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3.2. THE CHAIN RULE

35. f(x) = x2

37.f(x) = x4 − 4x2

39.f(x) = (x + 2)(x − 4)

41.y = (x2 + 1) sin x

43.y = (x2 + 1)(x10 − 5)

45.y = (x1/2 + 4)(x1/3 − 5)

47.y = x5 sin x

49.y = x2 cot x − 2x + 5

51.y = (sec x + tan x)(sin x + cos x)

x2 + 1 53. y = x2 + 4

x1/2 + 1

55. y = 3x3/2 + 2

57. y = t2 + 3t + 2 t3 + 1

59. y = 3 + sin t cos t

4 + sec t tan t

111

4		3		2
36. f(x) =		−		+		+ 1
	x3		x2		x

38.f(x) = (x2 + 2)(x2 + 1)

40.f(x) = (x3 + 1)(x3 − 1)

42.y = x2 cos x

44.y = x2 tan x

46. y = (2x + sin x)(x2 + 4)

48. y = x4(2 sin x − 3 cos x) 50. y = (x + sin x)(4 + csc x)

52. y = x2(2 cot x − 3 csc x)

54. y = 1 + sin x

1 + cos x

56. y = sin x − cos x sin x + cos x

x2ex

58. y = 1 + ex

t2 sin t 60. y = 4 + t2

3.2The Chain Rule

Suppose we have two functions, u and y, related by the equations: u = g(x) and y = f(u).

u→g(c)

f0(g(c))

f(u) − f(g(c)) u − g(c)

if u 6= g(c) if u = g(c)

112 CHAPTER 3. DIFFERENTIATION

Then y = (f ◦ g)(x) = f(g(x)).

The chain rule deals with the derivative of the composition and may be stated as the following theorem:

Theorem 3.2.1 (The Chain Rule). Suppose that g is deﬁned in an open interval I containing c, and f is deﬁned in an open interval J containing g(c), such that g(x) is in J for all x in I. If g is di erentiable at c, and f is di erentiable at g(c), then the composition (f ◦ g) is di erentiable at c and

(f ◦ g)0(c) = f0(g(c)) · g0(c).

In general, if u = g(x) and y = f(u), then dxdy = dudy · dxdu.

Proof. Let F be deﬁned on J such that

(

f(u)−f(g(c))

F (u) = u−g(c)

since f is di erentiable at g(c),

lim F (u) = lim

=f0(g(c))

=F (g(c)).

Therefore, F is continuous at g(c). By the deﬁnition of F ,

f(u) − f(g(c)) = F (u)(u − g(c))

for all u in J. For each x in I, we let y = g(x) on I. Then

◦

g)0(c) = lim

(f ◦ g)(x) − (f ◦ g)(c)

→

−

= lim

f(g(x)) − f(g(c))

g(x) − g(c)

→

g(x)

−

g(c)

−

g(x) − g(c)

lim

F (u)

lim

→

g(c)

→

−

= f0(g(c)) · g0(c).

3.2. THE CHAIN RULE

113

It follows that f ◦g is di erentiable at c. The general result follows by replacing c by the independent variable x. This completes the proof of Theorem 3.2.1.

Example 3.2.1 Let y = u2 + 1 and u = x3 + 4. Then

dudy = 2u and dudx = 3x2.

Therefore,

dxdy = dudy · dudx

=2u · 3x2

=6x2(x3 + 4) .

Using the composition notation, we get

y = (x3 + 4)2 + 1 = x6 + 8x3 + 17

and

dxdy = 6x5 + 24x2

= 6x2(x3 + 4) .

Using

(f ◦ g)0(x) = f0(g(x)) · g0(x),

we see that

(f ◦ g)(x) = (x3 + 4)2 + 1

and

(f ◦ g)0(x) = f0(g(x)) · g0(x)

=2(x3 + 4)1 · (3x2)

=6x2(x3 + 4) .

114	CHAPTER 3. DIFFERENTIATION

Example 3.2.2 Suppose that y = sin(x2 + 3).

We let u = x2 + 3, and y = sin u. Then

dxdy = dudy · dudx

=(cos u)(2x)

=(cos(x2 + 3)) · (2x).

Example 3.2.3 Suppose that y = w2, w = sin u + 3, and u = (4x + 1). Then

dxdy = dwdy · dwdu · dudx

=(2w) · (cos u) · 4

=8w cos u

=8[sin(4x + 1) + 3] · cos(4x + 1) · 4

=8(sin(4x + 1) + 3) · cos(4x + 1).

