The Chain Rule
LRT
02/15/2017
Recall that the composition oftwo functions f and g is
defined by (f ◦ g)(x) = f g(x) . We call f the outside
function and g the inside function.
The Chain Rule is the formula for computing the derivative
of f ◦ g.
(f ◦ g)0 (x) = f 0 g(x) · g 0 (x)
In English
The derivative of a composition is
the derivative of the outside function
with the inside function plugged in
multiplied by
the derivative of the inside function.
Here is yet another way to think of the formula. Suppose
c(x) = f g(x)
Then
c0 (x) = f 0 g(x) · g 0 (x)
In words, take the derivative of the outside function but
evaluate it at the same place you evaluated f in order to
compute c(x). Then multiply be the derivative of g.
The chief skill you need to develop, if you have not already
acquired it, is to recognize compositions in the wild.
√
x2 + 1
1
g(x)
f (x)
r
√
x2 + 1
outer :
√
x
inner : x2 + 1
1
outer0 : √
2 x
inner0 : 2x
1
√
(2x)
2 x2 + 1
1
g(x)
outer :
f (x)
1
x
inner : g(x)
outer0 :
−1
x2
inner0 : g 0 (x)
−1
2 g 0 (x)
g(x)
r
outer : xr
inner : f (x)
outer0 : rxr−1
inner0 : f 0 (x)
r f (x)
r−1
f 0 (x)
Notice that all three examples on the previous page are
examples of the Power plus Chain Rule. You can combine
them in the Generalized Power Rule:
d ur
du
= rur−1
dx
dx
(GP R)
√
1
g(x)
x2 + 1
u = x2 + 1, r =
1
2
r
f (x)
u = g(x), r = −1 u = f (x), r = r
Later in the semester we will need other versions of the
Chain Rule, but for now (GPR) will get you through most
problems.
Computing derivatives
The key to successful computations is to give up on trying
to solve the problem.
This seems counterintuitive but what you should try for is
to make the problem simpler. Eventually the needed
calculations will become so simple that you can do them
and then reassemble your work to answer the original
problem.
Example: If f (x) =
p
x+
√
x + 1, find f 0 (x).
Think about how you would compute f for some input.
First you would compute x + 1, then take the square root,
then add x and finally take one more square root.
Since taking
a square root is the
√ last thing you did,
√
1
f (x) = u = u 2 with u = x + x + 1.
du
The (GPR) is easy to plug into once you know
but
dx
du
suppose you are not prepared to write down
just yet:
dx
du
does look simpler than f 0 (x) so we continue.
dx
√
x + 1 is√a sum. I am prepared to differentiate x
d x+1
but I also need
.
dx
Since
root is the last thing I do here,
√ taking a square
1
d x+1
d w2
=
with w = x + 1. Now reassemble:
dx
dx
√
d x+1
1
1
1
= (x + 1)− 2 (1) = √
(GPR)
dx
2
2 x+1
√
dx + x + 1
1
=1+ √
(Sum Rule)
dx
2 x+1
√
1
1
− 21
0
f (x) = (x + x + 1)
1+ √
(GPR)
2
2 x+1
u=x+
√
x2 + 3x − 1
.
Find f (x) if f (x) =
x
x + x+1
0
The last thing we do is divide, so we will use the Quotient
Rule. To do this we need the derivatives of the numerator
and the denominator. I am not prepared to write down
either just yet.
u=
√
√
x2 + 3x − 1
f (x) =
x
x + x+1
x2 + 3x − 1
u=x+
x
x+1
x
1
u = w 2 , w = x2 + 3x − 1 u = x + w, w =
x+1
w0 = 2x + 3
1(x
+
1)
−
x(1)
2x + 3
w0 =
u0 = √
(x + 1)2
2
2 x + 3x − 1
1
w0 =
(x + 1)2
1
u0 = 1 +
(x + 1)2
f 0 (x) =
2x + 3
√
2 x2 + 3x − 1
x+
x
x+1
−
x+
√
x2
2
x
x+1
+ 3x − 1 1 +
1
(x + 1)2