# How do you find the derivative of arctan(1/x)?

Sep 21, 2016

$- \frac{1}{1 + {x}^{2}}$

#### Explanation:

Know ahead of time that $\frac{d}{\mathrm{dx}} {\cot}^{-} 1 \left(x\right) = - \frac{1}{1 + {x}^{2}}$:

We see that:

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

$\tan \left(y\right) = \frac{1}{x}$

$\frac{1}{\tan} \left(y\right) = \frac{1}{\frac{1}{x}}$

$\cot \left(y\right) = x$

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

Thus:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$