## Chain Rule Formula

The chain rule is used to help perform more complicated forms of differentiation. We use the chain rule when there is a function within a function. This may seem a little complicated but after a bit of practice it is easy to identify.

The formula for the chain rule is provided in the formula sheet: $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$

At first glance it may seem a little confusing but the basic idea is that we separate out the function within a function and differentiate them separately.

## Chain Rule Example

For example from the 2014 exam 1, Question 1b. Find f'(1) for: $f(x)=\sqrt{(x^{2}+3)}$

Step 1  We convert the function into a more friendly looking form, we will also replace our notation of f(x) with y so that our process will resemble the formula provided. There is no harm in doing this as long as when you write your final answer you use f ‘(x) instead of $\frac{dy}{dx}$. $y= (x^{2}+3)^{\frac{1}{2}}$

Step 2 This is the hardest part of the process as we need to choose which part of the function to set to u in order to simplify our differentiation. For most functions it will just be our bracket terms, for example: $u= x^{2}+3$

This also gives us: $y= (u)^{\frac{1}{2}}$

Step 3 Find $\frac{du}{dx}$ and $\frac{dy}{du}$

This is now just straightforward power differentiation. $\frac{du}{dx}=\frac{d}{dx}(x^{2}+3) \newline \frac{du}{dx}= 2x$ $\frac{dy}{du}=\frac{d}{du}((u)^{\frac{1}{2}}) \newline \frac{dy}{du}=\frac{1}{2} \times (u)^{-\frac{1}{2}}$

Step 4 multiply together the two terms to find $\frac{dy}{dx}$ $\frac{dy}{dx}=2x \times \frac{1}{2}(u)^{-\frac{1}{2}} \newline =x(u)^{-\frac{1}{2}}$

Step 5 we substitute in $u= x^{2}+3$ $\frac{dy}{dx} =x(x^{2}+3)^{-\frac{1}{2}}= \frac{x}{(x^{2}+3)^{\frac{1}{2}}}= f'(x)$

That completes the differentiation part of the question and we have converted it back to the form of f'(x). Now we just need to find f'(1) by substituting x = 1 into this function $f'(1) =\frac{1}{(1^{2}+3)^{\frac{1}{2}}} \newline = \frac{1}{4^{\frac{1}{2}}} \newline =\frac{1}{2}$