The Chain Rule

\( \frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx} \)

Example 1: Polynomial Function

Given that \(y=(2x^3+x)^4 \,, \quad find \quad \frac{dy}{dx} \)

\begin{align} Let \quad u & = 2x^3+x \\ then \quad y & =u^4 \\ \quad \frac{dy}{du} & =4u^3 \\ and \quad \frac{du}{dx} & =6x^2+1 \\ Using \, the \, & chain \, rule, \\ \frac{dy}{dx} & = \frac{dy}{du} \times \frac{du}{dx} \\ & = 4u^3(6x^2+1)\\ & = 4(2x^3+x)^3(6x^2+1) \end{align}

Example 2: Trigonometric Function

Given that \(y=cos^5(x) \,, \quad find \quad \frac{dy}{dx}\)

\begin{align} y = cos^5(x)& = (cos(x))^5 \\ Let \quad u & = cos(x) \\ then \quad y & = u^5 \\ \quad \frac{dy}{du} & = 5u^4 \\ and \quad \frac{du}{dx} & = -sin(x) \\ Using \, the \, & chain \, rule, \\ \frac{dy}{dx} & = \frac{dy}{du} \times \frac{du}{dx} \\ & = 5u^4 (-sin(x)) \\ & = -5cos^4(x)sin(x) \end{align}