Chain Rule

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

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

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

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
  - $\frac{d u}{d x} = a$ and $\frac{d y}{d u} = n u^{n - 1}$
- Put both parts into the chain rule
  - $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = a \times n u^{n - 1} = a n u^{n - 1}$
- Substitute u = ax + b back into your answer
  - $\frac{d y}{d x} = a n {(a x + b)}^{n - 1}$

The chain rule allows us to use substitution to differentiate any function in the form $y = \sqrt{a x + b}$
Rewrite $\sqrt{a x + b} = {(a x + b)}^{\frac{1}{2}}$
- Let u = ax + b, then y = u^½
- Differentiate both parts separately
  - $\frac{d u}{d x} = a$ and $\frac{d y}{d u} = \frac{1}{2} u^{- \frac{1}{2}}$
- Put both parts into the chain rule
  - $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = a \times \frac{1}{2} u^{- \frac{1}{2}} = \frac{a}{2} u^{- \frac{1}{2}}$
- Substitute u = ax + b back into your answer
  - $\frac{d y}{d x} = \frac{a}{2} {(a x + b)}^{- \frac{1}{2}} = \frac{a}{2 \sqrt{a x + b}}$
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

This method can be used for any linear or non – linear expression
- Let u = f(x) and follow the method above
- In general if $y = (f {(x))}^{n}$ then $\frac{d y}{d x} = n f^{'} (x) {(f (x))}^{n - 1}$
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

5-1-4-chain-rule-we-solution-part-1

5-1-4-chain-rule-we-solution-part-2

If using u as a substitution don't forget to substitute x back into your final answer.

Chain Rule (CIE AS Maths: Pure 1)