Chain Rule

Section 2.5 Chain Rule

So far, we can differentiate:

What is left?

compositions
functions defined implicitly by equations

(equations where we can't easily solve for \(y\))

The chain rule is used for derivatives of functions of the form:

\begin{equation*} y = f(g(x)) = (f \circ g)(x) \end{equation*}

We can solve this if we let \(u = g(x)\) and \(y = f(u)\) so that:

\begin{equation*} \dfrac{dy}{dx} = \dfrac{df}{du} \cdot \dfrac{du}{dx} \end{equation*}

This can also be written as:

\begin{equation*} \dfrac{d}{dx}\left( f( g(x) \right) = f'(g(x)) \cdot g'(x) \end{equation*}

Write as \(u = g(x)\) and \(y = f(u)\text{:}\)

\begin{equation*} f(x) = 4 \cos^5(x) = 4(\cos(x))^5 \end{equation*}

We can make \(u = \cos(x)\) and the leftovers as \(f(u) = 4u^5\text{.}\)

Steps:

composition -- \(f(x)\) is a composition of:
- \(\displaystyle f(u) = 4u^5\)
- \(\displaystyle u = g(x) = \cos(x)\)
differentiate each piece:
- outside: \(\dfrac{df}{du} = \dfrac{d}{du}(4x^5) = 20u^4\)
- inside:\(\dfrac{du}{dx} = \dfrac{d}{dx}(\cos(x)) = - \sin(x)\)
multiply and label:
- \(\displaystyle f'(x) = 20u^4 (- \sin(x)) = -20(\cos(x))^4(\sin(x))\)

Find \(f'(x)\) for \(f(x) = \dfrac{1}{(x^3 + 2x)^9}\text{:}\)