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dy
·
=
dx du dx.

Proof. Let F be de¬ned on J such that
f (u)’f (g(c))
if u = g(c)
u’g(c)
F (u) =
f (g(c)) if u = g(c)
since f is di¬erentiable at g(c),
f (u) ’ f (g(c))
lim F (u) = lim
u ’ g(c)
u’g(c) u’g(c)

= f (g(c))
= F (g(c)).
Therefore, F is continuous at g(c). By the de¬nition 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
(f —¦ g)(x) ’ (f —¦ g)(c)
(f —¦ g) (c) = lim
x’c
x’c
f (g(x)) ’ f (g(c)) g(x) ’ g(c)
·
= lim
g(x) ’ g(c) x’c
x’c

g(x) ’ g(c)
= lim F (u) · lim
x’c
x’c
u’g(c)

= f (g(c)) · g (c).
3.2. THE CHAIN RULE 113

It follows that f —¦g is di¬erentiable at c. The general result follows by replac-
ing 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
dy du
= 3x2 .
= 2u and
du dx
Therefore,
dy dy du
·
=
dx du dx
= 2u · 3x2
= 6x2 (x3 + 4) .

Using the composition notation, we get

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

and
dy
= 6x5 + 24x2
dx
= 6x2 (x3 + 4) .

Using
(f —¦ g) (x) = f (g(x)) · g (x),
we see that
(f —¦ g)(x) = (x3 + 4)2 + 1
and

(f —¦ g) (x) = f (g(x)) · g (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

dy dy du
·
=
dx du dx
= (cos u)(2x)
= (cos(x2 + 3)) · (2x).



Example 3.2.3 Suppose that y = w2 , w = sin u + 3, and u = (4x + 1).
Then
dy dw du
dy
· ·
=
dx dw du dx
= (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
dy
= 2(sin(4x + 1) + 3)1 · ((cos(4x + 1)) · 4 + 0)
dx
= 8(sin(4x + 1) + 3) cos(4x + 1).



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

dy
= 5(cos(3x + 1))4 · (’ sin(3x + 1)) · 3
dx
= ’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

dy
= 3(tan2 (2x2 + 1)) · (sec2 (2x2 + 1)) · 4x
dx
= 12x · tan2 (2x2 + 1) · sec2 (2x2 + 1).



x+1
Example 3.2.6 Suppose that y = cot . Then
x2 + 1

(x2 + 1) · 1 ’ (x + 1)2x
dy x+1
= ’ csc2
x2 + 1 (x2 + 1) · 2x
dx
x2 + 2x ’ 1 x+1
csc2
= .
2 + 1)2 x2 + 1
(x



3
x2 + 1
Example 3.2.7 Suppose that y = sec .
x4 + 2
Since the function y has a composition of several functions, let us de¬ne
some intermediate functions. Let
x2 + 1
3
y = sec w, w = u , and u = 4 .
x +2
Then
dy dy dw du
· ·
=
dx dw du dx
(x4 + 2) · 2x ’ (x2 + 1) · 4x3
2
= [sec(w) tan(w)] · [3u ] ·
(x4 + 2)2
4x ’ 4x3 ’ 2x5
2
= 3u (sec w tan w) ·
(x4 + 2)2
2 3 3
x2 + 1 x2 + 1 x2 + 1 4x ’ 4x3 ’ 2x5
·
=3 sec tan .
x4 + 2 x4 + 2 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 .


dy
Exercises 3.2 Evaluate for each of the following:
dx
3
x2 + 2
10
1. y = (2x ’ 5) 2. y =
x5 + 4

4. y = cos(x3 + 1)
3. y = sin(3x + 5)

5. y = tan5 (3x + 1) 6. y = sec2 (x2 + 1)

7. y = cot4 (2x ’ 4) 8. y = csc3 (3x2 + 2)
4
5
x2 + 1
3x + 1
9. y = 10. y =
x2 + 2 x3 + 2

11. y = sin(w), w = u3 , u = (2x ’ 1)

