# How do you differentiate (x-cosx)/ (sinx+x) using the quotient rule?

Jul 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + x \left(\sin x - \cos x\right) + \sin x + \cos x}{\sin x + x} ^ 2$

#### Explanation:

Let $y = \frac{x - \cos x}{\sin x + x}$ Using Quotient Rule for Diffn., we have,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sin x + x\right) \left(x - \cos x\right) ' - \left(x - \cos x\right) \left(\sin x + x\right) '}{\sin x + x} ^ 2$
$= \frac{\left(\sin x + x\right) \left(1 + \sin x\right) - \left(x - \cos x\right) \left(\cos x + 1\right)}{\sin x + x} ^ 2$
$= \frac{{\sin}^{2} x + x \sin x + \sin x + x + \left(\cos x - x\right) \left(\cos x + 1\right)}{\sin x + x} ^ 2$
$= \frac{{\sin}^{2} x + x \sin x + \sin x + x + {\cos}^{2} x - x \cos x + \cos x - x}{\sin x + x} ^ 2$
$= \frac{1 + x \left(\sin x - \cos x\right) + \sin x + \cos x}{\sin x + x} ^ 2$