Chain Rule (Cambridge (CIE) AS Maths: Pure 1): Revision Note

Chain Rule

What is the chain rule? 

• If y is a function of u, and u is a function of x, then the chain rule tells us that

$\frac{\mathrm{d}y}{\mathrm{d}x}= \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}x}$

• The chain rule allows us to differentiate more complicated expressions and composite functions

• You will often see and use the chain rule with different variables

  • This is particularly useful for connected rates of change

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

How do I differentiate (f(x))ⁿ?

• This method can be used for any linear or non – linear expression

  • Let u = f(x) and follow the method above

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

• With practice you will be able to carry out this method without the need for u

  • This is essential for learning the reverse chain rule later in the course