# How do you find the derivative of f(x)= (x+sinx)/(cosx) ?

Nov 3, 2016

$f ' \left(x\right) = \frac{\cos x + x \sin x + 1}{\cos} ^ 2 x$

#### Explanation:

using the quotient rule.

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

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

$u = x + \sin x \implies u ' = 1 + \cos x$

$v = \cos x \implies v ' = - \sin x$

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

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

$f ' \left(x\right) = \frac{\cos x + x \sin x + 1}{\cos} ^ 2 x$