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

1
13. y = tan(w), w = v 2 , v = u3 + 1, u =
x
x
14. y = sec w, w = v 3 , v = 2u2 ’ 1, u =
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.
3
x3 + 1
17. y = (x2 ’ 1)10
16. y =
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

23. y = cos(3x2 + 1)
22. y = sin(3x + 2)

24. y = sin(2x) cos(3x) 25. y = sec 2x + tan 3x

27. y = (x2 + 1)2 sin 2x
26. y = sec 2x tan 3x

29. y = sin2 (3x) + sec2 (5x)
28. y = x sin(1/x2 )

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

In exercises 31“60, assume that

d d d 1
(e’x ) = ’e’x
(ex ) = ex
(a) (b) (c) (ln x) =
dx dx dx x
d d 1
(bx ) = bx ln b
(d) (e) (logb x) = for b > 0 and b = 1.
dx dx 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

38. y = ln(1 ’ x)
37. y = ln(1 + x)


1’x
1
x2 + 1
39. y = ln 40. y = ln x +
2 1+x
√ 2
42. y = xe’x
x2 ’ 1
41. y = ln x +

43. y = esin 3x 44. y = e2x sin 4x
118 CHAPTER 3. DIFFERENTIATION


2 2
46. y = xe’x + 4e’x
45. y = ex (2 sin 3x ’ 4 cos 5x)
2 2 +4)
47. y = 4x 48. y = 10(x

49. y = 10sin 2x 50. y = 3cos 3x

51. y = log10 (x2 + 10) 52. y = log3 (x2 sin x + x)

54. y = ln(1 + e’x )
53. y = ln(sin(e2x ))

56. y = ln(ln(x2 + 4))
55. y = ln(cos x + 2)

3
x4 + 3
58. y = (1 + sin2 x)3/2
57. y = ln
x2 + 10

60. y = ln(csc 3x ’ cot 3x)
59. y = ln(sec 2x + tan 2x)




3.3 Di¬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

1
du
=
(i) dx
dx du

1 1
(ii) (f ’1 ) (x) = =
f (f ’1 (x)) f (u)

Proof. By comparison, x = f (f ’1 (x)) = x. Hence, by the chain rule

dx
= f (f ’1 (x)) · (f ’1 ) (x)
1=
dx
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 119

and
1
(f ’1 ) (x) = .
f (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.

π π
Example 3.3.1 Let u = arcsin x, ’1 ¤ x ¤ 1, and ’ ¤ u ¤ . Then
2 2
x = sin u and by the chain rule, we get

dx d(sin u) du
·
1= =
dx du dx
du
= cos u ·
dx
du 1
= .
dx cos u
Therefore,
d 1 π π
, ’ <u< ,
(arcsin x) =
dx cos u 2 2
1
= (Why?)
1 ’ sin2 u
1
=√ , ’1 < x < 1. (Why?)
1 ’ x2
Thus,
d 1
(arcsin x) = √ , ’1 < x < 1.
dx 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
dx du
= ’ sin u
1=
dx dx
du 1
=’ , 0<u<π
dx sin u
1
= ’√ , 0<u<π (Why?)
2u
1 ’ cos
1
= ’√ , ’1 < x < 1. (Why?)
1 ’ x2
We note again that x = ±1 are excluded.
Thus,
’1
d
(arccos x) = √ , ’1 < x < 1.
dx 2
1’x

π π
Example 3.3.3 Let u = arctan x, ’∞ < x < ∞, and ’ < u < . Then,
2 2
π π
x = tan u, ’ <u<
2 2
dx du π π
= (sec2 u), , ’ <u<
1=
dx dx 2 2
du 1
=
sec2 u
dx
1 π π
, ’ <u<
=
1 + tan2 u 2 2
1
, ’∞ < x < ∞
=
1 + x2
Therefore,
d 1
, ’∞ < x < ∞.
(arctan x) =
1 + x2
dx
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 121

