Category: Calculus

The Chain Rule is a handy tool for dealing with composite functions, making it easier to differentiate more complex functions. For instance, some functions can be differentiated straightforwardly without the Chain Rule, while others are best tackled using it. Without the Chain RuleWith the Chain Rule$latex y=x^3-2x^2+x+5$$latex y=\sqrt[3]{2x^2+x-1}$$latex y=\sin x+4x$$latex y=\sin\left(\cos x\right)$$latex y=\sec x+\tan x$$latex y=\left(\sin\left(2x+1\right)\right)^3$ If $latex y=f\left(u\right)$ is a differentiable function of $latex u$ and $latex u=g\left(x\right)$ is a differentiable function of $latex x$, then $latex y=f\left(g\left(x\right)\right)$ is a differentiable function of $latex x$ and $latex \displaystyle\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ or $latex \displaystyle\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f'\left(g\left(x\right)\right)\cdot g'\left(x\right)$ The Chain Rule Example 1 Find $latex \displaystyle F'\left(x\right)$ if $latex F\left(x\right)=\sqrt{x^2+1}$. Solution Let $latex u=x^2+1$ and $latex y=\sqrt{u}$: $latex F'\left(x\right)=\displaystyle\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2\sqrt{u}}\cdot 2x= \frac{1}{2\sqrt{x^2+1}}\cdot 2x=\frac{x}{\sqrt{x^2+1}}$ Example 2 Differentiate $latex y=\sin\left(x^2\right)$ Solution Let $latex u=x^2$ and $latex y=\sin u$: $latex \displaystyle\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\cos u\cdot 2x= \cos\left(x^2\right)\cdot 2x=2x\cos\left(x^2\right)$ Example 3 Differentiate $latex y=\sin^2x$ Solution Let $latex u=\sin x$ and $latex y=u^2$: $latex \begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ \displaystyle &=& 2u\cdot \cos x \\ \displaystyle &=&2\sin x\cos x \\ &=&\sin 2x \end{array}$ Example 4 Differentiate $latex y=\left(2x^5+2\right)^{50}$ Solution Let $latex u=2x^5+2$ and $latex y=u^{50}$: $latex \begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& \displaystyle 50u^{49}\cdot 2\cdot 5x^4 \\ &=& \displaystyle 50\left(2x^5+2\right)^{49}\cdot 10x^4 \\ &=& \displaystyle 500x^4\left(2x^5+2\right)^{49}\end{array}$ Example 5 Differentiate $latex y=\left(2x^5+2\right)^{50}$ Solution Let $latex u=2x^5+2$ and $latex y=u^{50}$: $latex \begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& \displaystyle 50u^{49}\cdot 2\cdot 5x^4 \\ &=& \displaystyle 50\left(2x^5+2\right)^{49}\cdot 10x^4 \\ &=& \displaystyle 500x^4\left(2x^5+2\right)^{49}\end{array}$ Example 6 Find the derivative of the function $latex f\left(x\right)=\displaystyle\left(\frac{x-1}{3x+2}\right)^8$ Solution Let $latex u=\displaystyle\frac{x-1}{3x+2}$ and $latex y=u^{8}$: $latex \begin{array}{rcl} \displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& 8u^7\cdot \displaystyle\frac{\left(x-1\right)'\cdot\left(3x+2\right)-\left(x-1\right)\cdot\left(3x+2\right)'}{\left(3x+2\right)^2} \\ &=& 8u^7\cdot\displaystyle\frac{3x+2-3\left(x-1\right)}{\left(3x+2\right)^2} \\ &=& 8u^7\cdot\displaystyle\frac{5}{\left(3x+2\right)^2} \\ &=& 40\displaystyle\cdot \left(\frac{x-1}{3x+2}\right)^7\cdot \frac{1}{\left(3x+2\right)^2} \\ &=& \displaystyle\frac{40\left(x-1\right)^7}{\left(3x+2\right)^9} \end{array}$ Example 7 Find the derivative of the function $latex f\left(x\right)=\displaystyle e^{\sin x}$. Solution Let $latex u=\sin x$ and $latex y=e^u$: $latex \begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& e^u\cdot \cos x \\ &=& e^{\sin x}\cdot \cos x \end{array}$ Example 8 Find the derivative of the function $latex f\left(x\right)=\displaystyle \sin\left(\cos\left(\tan x\right)\right)$. Solution $latex \begin{array}{rcl}\displaystyle\frac{dy}{dx}&=& \cos\left(\cos\left(\tan x\right)\right)\cdot \displaystyle\frac{d}{dx}\left(\tan x\right) \\ &=& \cos\left(\cos\left(\tan x\right)\right)\left(-\sin\left(\tan x\right)\right)\cdot \displaystyle\frac{d}{dx}\left(\tan x\right) \\ &=& -\cos\left(\cos\left(\tan x\right)\right)\sin\left(\tan x\right)\sec^2x \end{array}$ Example 9 Differentiate $latex y=\displaystyle e^{\sec 3\theta}$. Solution $latex \begin{array}{rcl}\displaystyle\frac{dy}{d\theta} &=& e^{\sec 3\theta}\cdot \displaystyle\frac{d}{d\theta}\left(\sec 3\theta\right) \\ &=& e^{\sec 3\theta}\cdot \sec 3\theta\tan 3\theta\cdot\displaystyle\frac{d}{d\theta}\left(3\theta\right) \\ &=& 3e^{\sec 3\theta}\sec 3\theta\tan 3\theta \end{array}$ More Practice Questions

Category: Calculus

AP Calculus – The Chain Rule and How to Use It

AP Calculus – Integration by Parts Made Easy