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

Jan 3, 2017

$f ' \left(x\right) = - {\csc}^{2} x$

#### Explanation:

Use the quotient rule.

Let $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, then $g \left(x\right) = \cos x$ and $h \left(x\right) = \sin x$. Then $g ' \left(x\right) = - \sin x$ and $h ' \left(x\right) = \cos x$.

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

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

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

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

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

$f ' \left(x\right) = - {\csc}^{2} x$

Hopefully this helps!