# How do you find the derivative of (tanx)^-1?

##### 1 Answer
May 29, 2016

I would rewrite the function before differentiating

#### Explanation:

${\left(\tan x\right)}^{-} 1 = \frac{1}{\tan} x = \cot x$

Now use the memorized $\frac{d}{\mathrm{dx}} \left(\cot x\right) = - {\csc}^{2} x$ or rewrite another step

$\cot x = \cos \frac{x}{\sin} x$ and use the quotient rule to get

$\frac{d}{\mathrm{dx}} \left(\cot x\right) = \frac{- {\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x = \frac{- 1}{\sin} ^ 2 x = - {\csc}^{2} x$