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

Dec 30, 2015

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

#### Explanation:

For a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$,

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

Here,

$g \left(x\right) = \cos x$
$h \left(x\right) = \sin x - \cos x$

and

$g ' \left(x\right) = - \sin x$
$h ' \left(x\right) = \cos x + \sin x$

so,

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

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

$\implies - \frac{1}{\sin x - \cos x} ^ 2$