Example 3.3.4 Let u = arcsec x, x ∈ (’∞, ’1] ∪ [1, ∞) and
π π
u ∈ 0, ∪ , π . Then,
2 2
x = sec u
dx du π π
= sec u tan u · u ∈ 0, ∪
1= , ,π
dx dx 2 2
du 1 π π
u ∈ 0, ∪
= , ,π
dx sec u tan u 2 2
1
√ (Why the absolute value?)
=
2u’1
| sec u| sec
1
=√ x ∈ (’∞, ’1) ∪ (1, ∞).
,
2’1
|x| x
Thus,
d 1
(arcsec x) = √ , x ∈ (’∞, ’1) ∪ (1, ∞).
dx 2’1
|x| x

Example 3.3.5 Let u = arccsc x, x ∈ (’∞, ’1] ∪ [1, ∞), and
π π
u ∈ ’ , 0 ∪ 0, . Then,
2 2
π π
x = csc u , u ∈ ’ , 0 ∪ 0,
2 2
dx du π π
= ’ csc u cot u · , u ∈ ’ , 0 ∪ 0,
1=
dx dx 2 2
’1 ’π
du π
, u∈ , 0 ∪ 0,
= , (Why?)
dx csc u cot u 2 2
1

= (Why?)
2u’1
| csc u| csc
1
=√ , x ∈ (’∞, ’1) ∪ (1, ∞).
2’1
|x| x
Note that x = ±1 are excluded.
Thus,
’1
d
(arccsc x) = √ , x ∈ (’∞, 1] ∪ (1, ∞).
dx x x2 ’ 1
122 CHAPTER 3. DIFFERENTIATION

Example 3.3.6 Let u = arccot x, x ∈ (’∞, 0] ∪ [0, ∞) and
π π
u ∈ 0, ∪ , π . Then
2 2
π π
u ∈ 0, ∪ ,π
x = cot u,
2 2
and
dx du π π
= ’ csc2 (u) · , u ∈ 0, ∪ ,π
1=
dx dx 2 2
’1
du π π
, u ∈ 0, ∪ ,π
=
csc2 u
dx 2 2
’1 π π
, u ∈ 0, ∪ ,π
=
1 + cot2 u 2 2
’1
, x ∈ (’∞, 0] ∪ [0, ∞).
=
1 + x2
Therefore,

’1
d
x ∈ (’∞, 0] ∪ [0, ∞).
(arccotx) = ,
1 + x2
dx
The results of these examples are summarized in the following theorem:


Theorem 3.3.2 (The Inverse Trigonometric Functions) The following dif-
ferentiation formulas are valid for the inverse trigonometric functions:
d 1
(arcsin x) = √ , ’1 < x < 1.
(i)
dx 1 ’ x2
’1
d
(arccos x) = √ , ’1 < x < 1.
(ii)
dx 2
1’x
1
d
, ’∞ < x < ∞.
(arctan x) =
(iii)
1 + x2
dx
’1
d
, ’∞ < x < ∞.
(iv) (arccot x) =
1 + x2
dx
d 1
(arcsec x) = √ , ’∞ < x < ’1 or 1 < x < ∞.
(v)
dx |x| x2 ’ 1
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 123

’1
d
(arccsc x) = √ , ’∞ < x < ’1 or 1 < x < ∞.
(vi)
dx |x| x2 ’ 1
Proof. Proof of Theorem 3.3.2 is outlined in Examples 76“80.

Theorem 3.3.3 (Logarithmic and Exponential Functions)
d 1
(i) (ln x) = for all x > 0.
dx x
d
(ex ) = ex for all real x.
(ii)
dx
d 1
(iii) (logb x) = for all x > 0 and b = 1.
dx x ln b
d
(bx ) = bx (ln b) for all real x, b > 0 and b = 1.
(iv)
dx
d u (x)
(u(x)v(x) = (u(x))v(x) v (x) ln(u(x)) + v(x)
(v) .
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 de¬nition for all x > 0, b > 0 and b = 1,

ln x
logb x = .
ln b
Then,

d 1
d
(logb x) = ln x
dx dx ln b
1 1
·
=
ln b x
1
= .
x ln b
Part (iv) By de¬nition, for real x, b > 0 and b = 1,

bx = ex ln b .
124 CHAPTER 3. DIFFERENTIATION

Therefore,

d d
(bx ) = (ex ln b )
dx dx
d
= ex ln b , (x ln b) (by the chain rule)
dx
= bx ln b. (Why?)



