# How do you differentiate f(x)=cosx/sinx?

Dec 1, 2016

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

#### Explanation:

differentiate f(x) using the $\textcolor{b l u e}{\text{quotient rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\text{If " f(x)=(g(x))/(h(x)) " then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

$h \left(x\right) = \sin x \Rightarrow h ' \left(x\right) = \cos x$

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

$= \frac{- {\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x = - \frac{{\sin}^{2} x + {\cos}^{2} x}{\sin} ^ 2 x$

$= - \frac{1}{\sin} ^ 2 x = - {\csc}^{2} x$

$\textcolor{b l u e}{\text{NOTE}}$

$\cos \frac{x}{\sin} x = \cot x$

$\text{ and } \frac{d}{\mathrm{dx}} \left(\cot x\right) = - {\csc}^{2} x$ is a standard result, well worth remembering.