The Chain Rule The Chain Rule

Introduction

We will prove that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

\frac{d}{dx}f(g(x))=f'(g(x))g'(x)
Here is an example:
f(x)=\sin(x) \quad g(x)=x^2 \quad f(g(x))=\sin(x^2)
f'(g(x))=\cos(x^2) \quad g'(x)=2x
\frac{d}{dx}\sin(x^2)=\cos(x^2)2x


The derivative of a function is the change in the function in relation to the change in the input.

For example,
\frac{d}{dx}f(x)
is the change in
f(x)
as
x
changes.

Proving
\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)
is proving the change in
f(g(x))
as
x
changes is the same as the change in
f(g(x))
as
g(x)
changes mulitplied with the change in
g(x)
as
x
changes. In notation, this is:
\frac{d}{dx}f(g(x))=\frac{d}{dg(x)}f(g(x)) \cdot \frac{d}{dx}g(x)

Proof of the Chain Rule

\begin{align*} \frac{d}{dx}f(x)&=\lim_{x \to a}\frac{f(x)-f(a)}{x-a} &\quad\quad \frac{d}{dx}g(x)&=\lim_{x \to a}\frac{g(x)-g(a)}{x-a} \\\\ \frac{d}{dx}f(g(x))&=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{x-a} &\quad\quad \frac{d}{dg(x)}f(g(x))&=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)} \\\\\\ \end{align*} \\ \begin{align*} \frac{d}{dg(x)}f(g(x)) \cdot \frac{d}{dx}g(x)&=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)} \cdot \lim_{x \to a}\frac{g(x)-g(a)}{x-a} \\\\ &=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{g(x)-g(a)} \cdot \frac{g(x)-g(a)}{x-a} \\\\ &=\lim_{x \to a}\frac{f(g(x))-f(g(a))}{x-a} \\\\ &=\frac{d}{dx}f(g(x)) \end{align*}
Therefore, the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied with the derivative of the inner function.

In other words, finding the rate that
f
changes as
g
changes and multiplying it with the rate
g
changes as
x
changes is the same as finding the rate
f
changes as
x
changes.

Examples

Example 1: Find the derivative of

y = (2x + 1)^5
.

\begin{align*} y &= (2x + 1)^5 \\\\ \frac{dy}{dx} &= 5(2x + 1)^4 \, \frac{d}{dx}(2x + 1) \\\\ &= 5(2x + 1)^4 \cdot 2 \\\\ &= 10(2x + 1)^4 \end{align*}


Example 2: Find the derivative of
f(x) = \sin(x^2)
.

\begin{align*} f(x) &= \sin(x^2) \\\\ f'(x) &= \cos(x^2) \cdot \frac{d}{dx}(x^2) \\\\ &= \cos(x^2) \cdot 2x \\\\ &= 2x\cos(x^2) \end{align*}


Example 3: Find the derivative of
g(t) = \sqrt{t^3 + 1}
.

\begin{align*} g(t) &= \sqrt{t^3 + 1} = (t^3 + 1)^{\frac{1}{2}} \\\\ g'(t) &= \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}} \cdot \frac{d}{dt}(t^3 + 1) \\\\ &= \frac{1}{2}(t^3 + 1)^{-\frac{1}{2}} \cdot 3t^2 \\\\ &= \frac{3t^2}{2\sqrt{t^3 + 1}} \end{align*}


Example 4: Find the derivative of
h(x) = e^{\tan(x)}
.

\begin{align*} h(x) &= e^{\tan(x)} \\\\ h'(x) &= e^{\tan(x)} \cdot \frac{d}{dx}(\tan(x)) \\\\ &= e^{\tan(x)} \cdot \sec^2(x) \\\\ &= \sec^2(x)e^{\tan(x)} \end{align*}