Part (v)

d
d
(u(x))v(x) = ev(x) ln(u(x))
dx dx
u (x)
= ev(x) ln(u(x)) v (x) ln(u(x)) + v(x)
u(x)
u (x)
= (u(x))v(x) v (x) ln u(x) + v(x)
u(x)



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

ln(x2 + 1)
d d
2
(log10 (x + 1)) =
dx dx ln 10
1 1
· 2x
= (by the chain rule)
ln 10 x2 + 1
2x
=2 .
(x + 1) ln 10


2 +1
Example 3.3.8 Let y = ex . Then, by the chain rule, we get

dy 2
= ex +1 · 2x
dx
2
= 2xex +1 .
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 125

3
Example 3.3.9 Let y = 10(x +2x+1) . By de¬nition and the chain rule, we
get
dy 3
= 10(x +2x+1) · (ln 10) · (3x2 + 2).
dx


Example 3.3.10

dx 2 2x
(x + 1)sin x = (x2 + 1)sin x cos x ln(x2 + 1) + sin x · 2
dx x +1
d d 2x
2 2
(x2 + 1)sin x = esin x ln(x +1) = 3sin x ln(x +1) · cos x ln(x2 + 1) + sin x · 2
dx dx x +1
2x sin x
= (x2 + 1)sin x cos x ln(x2 + 1) + 2 .
x +1



Theorem 3.3.4 (Di¬erentiation of Hyperbolic Functions)

d d
(i) (sinh x) = cosh x (ii) (cosh x) = sinh x
dx dx
d d
(tanh x) = sech2 x (cothx) = ’csch2 x
(iii) (iv)
dx dx
d d
(sech x) = ’sech x tanh x (csch x) = ’csch x coth x.
(v) (vi)
dx dx
Proof.
Part (i)

d d 1x
(e ’ e’x )
(sinh x) =
dx dx 2
1
= (ex ’ e’x (’1)) (by the chain rule)
2
1
= (ex + e’x )
2
= cosh x.
126 CHAPTER 3. DIFFERENTIATION

Part (ii)


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


Part (iii)


ex ’ e’x
d
d
(tanh x) =
dx ex + e’x
dx
(ex + e’x )(ex + e’x ) ’ (ex ’ e’x )(ex ’ e’x )
=
(ex + e’x )2
4
=x
(e + e’x )2
2
2
=
ex + e’x
= sech2 x.


Part (iv)


d 2
d
(sech x) =
dx ex + e’x
dx
(ex + e’x ) · 0 ’ 2(ex ’ e’x )
=
(ex + e’x )2
ex ’ e’x
2
=’ x ·
e + e’x ex + e’x
= ’sech x tanh x.
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 127

Part (v)

ex + e’x
d d
(coth x) = , x=0
dx ex ’ e’x
dx
(ex ’ e’x )(ex ’ e’x ) ’ (ex + e’x )(ex + e’x )
x=0
=
(ex ’ e’x )2
’4
=x , x=0
(e ’ e’x )2
2
2
=’ x , x=0
e ’ e’x
= ’csch2 x , x = 0.

Part (vi)

d 2
d
(csch x) = , x=0
dx ex ’ e’x
dx
(ex ’ e’x ) · 0 ’ 2(ex + e’x )
, x=0
=
(ex ’ e’x )2
ex + e’x
2
=’ x · , x=0
e ’ e’x ex ’ e’x
= ’csch x coth x, x = 0.



