5128_Ch03_pp098-184

Section 3.6 Chain Rule 149
Composite f ˚ g
Rate of change at
x is f'(g(x)) • g'(x)
Figure 3.42 Rates of change multiply:
the derivative of f g at x is the derivative
of f at the point gx) times the derivative
of g at x.
x
g
Rate of change
at x is g'(x)
u g(x)
f
Rate of change
at g(x) is f'(g(x))
y f(u) f(g(x))
RULE 8
The Chain Rule
If f is differentiable at the point u gx, and g is differentiable at x, then the
composite function f gx f gx is differentiable at x, and
f gx f gx • gx.
In Leibniz notation, if y f u and u gx, then
d y dy
• d u
,
dx
d u dx
where dydu is evaluated at u gx.
It would be tempting to try to prove the Chain Rule by writing
y
x
y
u
• u
x
(a true statement about fractions with nonzero denominators) and taking the limit as
x→0. This is essentially what is happening, and it would work as a proof if we knew
that Du, the change in u, was nonzero; but we do not know this. A small change in x could
conceivably produce no change in u. An air-tight proof of the Chain Rule can be constructed
through a different approach, but we will omit it here.
EXAMPLE 3
Applying the Chain Rule
An object moves along the x-axis so that its position at any time t 0 is given by
xt cos t 2 1. Find the velocity of the object as a function of t.
SOLUTION
We know that the velocity is dxdt. In this instance, x is a composite function:
x cos u and u t 2 1. We have
By the Chain Rule,
dx
sin u
d u
x cos (u)
d u
2t.
dt
u t 2 1
d x dx
• d u
dt
d u dt
sin u • 2t
sin t 2 1 • 2t
2t sin t 2 1. Now try Exercise 9.