# How do you differentiate f(x)=(x^2sinx)/(x^2cosx-sin^2x) using the quotient rule?

Jan 18, 2018

Here is the quotient rule:
If $f \left(x\right) = g \frac{x}{h \left(x\right)}$ then f'(x)=(g'(x)*h(x) - h'(x)*g(x))/((h(x))^2.
Below is the way you would go about solving a question like this.

#### Explanation:

Here is the quotient rule:
If $f \left(x\right) = g \frac{x}{h \left(x\right)}$ then f'(x)=(g'(x)*h(x) - h'(x)*g(x))/((h(x))^2.

So in this case:
$g \left(x\right) = {x}^{2} \sin x$
$h \left(x\right) = {x}^{2} \cos x - {\sin}^{2} x$

We differentiate both these parts with respect to $x$ partially using the product rule.
$g ' \left(x\right) = 2 x \sin x + {x}^{2} \cos x$
$h ' \left(x\right) = 2 x \cos x - {x}^{2} \sin x - 2 \sin x \cos x$

So $f ' \left(x\right)$ becomes quite complex. Now all you need do is work out the brackets. I won't do this for you, I am quite busy. Sorry.