Theorem 3.3.5 (Inverse Hyperbolic Functions)

d 1
(arcsinh x) = √
(i)
dx 1 + x2

d 1
(arccosh x) = √ , x>1
(ii)
dx x2 ’ 1

d 1
|x| < 1
(iii) (arctanh x) = ,
1 ’ x2
dx

Proof.
128 CHAPTER 3. DIFFERENTIATION

Part (i)

d d
ln(x + 1 + x2 )
(arcsinh x) =
dx dx
1 x
√ · 1+ √
= (by chain rule)
x + 1 + x2 1 + x2

1 + x2 + x
1
√ ·√
=
x + 1 + x2 1 + x2
1
=√ .
1 + x2
Part (ii)

d
d
ln(x + x2 ’ 1) , x ≥ 1
(arccosh x) =
dx dx
1 x
√ · 1+ √
= , x>0
2’1 2’1
x+ x x

x2 ’ 1 + x
1
√ ·√ ,x > 0
=
2’1 2’1
x+ x x
1
=√ , x > 0.
x2 ’ 1
Part (iii)

d d1 1+x
, |x| < 1
(arctanh x) = ln
1’x
dx dx 2
d1
ln(1 + x) ’ ln(1 ’ x) , |x| < 1
=
dx 2
’1
1 1
’ , |x| < 1
=
2 1+x 1’x
1 1 1
, |x| < 1
= +
2 1+x 1’x
1 1’x+1+x
, |x| < 1
=
1 ’ x2
2
1
, |x| < 1.
=
1 ’ x2
3.3. DIFFERENTIATION OF INVERSE FUNCTIONS 129

dy
Exercises 3.3 Compute for each of the following:
dx

1’x
1. y = ln(x2 + 1) , ’1 < x < 1
2. y = ln
1+x

4. y = log5 (x3 + 1)
3. y = log2 (x)

6. y = log10 (x2 + 4)
5. y = log10 (3x + 1)
2
7. y = 2e’x 8. y = ex

1 x2 1 x2
2 2
(e ’ e’x ) (e + e’x )
9. y = 10. y =
2 2
2 2
ex ’ e’x 2
11. y = x2 12. y =
e + e’x2 ex 2 + e’x2
2 2
13. y = 14. y =
ex 3 ’ e’x3 ex 4 + e’x4
x x
15. y = arcsin 16. y = arccos
2 3
x x
17. y = arctan 18. y = arccot
5 7
x x
19. y = arcsec 20. y = arccsc
2 3

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

23. y = e’x (4 sin 3x ’ 3 cos 3x) 24. y = 4 sinh 2x + 3 cosh 2x

25. y = 3 tanh(2x) ’ 7 coth (2x) 26. y = 3 sech (5x) + 4 csch (3x)
2 3 +1)
27. y = 10x 28. y = 2(x
4 +x2 )
29. y = 5(x 30. y = 3sin x
130 CHAPTER 3. DIFFERENTIATION

2) 3)
31. y = 4cos(x 32. y = 10tan(x

33. y = 2cot x 34. y = 10sec(2x)
2)
36. y = e’x (2 sin(x2 ) + 3 cos(x3 ))
35. y = 4csc(x

x x
37. y = arcsinh 38. y = arccosh
2 3
x x
39. y = arctan 40. y = x arcsinh
4 3

In exercises 41“50, use the following procedure to compute the derivative of
the given functions:
d g(x) ln(f (x))
d
[(f (x)g(x) ] = [e ]
dx dx
f (x)
= eg(x) ln(f (x)) · g (x) ln(f (x)) + g(x)
f (x)

f (x)
= (f (x))g(x) · g (x) ln(f (x)) + g(x) .
f (x)
41. y = (x2 + 4)3x 42. y = (2 + sin x)cos x
2 +1
43. y = (3 + cos x)sin 2x 44. y = (x2 + 4)x

45. y = (1 + x)1/x 46. y = (1 + x2 )cos 3x
3 2
47. y = (2 sin x + 3 cos x)x 48. y = (1 + ln x)1/x
2 +3
50. y = (sinh2 x + cosh2 x)x
49. y = (1 + sinh x)cosh x



3.4 Implicit 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 + y 2 = 16 (ii) x3 + y 3 = 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 y , noting any conditions
under which the derivative may or may not exist. This process is called
implicit di¬erentiation. We illustrate it by examples.

