# What is the derivative of arctan(1/x)?

Jul 31, 2015

The derivative is: $\frac{- 1}{{x}^{2} + 1}$

#### Explanation:

$\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$

So

$\frac{d}{\mathrm{dx}} \arctan \left(u\right) = \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$

And

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) = \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

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

$= {x}^{2} / \left({x}^{2} + 1\right) \cdot \frac{- 1}{x} ^ 2$

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

Faster Method?

Use the fact that $\arctan \left(\frac{1}{x}\right) = a r c \cot \left(x\right)$

and

$\frac{d}{\mathrm{dx}} a r c \cot \left(x\right) = - \frac{1}{1 + {x}^{2}}$

to go straight to the answer.
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