Section 4.4: The Chain Rule and Higher Derivatives
Theorem: (Chain Rule)
Suppose that f and g are differentiable functions and F (x) = (f ◦ g)(x) = f (g(x)). Then
F 0 (x) = f 0 (g(x))g 0 (x).
In Leibniz notation, if y = f (u) and u = g(x) are differentiable functions, then
dy
dy du
=
.
dx
du dx
Example: Differentiate each function.
(a) f (x) = (5x2 − 3x)3
(b) f (x) = (2x3 − 7x2 + 5x + 1)5
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(c) f (x) =
(d) f (x) =
(e) f (x) =
√
3x4 + 6x
√
3
4x2 − 2
2
(1 − 5x2 )3
2
(f) f (x) =
(g) f (x) =
2x2
x+1
4
(x − 3)2
x2 + (7x − 1)2
√
(h) f (x) = ( 1 − 2x2 + 1)3
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Implicit Differentiation
The functions we have considered thus far can be described by expressing one variable
explicitly in terms of the other, y = f (x). However, some functions are defined implicitly by
a relation between x and y. For example,
x2 + y 2 = 4
x2 − xy + y 3 = 8.
or
It is possible to find the derivative of y with respect to x without solving for y explicitly.
Instead, we use the method of implicit differentiation.
Example: If x2 + y 2 = 4, find
dy
.
dx
Note: For some implicitly defined functions, it is very difficult (or impossible) to solve for y
explicitly in terms of x. For instance, consider the equations
x2 − xy + y 3 = 8
and
x cos y + y sin x = 1.
It would be very difficult to solve these equations for y in terms of x. Thus, we rely on
implicit differentiation.
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Example: Use implicit differentiation to find
dy
for each equation.
dx
(a) y 3 = 2x2 + y 4
(b) x2 − xy + y 3 = 8
(c) (x2 + y 2 )3 = 2y 4 + 6x2
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Example: Find an equation of the tangent line to the circle x2 + y 2 = 25 at (4, −3).
Related Rates
Suppose that there is a relationship between several quantities. Then a change in one
quantity will produce a change in the other quantities. In particular, the rates at which
these quantities change are related. If we know how one quantity changes, we can find out
how the other quantity changes.
Example: If x2 + 3xy + y 2 = 1 and
dy
dx
= 2, then find
when y = 1 and x ≥ 0.
dt
dt
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Example: A rock is thrown into a circular pond and produces a circular ripple which moves
outward. The radius of the circular ripple increases at a rate of 2 feet per second. Find the
rate at which the area of the ripple is increasing when the radius is 20 feet.
Example: A spherical snowball is melting in such a way that its radius is decreasing by 1 cm
per minute. At what rate is the volume decreasing when the radius is 5 cm?
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Example: A ladder 10 feet long rests agaist a vertical wall. If the bottom of the ladder slides
away from the wall at a rate of 1 foot per second, how fast is the top of the ladder sliding
down the wall when the bottom of the ladder is 6 feet from the wall?
Example: Suppose that we pump water into an inverted conical tank at the rate of 5 cubic
feet per minute. The height of the tank is 6 feet and the radius at the top is 3 feet. What
is the rate at which the water level is rising when the water is 2 feet deep?
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Higher Derivatives
Definition: The second derivative of a function f (x) is the derivative of the first derivative.
That is,
d 0
d df
d2 f
00
f (x) =
[f (x)] =
= 2.
dx
dx dx
dx
Similarly, the third derivative of f (x) is the derivative of the second derivative,
d d2 f
d3 f
d 00
000
[f (x)] =
=
f (x) =
.
dx
dx dx2
dx3
In general, the nth derivative of f is denoted by f (n) (x).
Example: Find all higher derivatives of f (x) = x4 − 3x3 + 5x2 + 16x − 2.
Example: Find the first and second derivatives of f (x) = (2x2 + 4)3 .
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Example: Find the first three derivatives of f (x) =
1
.
x
Definition: Suppose that an object moves along a straight line and the displacement of the
object at time t is given by the position function s(t).
1. The velocity of the object at time t is given by
v(t) = s0 (t).
2. The acceleration of the object at time t is given by
a(t) = v 0 (t) = s00 (t).
Example: Suppose that the position of a car at time t is given by s(t) = 3t3 − 2t + 1 where
s is measured in meters and t in seconds. Find the car’s velocity and acceleration at time
t = 1.
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