dy
if x2 + y 2 = 16.
Example 3.4.1 Find
dx
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




d d
d
(x2 ) + (y 2 ) = (16)
dx dx dx
dy
2x + 2y =0 (Why?)
dx
dy
= ’2x
2y
dx
dy x
= ’ , provided y = 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,

dy 1 dy 1
= (16 ’ x2 )’1/2 (’2x) ; or = ’ (16 ’ x2 )’1/2 (’2x)
dx 2 dx 2
dy x x
=’ ; or
(16 ’ x2 )1/2 ’(16 ’ x2 )1/2
dx
x dy x
dy
= ’ , y = 0; or = ’ , y = 0.
dx y dx y


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



dy
for the equation x3 + y 3 = 4xy.
Example 3.4.2 Compute
dx
As in Example 2.4.1, we di¬erentiate each term with respect to x, assum-
ing that y is a function of x.


d d
dy 3
(x ) + (y 3 ) = (4xy)
dx dx dx
dy dx dy
3x2 + 3y 2 =4 y+x (Why?)
dx dx dx
dy dy
(3y 2 ) = 4y ’ 3x2
’ 4x (Why?)
dx dx
dy
(3y 2 ’ 4x) = 4y ’ 3x2 (Why?)
dx
4y ’ 3x2
dy
, if 3y 2 ’ 4x = 0.
=2 (Why?)
3y ’ 4x
dx

This di¬erentiation formula is valid for all points (x, y) on the given curve,
where 3y 2 ’ 4x = 0.



dy
Example 3.4.3 Compute for the equation x sin y + cos 3y = sin 2y. In
dx
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.
d d d
(x sin y) + (cos 3y) = (sin 2y)
dx dx dx
dx d dy dy
(sin y) + x (sin y) + (’3 sin 3y) = (cos 2y) 2
dx dx dx dx
dy dy dy
’ 3 sin(3y)
sin y + x(cos y) = (2 cos 2y) .
dx dx dx
dy
Upon collecting all terms containing on the left-side, we get
dx
dy
[x cos y ’ 3 sin 3y ’ 2 cos 2y] = ’ sin y
dx
dy sin y
=’
x cos y ’ 3 sin 3y ’ 2 cos 2y
dx
whenever
x cos y ’ 3 sin 3y ’ 2 cos 2y = 0.


(x ’ 2)2 (y ’ 3)2
dy
Example 3.4.4 Find for + = 1.
dx 9 16
On di¬erentiating each term with respect to x, we get



graph




(x ’ 2)2 (y ’ 3)2
d d
d
+ = (1)
dx 9 dx 16 dx
2 2 dy
(x ’ 2) + (y ’ 3) =0
9 16 dx
2(x ’ 2)/9
dy
=’ , if y = 3
2(y ’ 3)/16
dx
16(x ’ 2)
=’ , if y = 3.
9(y ’ 3)
134 CHAPTER 3. DIFFERENTIATION

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




dy
for the astroid x2/3 + y 2/3 = 16.
Example 3.4.5 Find
dx



graph




d
d
(x2/3 ) + (y 2/3 ) = 0
dx dx
2 ’1/3 2 ’1/3 dy
x +y = 0, if x = 0 and y = 0
3 3 dx
1/3
y ’1/3 x
dy
= ’ ’1/3 = ’ , if x = 0 and y = 0.
dx x y




dy
for the lemniscate with equation (x2 + y 2 )2 =
Example 3.4.6 Find
dx
4(x2 ’ y 2 ).



graph
3.4. IMPLICIT DIFFERENTIATION 135



d d
((x2 + y 2 )2 ) = 4 (x2 ’ y 2 )
dx dx
dy dy
2(x2 + y 2 ) 2x + 2y = 4 2x ’ 2y
dx dx
dy
[4y(x2 + y 2 ) + 8y] = 8x ’ 4x(x2 + y 2 ) (Why?)
dx
8x ’ 4x(x2 + y 2 )
dy
, if 4y(x2 + y 2 ) + 8y = 0, y = 0.
=
4y(x2 + y 2 ) + 8y
dx




