# How do you differentiate (1)/ (x^2cosx) using the quotient rule?

Jan 23, 2017

The answer is $= \frac{x \sin x - 2 \cos x}{x \cos x}$

#### Explanation:

The quotient rule is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

and the product rule is

$\left(u v\right) ' = u ' v + u v '$

Here, we have

$u = 1$, $\implies$, $u ' = 0$

$v = {x}^{2} \cos x$, $\implies$, $v ' = 2 x \cos x - {x}^{2} \sin x$

Therefore,

$\left(\frac{1}{{x}^{2} \cos x}\right) ' = \frac{0 \cdot \left({x}^{2} \cos x\right) - 1 \cdot \left(2 x \cos x - {x}^{2} \sin x\right)}{{x}^{2} \cos x}$

$= \frac{{x}^{2} \sin x - 2 x \cos x}{{x}^{2} \cos x}$

$= \frac{x \sin x - 2 \cos x}{x \cos x}$