Chain Rule (Cambridge (CIE) O Level Additional Maths): Revision Note

Chain rule

What is the chain rule?

•  The chain rule states if$\mathit{y}$ is a function of$\mathit{u}$ and$\mathit{u}$ is a function of$\mathit{x}$ then

$y=\mathrm{f}\left(u\right(x\left)\right)$

$\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{x}}\mathbf{=}\frac{\mathbf{d}\mathit{y}}{\mathbf{d}\mathit{u}}\mathbf{\times }\frac{\mathbf{d}\mathit{u}}{\mathbf{d}\mathit{x}}$

• In function notation this could be written

$y=\mathrm{f}\left(\mathrm{g}\right(x\left)\right)$

$\frac{\mathrm{d}y}{\mathrm{d}x}=\mathrm{f}\text{'}\left(\mathrm{g}\right(x\left)\right)\mathrm{g}\text{'}\left(x\right)$

How do I know when to use the chain rule?

•  The chain rule is used when we are trying to differentiate composite functions

  • “function of a function”

  • these can be identified as the variable (usually$x$) does not ‘appear alone’

    • $\sin x$ – not a composite function, $x$ ‘appears alone’

    • $\sin (3x+2)$ is a composite function; $x$ is tripled and has 2 added to it before the sine function is applied

How do I use the chain rule?

 STEP 1

 Identify the two functions

 Rewrite $y$ as a function of$u$; $y=\mathrm{f}\left(u\right)$

 Write $u$ as a function of$x$; $u=\mathrm{g}\left(x\right)$

 STEP 2

Differentiate $y$ with respect to $u$ to get$\frac{\mathrm{d}y}{\mathrm{d}u}$ Differentiate $u$ with respect to $x$ to get$\frac{\mathrm{d}u}{\mathrm{d}x}$

 STEP 3

Obtain $\frac{\mathrm{d}y}{\mathrm{d}x}$ by applying the formula$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times \frac{\mathrm{d}u}{\mathrm{d}x}$ and substitute$u$ back in for$\mathrm{g}\left(x\right)$

• In trickier problems chain rule may have to be applied more than once

How do I differentiate (ax + b)ⁿ?

• For n = 2 you will most likely expand the brackets and differentiate each term separately

• If n > 2 this becomes time-consuming and if n is not a positive integer we need a different method completely

• The chain rule allows us to use substitution to differentiate any function in the form y = (ax + b)ⁿ

  • Let u = ax + b, then y = uⁿ

  • Differentiate both parts separately

    • $\frac{du}{dx}=a$ and $\frac{dy}{du}=n{u}^{n-1}$

  • Put both parts into the chain rule

    • $\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}x}=a \times n{u}^{n-1} = an{u}^{n-1}$

  • Substitute u = ax + b back into your answer

    • $\frac{\mathrm{d}y}{\mathrm{d}x}=an{(ax+ b)}^{n-1}$

How do I differentiate √(ax+b)?

• The chain rule allows us to use substitution to differentiate any function in the form $y=\sqrt{ax+b}$

• Rewrite $\sqrt{ax+b}={(ax+b)}^{\frac{1}{2}}$ 

  • Let u = ax + b, then y = u^½

  • Differentiate both parts separately

    • $\frac{du}{dx}=a$ and $\frac{dy}{du}=\frac{1}{2}{u}^{-\frac{1}{2}}$

  • Put both parts into the chain rule

    • $\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}x}=a \times \frac{1}{2}{u}^{-\frac{1}{2}} = \frac{a}{2}{u}^{-\frac{1}{2}}$

  • Substitute u = ax + b back into your answer

    • $\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{a}{2}{\left(ax+b\right)}^{-\frac{1}{2}} =\frac{a}{2\sqrt{ax+b}}$

• This method can be used for any fractional power of any linear or non-linear expression

  • Provided you know how to differentiate the non-linear expression

Are there any standard results for using chain rule?

• The following general results are particularly useful

  • If$y={\left(\mathrm{f}\left(x\right)\right)}^n$ then$\frac{\mathrm{d}y}{\mathrm{d}x}=n{\mathrm{f}}^{\text{'}}\left(x\right)\mathrm{f}{\left(x\right)}^{n-1}$

• If$y={\mathrm{e}}^{\mathrm{f}\left(x\right)}$ then$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{f}}^{\mathrm{\text{'}}}\left(x\right)e^{\mathrm{f}\left(x\right)}$

  • If$y=\ln \left(\mathrm{f}\left(x\right)\right)$ then$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{{\mathrm{f}}^{\mathrm{\text{'}}}\left(x\right)}{\mathrm{f}\left(x\right)}$

  • If$y=\sin \left(\mathrm{f}\left(x\right)\right)$ then$\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{f}}^{\mathrm{\text{'}}}\left(x\right)\cos \left(\mathrm{f}\left(x\right)\right)$

  • If$y=\cos \left(\mathrm{f}\left(x\right)\right)$ then$\frac{\mathrm{d}y}{\mathrm{d}x}=-{\mathrm{f}}^{\text{'}}\left(x\right)\sin \left(\mathrm{f}\left(x\right)\right)$

  • If $y=\tan \left(\mathrm{f}\left(x\right)\right)$ then $\frac{\mathrm{d}y}{\mathrm{d}x}={\mathrm{f}}^{\text{'}}\left(x\right){\sec }^2\left(\mathrm{f}\left(x\right)\right)$