HW 8
21 – 1
Use the chain rule to find the following derivatives. Present your solution just like the
solution in Example 21.2.1 (i.e., write the given function as a composition of two functions
f and g, compute the quantities required on the right-hand side of the chain rule formula,
and finally show the chain rule being applied to get the answer).
(a)
d 4
(2x − 5x)7 .
dx
(b)
d
[cos(8t − 3)].
dt
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Solution:
(a) The outside function is the function that raises an input to the power 7:
4
7
(2x − 5x) = f (g(x)),
So
f (x) = x
0
where
f (x) = x
7
f (x) = 7x
7
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4
and g(x) = 2x − 5x.
4
g(x) = 2x − 5x
g 0 (x) = 8x3 − 5
6
0
4
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6
f (g(x)) = 7(2x − 5x)
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giving
d
dx
d
dx
[f (g(x))]
↓
4
(2x − 5x)7
=
f 0 (g(x))
·
↓
=
4
g 0 (x)
↓
6
7(2x − 5x)
·
3
(8x − 5)
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(b) The outside function is the cosine function:
cos(8t − 3) = f (g(t)),
where
f (t) = cos t and g(t) = 8t − 3.
So
g(t) = 8t − 3
f (t) = cos t
0
g 0 (t) = 8
f (t) = − sin t
0
f (g(t)) = − sin(8t − 3)
giving
21 – 2
d
dt
[f (g(t))]
d
dt
[cos(8t − 3)]
=
↓
f 0 (g(t))
·
↓
=
− sin(8t − 3)
g 0 (t)
↓
·
8
Find the derivatives of the following functions. (You may omit details and proceed as in
Example 21.2.5.)
√
(a) f (x) = 3 8 − x3 .
(b) f (t) = esin 2t .
Solution:
(a) The outside function is the cube root function, so
i
d hp
3
f 0 (x) =
8 − x3 = 13 (8 − x3 )−2/3 · (−3x2 ).
dx
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(b) The outside function is the exponential function, so
d sin 2t e
dt
d
= esin 2t ·
[sin 2t]
dt
sin 2t
=e
· (cos 2t) · 2,
f 0 (t) =
where we have applied the chain rule a second time with outside function the sine
function.
21 – 3
Find the derivative of f (x) = 2x
Solution:
3
+4x
cos(3x).
The first rule applied is the product rule:
i
d h x3 +4x
2
cos(3x)
dx
3
d
d h x3 +4x i
2
cos(3x) + 2x +4x
[cos(3x)]
=
dx
dx
3
3
= 2x +4x (ln 2)(3x2 + 4) cos(3x) + 2x +4x (− sin(3x))(3).
f 0 (x) =
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21 – 4
Find the derivative of f (t) = cos
e5t
.
t − t8
Solution:
5t d
e
cos
dt
t − t8
5t 5t e
e
d
= − sin
·
t − t8
dt t − t8
5t (t − t8 ) d e5t − e5t d t − t8 e
dt
dt
= − sin
·
t − t8
(t − t8 )2
5t e
(t − t8 )e5t (5) − e5t (1 − 8t7 )
= − sin
·
t − t8
(t − t8 )2
f 0 (t) =
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22 – 1
Find the derivatives of each of the following functions:
(a) f (x) = 3 log3 x − 4 ln x,
(b) f (t) = ln 1 + 3e2t .
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Solution:
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(a)
f 0 (x) = 3
1
x ln 3
1
−4
.
x
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(b)
d ln 1 + 3e2t
dt
1
d =
·
1 + 3e2t
2t
1 + 3e
dt
1
=
· 3e2t (2).
1 + 3e2t
f 0 (t) =
22 – 2
Find the derivative of the function f (x) = ln (ln (ln x)).
Solution:
logarithm:
The chain rule is used twice, each time with outside function the natural
d
[ln (ln (ln x))]
dx
1
d
=
·
[ln (ln x)]
ln (ln x) dx
1
1
d
=
·
·
[ln x]
ln (ln x) ln x dx
1
1
1
=
·
· .
ln (ln x) ln x x
f 0 (x) =
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22 – 3
Find the derivative of the function f (x) = sin−1
√
1 − x2 .
Solution:
i
d h −1 p
1 − x2
sin
dx
i
1
d hp
2
1
−
x
=q
·
√
dx
1 − ( 1 − x2 ) 2
f 0 (x) =
1 1
· (1 − x2 )−1/2 (−2x)
|x| 2
x
=− √
,
|x| 1 − x2
=
where we have used that
√
x2 = |x|.
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