If we express y in terms of x explicitly, then we get

y = (sin(4x + 1) + 3)2

and

dxdy = 2(sin(4x + 1) + 3)1 · ((cos(4x + 1)) · 4 + 0) = 8(sin(4x + 1) + 3) cos(4x + 1).

Example 3.2.4 Suppose that y = (cos(3x + 1))5. Then

dxdy = 5(cos(3x + 1))4 · (− sin(3x + 1)) · 3 = −15(cos(3x + 1))4 sin(3x + 1).

3.2. THE CHAIN RULE

115

Example 3.2.5 Suppose that y = tan3(2x2 + 1). Then

= 3(tan2(2x2 + 1)) · (sec2(2x2 + 1)) · 4x

= 12x · tan2(2x2 + 1) · sec2(2x2 + 1).

x + 1

Example 3.2.6 Suppose that y = cot

. Then

x2 + 1

csc2

x + 1

(x2

+ 1) 1 − (x + 1)2x

x2 + 1

(x2 ·+ 1) · 2x

−

x2 + 2x − 1

csc2

x + 1

(x2 + 1)2

x2 + 1

x2 + 1 3

Example 3.2.7 Suppose that y = sec x4 + 2 .

Since the function y has a composition of several functions, let us deﬁne some intermediate functions. Let

y = sec w, w = u3

and

u =

+ 1

+ 2

Then

= [sec(w) tan(w)]

[3u2]

(x4

+ 2) · 2x − (x2 + 1) · 4x3

(x4 + 2)2

= 3u2(sec w tan w)

4x − 4x3 − 2x5

(x4 + 2)2

= 3

x2 + 1

2 sec

x2 + 1

tan

x2 + 1

4x − 4x3 − 2x5

x4 + 2

(x4 + 2)5

116		CHAPTER 3. DIFFERENTIATION
Example 3.2.8 Suppose that y = csc(2x + 5)4. Then
	dy	= [− csc(2x + 5)4 cot(2x + 5)4] · 4(2x + 5)3 · 2

	dx

= −8(2x + 5)3 csc(2x + 5)4 cot(2x + 5)4.

Exercises 3.2 Evaluate dxdy for each of the following:

1.y = (2x − 5)10

3.y = sin(3x + 5)

5.y = tan5(3x + 1)

7.y = cot4(2x − 4)

9.y = 3x + 1 5 x2 + 2

2. y =

x2 + 2 3

x5 + 4

4. y = cos(x3 + 1)

6. y = sec2(x2 + 1)

8. y = csc3(3x2 + 2)

10. y =

x2 + 1 4

x3 + 2

11.y = sin(w), w = u3, u = (2x − 1)

12.y = cos(w), w = u2 + 1, u = (3x + 5)


13.	y = tan(w), w = v2, v = u3 + 1, u = x
	1
14.	y = sec w, w = v3, v = 2u2 − 1, u =	x

		x2 + 1
15.	y = csc w, w = 3v + 2, v = (u + 1)3, u = (x2 + 3)2

In exercises 16–30, compute the derivative of the given function.

16. y =	x3	+ 1	3
			17. y = (x2 − 1)10
	x2	+ 4

3.2.	THE CHAIN RULE		117
18.	y = (x2 + x + 2)100	19.	y = (2 sin t − 3 cos t)3
20.	y = (x2/3 + x4/3)2	21.	y = (x1/2 + 1)50
22.	y = sin(3x + 2)	23.	y = cos(3x2 + 1)
24.	y = sin(2x) cos(3x)	25.	y = sec 2x + tan 3x
26.	y = sec 2x tan 3x	27.	y = (x2 + 1)2 sin 2x
28.	y = x sin(1/x2)	29.	y = sin2(3x) + sec2(5x)

30.

y = cot(x2) + csc(3x)

In exercises 31–60, assume that

(a)

(ex) = ex

(b)

(e−x) = −e−x

(c)

(ln x) =

(bx) = bx ln b

for b > 0 and b 6= 1.

(d)

(e)

(logb x) =

x ln b

Compute the derivative of the given function.

