## Quotient Rule Formula

We use the quotient rule when we have a function being divided by another function. The question will usually be in the form of the fraction.

The formula for the quotient rule is provided in the formula sheet: $\frac{d}{dx}(\frac{u}{v})=\frac{(u\frac{dv}{dx}-v\frac{du}{dx})}{v^{2}}$

For the quotient rule u has to be equal to the numerator and v has to be equal to the denominator. Otherwise the formula supplied in the formula sheet will not work!

The differentiation of these functions can become quite messy so it is always good to set out your calculations labelled clearly by setting out clearly which function is u and which is v, so that any silly mistakes can be picked up upon.

## Quotient Rule Differentiation

2012 Exam 1 Q1b

For $f(x)=\frac{x}{sin(x)}$

Find $f'(\frac{\pi}{2})$

Here we have two functions: $u=x$ $v=sin(x)$

Again we follow our power and trigonometric differentiation rules $\frac{du}{dx}=1$ $\frac{dv}{dx}=cos(x)$

We can now simply substitute these values into the formula provided: $\frac{d}{dx}(\frac{u}{v})=\frac{(1\times sin(x)-cos(x) \times x)}{sin^{2}(x)}\newline =\frac{(sin(x)-xcos(x))}{sin^{2}(x)} = f'(x)$

You’ll find that exam questions don’t test topics separately. This question for example tests you understanding of differentiation and trigonometric functions. This is significant because the whole methods course builds on itself. You can’t forget a topic once you’ve learnt it because you will use it again in a different topic.

So let’s use our trig knowledge to find $f'(\frac{\pi}{2})$ $f'(\frac{\pi}{2})=\frac{(1\times sin(\frac{\pi}{2})-cos(\frac{\pi}{2}) \times \frac{\pi}{2})}{sin^{2}(\frac{\pi}{2})}$

Let’s calculate our sin and cos values separately $sin(\frac{\pi}{2})=1$ $cos(\frac{\pi}{2})=0$

So if we sub back in these values $f'(\frac{\pi}{2}) = \frac{(1-\frac{\pi}{2} \times 0)}{1^{2}} \newline f'(\frac{\pi}{2})=1$