Warm-up
Find the derivative of the following:
1)
2)
3)
13
2
x
 2
23 x
x2
sin
Lesson 4.1
The Chain
Rule
The Chain Rule
Derivatives become complicated when we have composite
functions
Use a substitution, u = “the inside function”
then
Break up functions using the chain rule:
253 2
 xxu
dx
du
du
dy
dx
dy

“derivative of y in terms of u”
“derivative of u in terms of x”
 42
253  xxy
4
uy 
56
253 2


x
dx
du
xxu
3
4
4u
du
dy
uy


𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
∙
𝑑𝑢
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 4𝑢3
∙ 6𝑥 − 5
𝑑𝑦
𝑑𝑥
= 4 3𝑥2
− 5𝑥 + 2 3
∙ 6𝑥 − 5
Steps:
1) Set u equal to an “inner” function
2) Rewrite original function in terms of u
3) Derive u in terms of x and y in terms of u
4) Plug derivatives into chain rule
5) Simplify
Function form:
        xgxgfxgf
dx
d

 42
253  xxy

dx
dy
Example 1 Find the derivatives of the warm-ups using the chain rule
13
2
x
 2
23 x
x2
sin
Since polynomials are prevalent we come up with a
general power rule:
derivative of outside function times derivative of
inside function

dx
dy
Example 1 Find the derivative
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall
Slopes of Parametrized Curves
    A parametrized curve , is if
and are differentiable at .
x t y t differentiable at t
x y t
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall Slide 3- 15
Finding dy/dx Parametrically
If all three derivatives exist and 0,
dx
dt
dy
dy dt
dxdx
dt



4.1 the chain rule