31.

y = sinh x

32.

y = cosh x

33.

y = tanh x

34.

y = coth x

35.

y = sech x

36.

y = csch x

37.

y = ln(1 + x)

38.

y = ln(1 − x)

1 − x

x + √

39.

y =

40.

y = ln

x2 + 1

1 + x

x + √

y = xe−x2

41.

y = ln

x2 − 1

42.

43. y = esin 3x

44. y = e2x sin 4x

118

45.y = ex2 (2 sin 3x − 4 cos 5x)

47.y = 4x2

49.y = 10sin 2x

51.y = log10(x2 + 10)

53.y = ln(sin(e2x))

55.	y = ln(cos x + 2)
	y = ln	x4 + 3	3
57.
		x2 + 10
59.	y = ln(sec 2x + tan 2x)

CHAPTER 3. DIFFERENTIATION

46.y = xe−x2 + 4e−x

48.y = 10(x2+4)

50.y = 3cos 3x

52.y = log3(x2 sin x + x)

54.y = ln(1 + e−x)

56.y = ln(ln(x2 + 4))

58.y = (1 + sin2 x)3/2

60.y = ln(csc 3x − cot 3x)

3.3Di erentiation of Inverse Functions

One of the applications of the chain rule is to compute the derivatives of inverse functions. We state the exact result as the following theorem:

Theorem 3.3.1 Suppose that a function f has an inverse, f−1, on an open interval I. If u = f−1(x), then

(i)	du		=	1
(i)			=	dx
		dx		du	1		1
(ii)	(f−1)0(x) =					=
(ii)	(f−1)0(x) =				f0(f−1(x))	=	f0(u)
Proof.			By comparison, x = f(f−1(x)) = x. Hence, by the chain rule
					1 =	dx	= f0(f−1(x)) · (f−1)0(x)

						dx

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS						119
and				1
(f−1)0(x) =				1	.

			f0(f−1(x))
In the u = f−1(x) notation, we have
	du	=	1	.
		=	dx	.
	dx		du

Remark 11 In Examples 76–81, we assume that the inverse trigonometric functions are di erentiable.

Example 3.3.1 Let u = arcsin x, −1	≤ x ≤ 1, and −	π	≤ u ≤	π	. Then

		2		2
x = sin u and by the chain rule, we get

1 = dx = d(sin u) · du dx du dx

= cos u · dudx dudx = cos1 u.

Therefore,

, −

(arcsin x) =

< u <

cos u

(Why?)

11− sin2 u

√

, −1 < x < 1. (Why?)

1 − x2

Thus,

(arcsin x) =

√

, −1

< x < 1.

− x2

We note that x = ±1 are excluded.

120 CHAPTER 3. DIFFERENTIATION

Example 3.3.2 Let u = arccos x, −1 ≤ x ≤ 1, and 0 ≤ u ≤ π. Then x = cos u and

1 =

= − sin u

= −

, 0 < u < π

sin u

= −√

, 0 < u < π

(Why?)

1 − cos2 u

= −√

, −1 < x < 1.

(Why?)

1 − x2

We note again that x = ±1 are excluded.

Thus,

−1

(arccos x) =

1 < x < 1.

√1 − x2

−

Example 3.3.3 Let u = arctan x, −∞ < x < ∞, and −

< u <

. Then,

x = tan u, −

< u <

= (sec2 u),

−

1 =

< u <

sec2 u

−

< u <

1 + tan2 u

−∞ < x < ∞

1 + x2

Therefore,

, −∞ < x < ∞.

(arctan x) =

1 + x2

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS

121

Example 3.3.4 Let u = arcsec x, x (−∞, −1] [1, ∞) and

uh0, π2 π2 , πi. Then, x = sec u

1 =

= sec u tan u ·

u h0,

, πi

, π

sec u tan1

| sec u|√

(Why the absolute value?)

sec2 u − 1

|x|√

(−∞, −1) (1, ∞).

x2 − 1

Thus,

(arcsec x) =

|x|√

, x (−∞, −1) (1, ∞).

x2 − 1

Example 3.3.5 Let u = arccsc x,

x (−∞, −1] [1, ∞), and

u h−

, 0 0,

i. Then,

x = csc u , u h−

, 0 0,

1 =

− csc u cot u ·

u h−

, 0

csc u cot u

−2

−1

, u

, 0

(Why?)

| csc u|√

(Why?)

csc2 u − 1

|x|√

, x

(−∞, −1) (1, ∞).

x2 − 1

Note that x = ±1 are excluded.

