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

Jun 13, 2018

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

Explanation:

We seek:

$\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right)$

Using the standard result:

$\frac{d}{\mathrm{dx}} \tan x = \frac{1}{1 + {x}^{2}}$

In conjunction with the power rule and the chain rule we get:

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

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

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

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

Observation:

The astute reader will notice that:

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

From which we conclude that:

$\arctan \left(\frac{1}{x}\right) = - \arctan x + C \implies \arctan \left(\frac{1}{x}\right) + \arctan x = C$

Although this result may look like an error, it is in fact correct, and a standard trigonometric result: