Intuitive “Proof” of the Chain Rule:
Intuitive “Proof” of the Chain Rule:
Let ¢u
be the change in u corresponding to a change of ¢x in x, that is
¢u = gsx + ¢xd - gsxd Then the corresponding change in y is ¢y = ƒsu + ¢ud - ƒsud. It would be tempting to write
¢y
# ¢u ¢x (1) ¢u ¢x
¢y
and take the limit as ¢x : 0:
dy
¢y
dx = lim ¢x:0 ¢x
= lim # ¢u ¢x:0 ¢u ¢x
¢y
¢y
= lim # lim ¢u
¢x:0 ¢u ¢x:0 ¢x
¢y
= lim #
lim as
¢u
(Note that
¢u : 0 ¢x : 0
¢u:0 ¢u ¢x:0 ¢x
since g is continuous.)
dy
du .
du dx
3.5 The Chain Rule and Parametric Equations
The only flaw in this reasoning is that in Equation (1) it might happen that ¢u = 0 (even when ¢x Z 0 ) and, of course, we can’t divide by 0. The proof requires a different ap- proach to overcome this flaw, and we give a precise proof in Section 3.8.
EXAMPLE 3 Applying the Chain Rule An object moves along the x-axis so that its position at any time tÚ 0 is given by
xstd = cos st 2 + 1d. Find the velocity of the object as a function of t.
Solution
We know that the velocity is dx >dt . In this instance, x is a composite function:
x= cos sud and We u=t 2 + 1. have
dx = - sinsud x= du cossud
du = 2t.
u=t 2 +1
dt
By the Chain Rule,
dx = dx # du
dt
du dt
= - sin sud # 2t
dx du evaluated at u
= - sin st 2 + 1d # 2t
= - 2t sin st 2 + 1d.
As we see from Example 3, a difficulty with the Leibniz notation is that it doesn’t state specifically where the derivatives are supposed to be evaluated.
“Outside-Inside” Rule
It sometimes helps to think about the Chain Rule this way: If y= ƒsg sxdd , then
dy
= ƒ¿sgsxdd # g¿sxd .
dx
In words, differentiate the “outside” function ƒ and evaluate it at the “inside” function g (x) left alone; then multiply by the derivative of the “inside function.”
EXAMPLE 4 Differentiating from the Outside In
Differentiate sin sx 2 + xd with respect to x.
Solution
d sin (x
2 + x) = cos (x 2 + x) # (2x + 1)
dx
derivative of
left alone
the inside
Repeated Use of the Chain Rule
We sometimes have to use the Chain Rule two or more times to find a derivative. Here is an example.
Chapter 3: Differentiation
H ISTORICAL B IOGRAPHY
EXAMPLE 5
A Three-Link “Chain”
Johann Bernoulli
Find the derivative of g std = tan s5 - sin 2td .
Solution
Notice here that the tangent is a function of 5 - sin 2t , whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule,
g¿std = d A tan
A 5 - sin 2t BB
dt
= sec 2 d s5 - sin Derivative of tan u with 2td # 5 - sin 2t
dt A B u= 5 - sin 2t
= sec 2 2td #
2t # d Derivative of 5 - sin u s5 - sin a0 - cos A 2t
B b with u = 2t
dt
= sec 2 s5 - sin 2td # s - cos 2td # 2
= - 2scos 2td sec 2 s5 - sin 2td .