Thus,

x√x2

− 1

−∞

∞

(arccsc x) =

−1

, x

(

, 1]

(1,

122

CHAPTER 3.

DIFFERENTIATION

Example 3.3.6 Let u = arccot x,

x (−∞, 0] [0, ∞) and

u 0,

, π . Then

x = cot u,

u 0,

i h

, π

and

1 =

− csc2(u) ·

, u 0,

, π

−1

, u

, π

csc2 u

2 i h

−1

, u

, π

1 + cot2 u

2 i h

−1

(

, 0]

[0,

∞

1 + x2

−∞

Therefore,

1 + x2

−∞

∞

(arccotx) =

−1

(

, 0]

[0,

The results of these examples are summarized in the following theorem:

Theorem 3.3.2 (The Inverse Trigonometric Functions) The following differentiation formulas are valid for the inverse trigonometric functions:

(i)dxd

(ii)dxd

(iii)dxd

(iv)dxd

(v)dxd

(arcsin x) =

√

, −1 < x < 1.

1 − x2

(arccos x) =

√

−1

, −1 < x < 1.

1 − x

(arctan x) =

−∞ < x < ∞.

1 + x2

−∞

∞

(arccot x) =

−1

< x <

(arcsec x) =

|x|√

, −∞ < x < −1 or 1 < x < ∞.

x2 − 1

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS

123

|x|√x2 − 1

−∞

−

∞

(vi)

(arccsc x) =

−1

< x <

1 or 1 < x <

Proof. Proof of Theorem 3.3.2 is outlined in Examples 76–80.

Theorem 3.3.3 (Logarithmic and Exponential Functions)

(i)

(ln x) =

for all x > 0.

(ii)

(ex) = ex for all real x.

for all x > 0 and b 6= 1.

(iii)

(logb x) =

x ln b

(iv)dxd (bx) = bx(ln b) for all real x, b > 0 and b 6= 1.

(v)d (u(x)v(x) = (u(x))v(x) v0(x) ln(u(x)) + v(x) u0(x) . dx u(x)

Proof. Proof of Theorem 3.3.3 is outlined in the proofs of Theorems 5.5.1– 5.5.5. We illustrate the proofs of parts (iii), (iv) and (v) here.

Part (iii) By deﬁnition for all x > 0, b > 0 and b 6= 1,

ln x logb x = ln b .

Then,

dx	(logb x) = dx		ln b ln x
d		d	1

=	1			·	1

		ln b			x
=	1		.

	x ln b

Part (iv) By deﬁnition, for real x, b > 0 and b 6= 1,

bx = ex ln b.

124						CHAPTER 3. DIFFERENTIATION
Therefore,
	d	(bx) =	d	(ex ln b)
	dx	(bx) =		(ex ln b)
	dx		dx
		= ex ln b,			d	(x ln b)	(by the chain rule)
		= ex ln b,			dx	(x ln b)	(by the chain rule)
					dx
		= bx ln b.					(Why?)

Part (v)

dxd (u(x))v(x) = dxd ev(x) ln(u(x))

= ev(x) ln(u(x)) v0(x) ln(u(x)) + v(x) u0(x) u(x)

= (u(x))v(x) v0(x) ln u(x) + v(x) u0(x) u(x)

Example 3.3.7 Let y = log10(x2 + 1). Then

ln(x2 + 1)

(log10(x2 + 1)) =

ln 10

· 2x

(by the chain rule)

ln 10

x2 + 1

(x2 + 1) ln 10

Example 3.3.8 Let y = ex2+1. Then, by the chain rule, we get

dxdy = ex2+1 · 2x = 2xex2+1.

dxdy = 10(x3+2x+1) · (ln 10) · (3x2 + 2).

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS

125

Example 3.3.9 Let y = 10(x3+2x+1). By deﬁnition and the chain rule, we get

Example 3.3.10

+ 1)sin x cos x ln(x2

(x2

+ 1)sin x = (x2

+ 1) + sin x ·

x2 + 1

+1) · cos x ln(x2 + 1) + sin x ·

(x2

+ 1)sin x =

hesin x ln(x

+1)i = 3sin x ln(x

x2 + 1

+ 1)sin x cos x ln(x2

2x sin x

= (x2

+ 1) +

x2 + 1

Theorem 3.3.4 (Di erentiation of Hyperbolic Functions)

(i)

(sinh x) = cosh x

(ii)

(cosh x) = sinh x

(iii)

(tanh x) = sech2x

(iv)

(cothx) = −csch2x

(v)

(sech x) = −sech x tanh x

(vi)

(csch x) = −csch x coth x.

Proof.

Part (i)

d	(sinh x) =	d	1	(ex

dx		dx	2

=21 (ex − e−x

=21 (ex + e−x

=cosh x.

− e−x)

(−1)) (by the chain rule)

)

126	CHAPTER 3. DIFFERENTIATION

Part (ii)


d	(cosh x) =	d		1	(ex + e−x)
dx		dx 2
	=	1	(ex + e−x(−1))			(by the chain rule)

		2
	=	1	(ex − e−x)

		2

		= sinh x.
Part (iii)
	d	(tanh x) =	d	ex − e−x
	dx	(tanh x) =	dx	ex + e−x
	dx		dx	ex + e−x
		=	(ex + e−x)(ex + e−x) − (ex − e−x)(ex − e−x)
					(ex + e−x)2

=(ex + e−x)2

=ex + e−x

=sech2x.

Part (iv)

d	(sech x) =	d	2

dx		dx	ex + e−x

= (ex + e−x) · 0 − 2(ex − e−x) (ex + e−x)2

2ex − e−x

=−ex + e−x · ex + e−x

=−sech x tanh x.

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS

127

Part (v)

ex + e−x

(coth x) =

, x 6= 0

ex − e−x

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

x = 0

−4

(ex − e−x)2

x = 0

(ex − e−x)2

= −

, x 6= 0

ex e−x

−

= −csch2x ,

x 6= 0.

Part (vi)

(csch x) =

, x 6= 0

ex − e−x

= (ex − e−x) · 0 − 2(ex + e−x)

, x = 0

(ex − e−x)2

ex + e−x

= −

, x 6= 0

ex − e−x

= −csch x coth x,

x 6= 0.

Theorem 3.3.5 (Inverse Hyperbolic Functions)

(i)

(arcsinh x) =

√

1 + x2

(ii)

(arccosh x) =

√

x > 1

x2 − 1

(iii)

(arctanh x) =

|x| < 1

1 − x2

Proof.

128

CHAPTER 3.

DIFFERENTIATION

Part (i)

ln(x + √

)

(arcsinh x) =

1 + x2

x + √

1 +

√

(by chain rule)

1 + x2

√

+ x

1 + x2

x + √

√

1 + x2

√

1 + x2

Part (ii)

√

− 1) ,

≥ 1

(arccosh x) =

ln(x +

x + √1x2

· 1 +

√

, x > 0

−

√

−

x2 − 1

+ x

, x > 0

x + √x2 − 1 · √x2 − 1

√

, x > 0.

x2 − 1

Part (iii)

1 − x ,

(arctanh x) = dx

|x| < 1

1 + x

= dx

ln(1 + x) − ln(1 − x) , |x| < 1

= 2

1 + x − 1−1x , |x| < 1

−


1				1		1		, \|x\| < 1
=						+
	2		1 + x				1 − x
= 2				−1 − x2				, \|x\| < 1
	1		1		x + 1 + x
=		1			,	\|x\| < 1.

	1 − x2

3.3. DIFFERENTIATION OF INVERSE FUNCTIONS								129
Exercises 3.3 Compute		dy	for each of the following:
Exercises 3.3 Compute		dx	for each of the following:
		dx
					1 + x		−
1.	y = ln(x2 + 1)		2.	y = ln	1 − x	,		1 < x < 1
1.	y = ln(x2 + 1)		2.	y = ln		,		1 < x < 1
3.	y = log2(x)		4.	y = log5(x3 + 1)

5.y = log10(3x + 1)

7.y = 2e−x

9.y = 21 (ex2 − e−x2 )

ex2 − e−x2

11. y = ex2 + e−x2

2 13. y = ex3 − e−x3

15.y = arcsin x2

17.y = arctan x5

19.y = arcsec x2

21.y = 3 sinh(2x) + 4 cosh 3x

23.y = e−x(4 sin 3x − 3 cos 3x)

25.y = 3 tanh(2x) − 7 coth (2x)

27.y = 10x2

29.y = 5(x4+x2)

6.y = log10(x2 + 4)

8.y = ex2

10.y = 21 (ex2 + e−x2 )

2 12. y = ex2 + e−x2

2 14. y = ex4 + e−x4

16. y = arccos x3

18. y = arccot x7

20. y = arccsc x3

22. y = ex(3 sin 2x + 4 cos 2x) 24. y = 4 sinh 2x + 3 cosh 2x

26.y = 3 sech (5x) + 4 csch (3x)

28.y = 2(x3+1)

30.y = 3sin x

130				CHAPTER 3. DIFFERENTIATION
31.	cos(x2)			32.	tan(x3)
31.	y = 4			32.	y = 10
33.	y = 2cot x			34.	y = 10sec(2x)
35.	y = 4csc(x2)			36.	y = e−x(2 sin(x2) + 3 cos(x3))
	y = arcsinh		x			x
37.				38.	y = arccosh
37.			2	38.	y = arccosh	3
39.	y = arctan	x		40.	y = x arcsinh		x

		4					3

In exercises 41–50, use the following procedure to compute the derivative of the given functions:

[(f(x)g(x)] =

[eg(x) ln(f(x))]

= eg(x) ln(f(x)) · g0(x) ln(f(x)) + g(x) f0((x))

f x

= (f(x))g(x) · g0(x) ln(f(x)) + g(x) f0((x)) .

f x

41.

y = (x2 + 4)3x

42.

y = (2 + sin x)cos x

43.

sin 2x

44.

y = (x

x2+1

y = (3 + cos x)

+ 4)

45.

y = (1 + x)1/x

46.

y = (1 + x2)cos 3x

47.

y = (2 sin x + 3 cos x)

48.

1/x2

y = (1 + ln x)

49.

cosh x

50.

y = (1 + sinh x)

y = (sinh

x + cosh

3.4Implicit Di erentiation

So far we have dealt with explicit functions such as x2, sin x, cos x, ln x, ex, sinh x and cosh x etc. In applications, two variables can be related by an equation such as

3.4. IMPLICIT DIFFERENTIATION		131
(i) x2 +y2 = 16	(ii) x3 +y3 = 4xy	(iii) x sin y +cos 3y = sin 2y.

In such cases, it is not always practical or desirable to solve for one variable explicitly in terms of the other to compute derivatives. Instead, we may implicitly assume that y is some function of x and di erentiate each term of the equation with respect to x. Then we solve for y0, noting any conditions under which the derivative may or may not exist. This process is called implicit di erentiation. We illustrate it by examples.

Example 3.4.1 Find dxdy if x2 + y2 = 16.

Assuming that y is to be considered as a function of x, we di erentiate each term of the equation with respect to x.

graph

dxd (x2) + dxd (y2) = dxd (16)


2x + 2y		dy		= 0		(Why?)
		dx

	2y		dy	= −2x

			dx
			dy	= −	x	, provided y 6= 0.

			dx		y

We observe that there are two points, namely (4, 0) and (−4, 0) that satisfy the equation. At each of these points, the tangent line is vertical and hence, has no slope.

If we solve for y in terms of x, we get two solutions, each representing a function of x:

y = (16 − x2)1/2 or y = −(16 − x2)1/2.

132

CHAPTER 3. DIFFERENTIATION

On di erentiating each function with respect to x, we get, respectively,

(16 − x2)−1/2(−2x) ; or

= −

(16 − x2)−1/2(−2x)

= −

;

(16 − x2)1/2

−(16 − x2)1/2

= −

, y 6= 0; or

= −

, y 6= 0.

In each case, the ﬁnal form is the same as obtained by implicit di erentiation.

Example 3.4.2 Compute dxdy for the equation x3 + y3 = 4xy.

As in Example 2.4.1, we di erentiate each term with respect to x, assum-

ing that y is a function of x.

(x3) +

(y3) =

(4xy)

(Why?)

3x2 + 3y2

4 dx y + x

(3y2)

4y − 3x2

− 4x

(Why?)

(3y2 −

4x)

4y − 3x2

(Why?)

4y − 3x2

, if 3y2

−

4x = 0. (Why?)

3y2 − 4x

This di erentiation formula is valid for all points (x, y) on the given curve, where 3y2 − 4x 6= 0.

Example 3.4.3 Compute dxdy for the equation x sin y + cos 3y = sin 2y. In this example, it certainly is not desirable to solve for y explicitly in terms of

3.4. IMPLICIT DIFFERENTIATION

133

x. We consider y to be a function of x, di erentiate each term of the equation with respect to x and then algebraically solve for y in terms of x and y.

