3.6 – Chain Rule

The composite function $f(g(x))$ is differentiable at $x$ if $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$.

The Chain Rule

$$\frac{d}{dx}f(g(x))=f^{\prime }(g(x))\cdot g^{\prime }(x)$$ $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$

Examples

Differentiate $y=(x^4+10x{)}^3$.

Solution

We let $x^3$ be the outside function $f(x)$ and $x^4+10x$ be the inside function $g(x)$.

$$f^{\prime }(x)=3x^2$$ $$g^{\prime }(x)=4x^3+10$$

Applying the chain rule, we find the derivative to be:

$$\frac{dy}{dx}=3(x^3+10x{)}^2\cdot (4x^3+10)$$

Differentiate $y=\sqrt{2x^3-4x}$

Solution

Let $u=2x^3-4x$

$$y=\sqrt{u},u=2x^3-4x$$ $$\frac{dy}{du}=\frac{1}{2\sqrt{u}},\frac{du}{dx}=6x^2-4$$

Applying the chain rule:

$$\frac{dy}{dx}=\frac{1}{2\sqrt{2x^3-4x}}\cdot (6x^2-4)$$ $$=\frac{6x^2-4}{2\sqrt{2x^3-4x}}$$

Differentiate $y=\sin (\cos (5x^2))$.

Solution

The chain rule can be used multiple times. We let $\sin (x)$ be the outside function $f(x)$ and $\cos (5x^2)$ be the inside function $g(x)$.

$$f^{\prime }(x)=\frac{d}{dx}\sin (x)=\cos (x)$$

To find $g^{\prime }(x)$, we can use the chain rule again with $\cos (x)$ as the outside function and $5x^2$ as the inside function.

$$g^{\prime }(x)=-\sin (5x^2)\cdot 10x$$

Applying the chain rule, we find the derivative of $y$ to be:

$$\frac{dy}{dx}=\cos (\cos (5x^2))\cdot (-10x\cdot \sin (5x^2))$$ $$=-10x\cdot \cos (\cos (5x^2))\cdot \sin (5x^2)$$

Differentiate $y=\csc (x)$

Solution

$\csc (x)$ is equal to $(\sin (x){)}^{-}1$.

$$\frac{dy}{dx}=-1(\sin (x){)}^{-2}\cdot \cos (x)$$ $$=-\frac{1}{(\sin (x){)}^2}\cdot \cos (x)$$ $$=-\frac{1}{\sin (x)}\cdot \frac{\cos (x)}{\sin (x)}$$ $$=-\csc (x)\cot (x)$$

Parametric Functions

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

For a parametric curve $(x(t),y(t))$.

Example

Differentiate the parametric curve defined by the functions $x(t)=\sin (t),y(t)=-10t^2+5t$.

$$\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}=\frac{-20t+5}{\cos (t)}$$ $$=\frac{5-20t}{\cos (t)}$$