# How do you find the derivative of (sin x + 2x) / (cos x - 2)?

Dec 13, 2015

$f ' \left(x\right) = \frac{2 x \sin x - 3}{\cos x - 2} ^ 2$

#### Explanation:

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

This can be differentiate using quotient rule

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

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

$\implies f ' \left(x\right) = \frac{{\cos}^{2} x - 4 + {\sin}^{2} x + 2 x \sin x}{\cos x - 2} ^ 2$

$\implies f ' \left(x\right) = \frac{\textcolor{red}{{\cos}^{2} x + {\sin}^{2} x} - 4 + 2 x \sin x}{\cos x - 2} ^ 2$

$\implies f ' \left(x\right) = \frac{\textcolor{red}{1} - 4 + 2 x \sin x}{\cos x - 2} ^ 2$

$f ' \left(x\right) = \frac{2 x \sin x - 3}{\cos x - 2} ^ 2$

*${\sin}^{2} x + {\cos}^{2} x = 1$