# How do you differentiate (sinx+sinxcosx)/x?

Nov 22, 2017

$\frac{x \left(\cos x + {\cos}^{2} x - {\sin}^{2} x\right) - \sin x - \sin x \cos x}{x} ^ 2$

#### Explanation:

we need the quotient rule and the product rule

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

$\frac{d}{\mathrm{dx}} \left(u v\right) = v u ' + u v '$

$\frac{d}{\mathrm{dx}} \left(\frac{\left(\sin x + \sin x \cos x\right)}{x}\right)$

$= \frac{x \frac{d}{\mathrm{dx}} \left(\sin x + \sin x \cos x\right) - \left(\sin x + \sin x \cos x\right) \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2$

we will use the product rule on $\sin x \cos x$

$= \frac{x \left(\cos x + \cos x \cos x - \sin x \sin x\right) - \left(\sin x + \sin x \cos x\right)}{x} ^ 2$

$= \frac{x \left(\cos x + {\cos}^{2} x - {\sin}^{2} x\right) - \sin x - \sin x \cos x}{x} ^ 2$