)

12. x

y 1 ln(y) 5 1 P

5 (21, 1)

13. xe

y21

1 y

5 4 P

5 (3, 1)

14. 5x

x 1 2y

x 2 y

5 0 P

5 (1, 2)

c In Exercises 1524, use implicit differentiation to ﬁnd the

tangent

line to the given curve at the given point P

. b

15. xy21 5

ﬃﬃ

ﬃ

ﬃﬃﬃ

5 (4, 1)

16. x

2 y

1 4y 5 5 P

5 (2,2 1)

3.8 Implicit Differentiation 243

17. sin

(x) 1 cos

(y) 5 5/4 P

5 (π/3, π/4)

18. 7x

1/2

2 xy

1/2

5 3 P

5 (9, 4)

19. x 1 3/y 1 ln(y) 5 5 P

5 (2, 1)

20. x 1 y/x 2 exp (y

) 5 2 P

5 (3, 0)

21. e

5 2y

2 1 P

5 (0, 21)

22. 2

x2y

5 xy

5 (2, 1)

23. ln (xy 2 1)1 y

5 4 P

5 (1, 2)

24. xe

x21

2 ln (xy) 5 1 P

5 (1, 1)

c In Exercises 2528, use implicit differentiation to ﬁnd the

normal

line to the given curve at the given point P

. b

25. xy

2 x

y 5 16 P

5 (2, 2)

26. x

2 8y 1 y

5 9 P

5 (2, 1)

27. x 2 y 1 sin (2y) 5 1 P

5 (π/4, π/4)

28. x

2 3xy

1 1 5 1/yP

5 (3, 1)

c In Exercises 2934, ﬁnd dy/dx and d

y/dx

at the point P

by implicit differentiation. b

29. y

1 y 1 x 5 4 P

5 (2, 1)

30. y

2 2xy 5 20 P

5 (23, 2)

31. xy 2 6/y 5 4 P

5 (2, 3)

32. y 2 sin (2y) 5 x ln( x) P

5 (1, 0)

33. 2e

5 3 2 x 1 yP

5 (1, 0)

34. 2x 1 y

1/3

1 y 5 0 P

5 (1,21)

c In Exercises 3546, ﬁnd the tangent line to the parametric

curve x 5 ϕ

(t), y 5 ϕ

(t) at the point corresponding to the

given value t

of the parameter. b

35. ϕ

(t) 5 t

1 1, ϕ

(t) 5 t

1 1 t

5 1

36. ϕ

(t) 5 t

1 t, ϕ

(t) 5 t

1 1 t

5 1

37. ϕ

(t) 5 t

1 t, ϕ

(t) 5 t

2 tt

5 2/3

38. ϕ

(t) 5 cos (2t), ϕ

(t) 5 sin (3t) t

5 π/4

39. ϕ

(t) 5 t 1 cos (t), ϕ

(t) 5 t 1 sin (t) t

5 π/4

40. ϕ

(t) 5 t/(1 1 t

), ϕ

(t) 5 t

/(1 1 t

) t

5 1/2

41. ϕ

(t) 5 t/(1 1 t

), ϕ

(t) 5 t

/(1 1 t

) t

5 2/3

42. ϕ

(t) 5 te

, ϕ

(t) 5 t 1 e

5 0

43. ϕ

(t) 5 exp (cos (t)), ϕ

(t) 5 exp (sin (t)) t

5 π/6

44. ϕ

(t) 5 1/ (1 1 t

), ϕ

(t) 5 t (1 1 ln (t)) t

5 1

45. ϕ

(t) 5 2

1 3, ϕ

(t) 5 3

1 2 t

5 1

46. ϕ

(t) 5 log

(t), ϕ

(t) 5 t log

(t). t

5 3

Further Theory and Practice

47. Show that the tangent lines to the curve x

2 4xy 1 y

5 9

at the points where the curve crosses the x-axis are

parallel.

48. Find all points on the curve xy 2 2x

1 y

5 8 where the

tangent line has slope 22, 21, 0, 1, or 2. Show that no

tangent line to the curve is vertical.

49. The curve x

2 xy 1 y

5 4 is an ellipse. Notice that

P 5 (a, b) is on the ellipse if and only if Q 5 (2a, 2b)ison

the ellipse. Assuming that P and Q are both points on the

ellipse, show that the tangent lines at P and Q are parallel.

50. Suppose that (x

) is a point on the ellipse x

5 1. Show that the tangent line to the ellipse at (x

) is given by

5 1:

51. For each positive value of α, the equation

2 3αxy 1 y

5 0

deﬁnes a curve known as a Folium of Descartes (see

Figure 4 of this chapter’s Genesis and Development for a

plot). Suppose that the point P 5 (a, b) is on the Folium

of Descartes. Because the equation is symmetric in x and

y, it follows that Q 5 (b, a) is also on the curve. What is

the product of the slopes of the tangent lines at P and Q?

52. Let S and T be open intervals in R. Let f : S - T be a

function that is differentiable and invertible. Show how to

derive the formula for

that we learned in Theorem

2 of Section 3.6 by applying implicit differentiation to the

equation t 5 f (s).

53. The graph of the curve x

5 y

(x, y . 0) consists of the

ray x 5 y . 0 and a curve C. Find dy/dx at the point P on

C whose abscissa is 2. (Hint: Take the natural logarithm

of each side ﬁrst.)

54. Use the method of implicit differentiation to ﬁnd the

slope of the curve x

2/3

1 y

2/3

5 25 at the points (27, 64)

and at (27,264). Verify your answers by explicitly solving

for y and differentiating the explicit expression. Why

must you use two explicit formulas for y?

55. Let a be a positive constant. Let T be any tangent line to

ﬃﬃﬃ

5 a:

Compute the sum of the intercepts of T on the coordinate

axes. (The answer does not depend on the choice of T.)

56. Suppose that A 6¼0. The locus of

2=3

1 y

2=3

5 a

2=3

is called an astroid A

. Let T be a tangent line to the

astroid A

. Find a formula for the distance between the

two intercepts of T. How does this distance depend on T ?

57. Saha’s equation

1 2 y

A exp ðb=xÞ

3=2

describes the degree of ionization within stellar interiors.

In this equation, A and b are constants, y represents the

fraction of ionized atoms in the star, and x represents

stellar temperature in degrees Kelvin. Find dy/dx.

58. Let p and q be nonzero integers. Apply implicit differ-

entiation to the equation x

2 y

5 0 to obtain a proof of

the Power Rule for rational exponents:

ðx

p=q

Þ5

 x

ðp=qÞ21

244 Chapter 3 The Derivative