(x sin y) +

(cos 3y) =

(sin 2y)

2dx

dx (sin y) + x dx (sin y) + (−3 sin 3y) dx = (cos 2y)

sin y + x(cos y)

− 3 sin(3y)

= (2 cos 2y)

Upon collecting all terms containing

on the left-side, we get

[x cos y − 3 sin 3y − 2 cos 2y]

= − sin y

sin y

= −

x cos y

−

3 sin 3y

−

2 cos 2y

whenever

x cos y − 3 sin 3y − 2 cos 2y 6= 0.

Example 3.4.4 Find

for

(x − 2)2

(y − 3)2

= 1.

On di erentiating each term with respect to x, we get

graph

(x − 2)2

(y − 3)2

(1)

(x − 2) +

(y − 3)

= 0

2(x − 2)/9

, if y = 3

−2(y

−

3)/16

−

16(x − 2)

if y = 3.

9(y

−

134	CHAPTER 3. DIFFERENTIATION

The tangent lines are vertical at (−1, 3) and (5, 3). The graph of this equation is an ellipse.

Example 3.4.5 Find dxdy for the astroid x2/3 + y2/3 = 16.

graph

dxd (x2/3) + dxd (y2/3) = 0

23 x−1/3 + 23 y−1/3 dxdy = 0, if x 6= 0 and y 6= 0

dy		y−1/3	= −	x	1/3
	= −				, if x 6= 0 and y 6= 0.
dx	= −	x−1/3		y	, if x 6= 0 and y 6= 0.

Example 3.4.6 Find dxdy for the lemniscate with equation (x2 + y2)2 = 4(x2 − y2).

graph

3.4. IMPLICIT DIFFERENTIATION

135

((x2 + y2)2) = 4

(x2 − y2)

2(x2 + y2) 2x + 2y dx

= 4

2x − 2y dx

[4y(x2 + y2) + 8y]

= 8x − 4x(x2 + y2) (Why?)

8x − 4x(x2 + y2)

, if 4y(x2 + y2) + 8y = 0, y = 0.

4y(x2 + y2) + 8y

6 6

Example 3.4.7 Find the equations of the tangent and normal lines at (x0, y0) to the graph of an ellipse of the form

(x − k)2 + (y − k)2 = 1. a2 b2

First, we ﬁnd dxdy by implicit di erentiation as follows:

(x − h)2

(y − k)2

(1)

− h) +

(y − k)

= 0

= −

(x − h) ·

, if y 6= k

2(y

−

−b2

, y = k.

y − k

−

It is clear that at (a + h, k) and (−a + h, k), the tangent lines are vertical and have the equations

x = a + h and x = −a + h.

Let (x0, y0) be a point on the ellipse such that y0 6= k. Then the equation of the line tangent to the ellipse at (x0, y0) is

y		y		=	−b2	x0 − h	(x		x	).
	−					y0 − k		−
			0		a2				0

136			CHAPTER 3.	DIFFERENTIATION
We may express this in the form
	(y − y0)(y0 − k)	+	(x − x0)(x0 − h)	= 0.
	b2	+	a2	= 0.
	b2		a2

By rearranging some terms, we can simplify the equation in the following traditional form:

(y − k) + (k − y0)

0 −

k) +

(x − h) + (h − x0)

−

h) = 0

(y − k)(y0 − k)

(x − h)(x0 − h)

(x0 − h)2

(y0 − k)2

= 1.

(y − k)(y0 − k)

(x − h)(x0 − h)

= 1

Exercises 3.4 In each of the following, ﬁnd dxdy by implicit di erentiation.

1.	y2 + 3xy + 2x2 = 16					2. x3/4 + y3/4 = 103/4
3.	x5 + 4x3y2 + 3y4 = 8					4.	sin(x − y) = x2y cos x
	x2			y2				x2		y2
5.			−		= 1	6.			+		= 1
		4		9				16		9

Find the equation of the line tangent to the graph of the given equation at the given point.

7.	9			+ y4		= 1 at 2,		2 3			!
										5
		x2		2					√
			x2		y2	3

8.				−		= 1 at		√5, 1
	9				4		2
9.	x2y2 = (y + 1)2(9 − y2) at											2 √5, 2
10.			y2 = x3(4 − x) at (2, 4)									3

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