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The chain rule is a fundamental concept in calculus that allows finding the derivative of a composite function. This rule states that the derivative of a function that is a composition of another function and a constant or variable function is the product of the derivatives of each function. examples and explanations of how to apply the chain rule to find derivatives of composite functions.
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The Chain Rule:
d dx (f (g(x)) = f ′(g(x)) · g′(x).
Example 1. Find the derivative of y =
x^3 + x + 2 = f (g(x)), where
f (u) =
u = u^1 /^2 and g(x) = x^3 + x + 2, so...
d dx
x^3 + x + 2 = d dx
x^3 + x + 2
f ′(g(x)) ︷ ︸︸ ︷ 1 2
x^3 + x + 2
(3x^2 + 1) ︸ ︷︷ ︸ g′(x)
(*) The chain rule can also be expressed as follows. If y = f (u) and u = g(x), then y = f (g(x)) and
dy dx
f ′(g(x)) ︷︸︸︷ dy du
g′(x) ︷︸︸︷ du dx
Explanation: With y = f (u) and u = g(x), linear approximation says that if ∆x ≈ 0, then
∆u = g(x + ∆x) − g(x) ≈ g′(x)∆x. (1)
Likewise, if ∆u ≈ 0, then
∆y = f (u + ∆u) − f (u) ≈ f ′(u)∆u. (2)
Now, if ∆x is close to 0, then so is ∆u (because g′(x) is fixed), so if ∆x ≈ 0, then using both approximations (1) and (2) gives
∆y ≈ f ′(u)∆u ≈ f ′(u)
︷ ≈︸︸∆u ︷ g′(x)∆x = f ′(g(x))g′(x)∆x,
so ∆y ∆x
f ′(g(x))g′(x)∆x ∆x^
= f ′(g(x))g′(x).
These approximations all becomes more and more accurate as ∆x → 0, and therefore dy dx = (^) ∆limx→ 0 ∆y ∆x = f ′(g(x))g′(x).
-3 -2 -1 0 1 2 3 4 5 6
4
8
12
y=(x^2 -3x-4)^3
Figure 1: The graph of y = (x^2 − 3 x − 4)^3.
Differentiating without the chain rule...
h(x) = (x^2 − 3 x − 4)^3 = (x^2 − 3 x − 4)^2 (x^2 − 3 x − 4) = (x^4 − 6 x^3 + x^2 + 24x + 16)(x^2 − 3 x − 4) = x^6 − 9 x^5 + 15x^4 + 45x^3 − 60 x^2 − 144 x − 64
So
h′(x) = 6x^5 − 45 x^4 + 60x^3 + 135x^2 − 120 x − 144.
Now all we have to do is to solve the equation
6 x^5 − 45 x^4 + 60x^3 + 135x^2 − 120 x − 144 = 0...
Observation: Using the chain rule in this example has two advantages:
(*) No messy arithmetic.
(*) The chain rule gives h′(x) in a (partially) factored form, which makes solving the equation h′(x) = 0 is much easier.
-2 -1 0 1 2 3 4 5
1
2
3
y=2(x^2 +4)-1/
y=4/3-x/
Figure 2: The graphs of y = 2(x^2 + 4)−^1 /^3 and the tangent line at (2, 1).
Observation: f (x)/g(x) = f (x)g(x)−^1 , so...
d dx
f (x) g(x)
d dx
f (x)g(x)−^1
= f ′(x)g(x)−^1 + f (x) d dx
g(x)−^1
= f ′(x)g(x)−^1 + f (x)
(−1)g(x)−^2 g′(x)
f ′(x) g(x)
f (x)g′(x) g(x)^2
f ′(x)g(x) − f (x)g′(x) g(x)^2
I.e., the quotient rule follows from combining the product rule and chain
rule.
Example: The demand equation for a firm’s product is
p = 100 − 0. 8 q
and the firm’s production function is
q = 5
4 l − 15 ,
where labor input l is measured in 40-hour work-weeks.
(*) Find the firm’s marginal revenue product when l = 10.
dq dl
d dl
5(4l − 15)^1 /^2
= 5 · 12 (4l − 15)−^1 /^2 · 4 = 10(4l − 15)−^1 /^2
dr dl
l=
dr dq
q=
dq dl
l=
