The Chain Rule:
d
(f (g(x)) = f 0 (g(x)) · g 0 (x).
dx
√
Example 1. Find the derivative of y = x3 + x + 2 = f (g(x)), where
√
f (u) = u = u1/2 and g(x) = x3 + x + 2, so...
f 0 (g(x))
1/2
d p 3
d
3
x +x+2=
x +x+2
dx
dx
z
}|
{
−1/2
1 3
x +x+2
=
(3x2 + 1)
| {z }
2
g 0 (x)
(*) The chain rule can also be expressed as follows. If y = f (u) and
u = g(x), then y = f (g(x)) and
f 0 (g(x)) g 0 (x)
z}|{ z}|{
dy
dy
du
=
·
dx
du
dx
1
Explanation: With y = f (u) and u = g(x), linear approximation says
that if ∆x ≈ 0, then
∆u = g(x + ∆x) − g(x) ≈ g 0 (x)∆x.
(1)
Likewise, if ∆u ≈ 0, then
∆y = f (u + ∆u) − f (u) ≈ f 0 (u)∆u.
(2)
Now, if ∆x is close to 0, then so is ∆u (because g 0 (x) is fixed), so if
∆x ≈ 0, then using both approximations (1) and (2) gives
≈∆u
z }| {
∆y ≈ f 0 (u)∆u ≈ f 0 (u)g 0 (x)∆x = f 0 (g(x))g 0 (x)∆x,
so
∆y
f 0 (g(x))g 0 (x)
∆x
0
0
=
f
(g(x))g
(x).
≈
∆x
∆x
These approximations all becomes more and more accurate as ∆x → 0,
and therefore
dy
∆y
= lim
= f 0 (g(x))g 0 (x).
dx ∆x→0 ∆x
2
3
Example 4. Find the point(s) on the graph of h(x) = x − 3x − 4
where the tangent line is horizontal.
2
(*) We need to find the point(s) where h0 (x) = 0, which means that we
first have to differentiate h(x).
(*) Differentiation: use the chain rule on h(x) = f (g(x)),
with f (u) = u3 and g(x) = x2 − 3x − 4:
u
z
}|
{ zdg/dx
}| {
0
2
2
h (x) = 3(x − 3x − 4) (2x − 3)
|
{z
}
df /du
(*) Recall: A product AB = 0 if and only if A = 0 or B = 0, so
h0 (x) = 0 ⇐⇒ x2 − 3x − 4 = 0 or 2x − 3 = 0
and x2 − 3x − 4 = (x − 4)(x + 1), so h0 (x) = 0 when x = −1, x = 3/2
and x = 4.
3
12
y=(x2-3x-4)3
8
4
-3
-2
-1
0
1
2
3
4
-4
-8
-12
Figure 1: The graph of y = (x2 − 3x − 4)3 .
4
5
6
Differentiating without the chain rule...
h(x) = (x2 − 3x − 4)3 = (x2 − 3x − 4)2 (x2 − 3x − 4)
= (x4 − 6x3 + x2 + 24x + 16)(x2 − 3x − 4)
= x6 − 9x5 + 15x4 + 45x3 − 60x2 − 144x − 64
So
h0 (x) = 6x5 − 45x4 + 60x3 + 135x2 − 120x − 144.
Now all we have to do is to solve the equation
6x5 − 45x4 + 60x3 + 135x2 − 120x − 144 = 0...
Observation: Using the chain rule in this example has two advantages:
(*) No messy arithmetic.
(*) The chain rule gives h0 (x) in a (partially) factored form, which
makes solving the equation h0 (x) = 0 is much easier.
5
Example 5. Find the equation of the tangent line to the graph
y= √
3
2
x2 + 4
at the point (2, 1).
We can use the quotient rule combined with the chain rule to find the
derivative dy/dx, or we can just use the chain rule and the observation
that
2
2
−1/3
y= √
=
2(x
+
4)
3
x2 + 4
1
4x
dy
=2· −
(x2 + 4)−4/3 · (2x) = − (x2 + 4)−4/3
=⇒
dx
3
3
dy 8 −4/3
1
=⇒
=− ·8
=− .
dx x=2
3
6
Now we use the point-slope formula to find the equation of the tangent
line:
1
1
4 x
y − 1 = − (x − 2) =⇒ y = 1 − (x − 2)
or y = −
6
6
3 6
6
3
2
y=4/3-x/
6
1
-2
-1
0
y=2(x2+4)-1/3
1
2
3
4
5
-1
Figure 2: The graphs of y = 2(x2 + 4)−1/3 and the tangent line at (2, 1).
7
Observation: f (x)/g(x) = f (x)g(x)−1 , so...
d
dx
f (x)
g(x)
d
−1
=
f (x)g(x)
dx
0
−1
0
−1
= f (x)g(x)
= f (x)g(x)
d
−1
+ f (x)
g(x)
dx
−2 0
+ f (x) (−1)g(x)
g (x)
f 0 (x) f (x)g 0 (x)
f 0 (x)g(x) − f (x)g 0 (x)
=
−
=
g(x)
g(x)2
g(x)2
I.e., the quotient rule follows from combining the product rule and chain
rule.
8
Marginal revenue product.
Suppose that a firm’s revenue function is r = f (q) (where q is output),
and their production function is q = g(l) (where l is labor input). In
this case, the firm’s revenue depends on its labor input
r = f (g(l)).
The derivative dr/dl is called the firm’s marginal revenue product.
(*) By the chain rule
dr dq
dr
=
· ,
dl
dq dl
i.e.,
marginal revenue product = (marginal revenue) × (marginal product).
9
Example: The demand equation for a firm’s product is
p = 100 − 0.8q
and the firm’s production function is
√
q = 5 4l − 15,
where labor input l is measured in 40-hour work-weeks.
(*) Find the firm’s marginal revenue product when l = 10.
1. r = pq = 100q − 0.8q
2
=⇒
dr
= 100 − 1.6q
dq
dq
d
1/2
=
5(4l − 15)
= 5 · 12 (4l − 15)−1/2 · 4 = 10(4l − 15)−1/2
2.
dl
dl
√
3. q(10) = 5 40 − 15 = 25.
2
60
z
}|
{
z
}|
{
dr dr dq −1/2
4.
(100
−
1.6
·
25)
×
10
·
25
= 120
=
×
=
dl l=10
dq q=25 dl l=10
10