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
dy
First, we ¬nd by implicit di¬erentiation as follows:
dx
(x ’ h)2 (y ’ k)2
d d d
(1)
+ =
a2 b2
dx dx dx
2 2 dy
(x ’ h) + 2 (y ’ k) =0
a2 b dx
b2
dy 2
= ’ 2 (x ’ h) · , if y = k
2(y ’ k)
dx a
’b2 x ’ h
=2 , y = k.
y’k
a
It is clear that at (a + h, k) and (’a + h, k), the tangent lines are vertical
and have the equations
x = ’a + h.
x=a+h and
Let (x0 , y0 ) be a point on the ellipse such that y0 = k. Then the equation of
the line tangent to the ellipse at (x0 , y0 ) is
’b2 x0 ’ h
(x ’ x0 ).
y ’ y0 = 2
y0 ’ k
a
136 CHAPTER 3. DIFFERENTIATION

We may express this in the form
(y ’ y0 )(y0 ’ k) (x ’ x0 )(x0 ’ h)
+ = 0.
b2 a2
By rearranging some terms, we can simplify the equation in the following
traditional form:
(x ’ h) + (h ’ x0 )
(y ’ k) + (k ’ y0 )
· (y0 ’ k) + (x0 ’ h) = 0
b2 a2
(x0 ’ h)2 (y0 ’ k)2
(y ’ k)(y0 ’ k) (x ’ h)(x0 ’ h)
+ = + = 1.
b2 a2 a2 b2
(y ’ k)(y0 ’ k) (x ’ h)(x0 ’ h)
+ =1 .
b2 a2


dy
Exercises 3.4 In each of the following, ¬nd by implicit di¬erentiation.
dx
1. y 2 + 3xy + 2x2 = 16 2. x3/4 + y 3/4 = 103/4

3. x5 + 4x3 y 2 + 3y 4 = 8 4. sin(x ’ y) = x2 y cos x

x2 y 2 x2 y 2

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.

2 2
x y 25
7. + = 1 at 2,
9 4 3

3√
x2 y 2
’ = 1 at 5, 1
8.
9 4 2

3√
22 2 2
9. x y = (y + 1) (9 ’ y ) at 5, 2
2

10. y 2 = x3 (4 ’ x) at (2, 4)
3.5. HIGHER ORDER DERIVATIVES 137

Two curves are said to be orthogonal at each point (x0 , y0 ) of their intersection
if their tangent lines are perpendicular. Show that the following families of
curves are orthogonal.

11. x2 + y 2 = r2 , y + mx = 0

12. (x ’ h)2 + y 2 = h2 , x2 + (y ’ k)2 = k 2

Compute y and y in exercises 13“20.

13. 4x2 + 9y 2 = 36 14. 4x2 ’ 9y 2 = 36

15. x2/3 + y 2/3 = 16 16. x3 + y 3 = a3

17. x2 + 4xy + y 2 = 6 18. sin(xy) = x2 + y 2

19. x4 + 2x2 y 2 + 4y 4 = 26 20. (x2 + y 2 )2 = x2 ’ y 2


3.5 Higher Order Derivatives
If the vertical height y of an object is a function f of time t, then y (t) is
called its velocity, denoted v(t). The derivative v (t) is called the acceleration
of the object and is denoted a(t). That is,
y(t) = f (t), y (t) = v(t), v (t) = a(t).
We say that a(t) is the second derivative of y, with respect to t, and write
d2 y
= a(t).
y (t) = a(t) or
dt2
Derivatives of order two or more are called higher derivatives and are repre-
sented by the following notation:
d2 y d3 y dn y
dy (n)
y (x) = , y (x) = 2 , y (x) = 3 , . . . , y (x) = n .
dx dx dx dx
The de¬nition is given as follows by induction:
d2 f dn f dn’1 f
d df d
, n = 2, 3, 4, · · · .
= and =
dx2 dxn dxn’1
dx dx dx
138 CHAPTER 3. DIFFERENTIATION

A convenient notation is
dn f
(n)
f (x) = n
dx
which is read as “the nth derivative of f with respect to x.”

Example 3.5.1 Compute the second derivative y for each of the following
functions:

(ii) y = cos(4x2 )
(i) y = sin(3x) (iii) y = tan(3x)

(vi) y = csc(x2 )
(iv) y = cot(5x) (v) y = sec(2x)

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