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

Jun 11, 2016

$0$

#### Explanation:

Alternatively, we can simplify the original function.

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

Take the tangent of both sides.

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

Use the tangent addition formula: $\tan \left(\alpha + \beta\right) = \frac{\tan \left(\alpha\right) + \tan \left(\beta\right)}{1 - \tan \left(\alpha\right) \tan \left(\beta\right)}$

Here, for $\tan \left(\arctan \left(x\right) + \arctan \left(\frac{1}{x}\right)\right)$, we see that $\alpha = \arctan \left(x\right)$ and $\beta = \arctan \left(\frac{1}{x}\right)$, so we obtain:

tan(y)=(tan(arctan(x))+tan(arctan(1/x)))/(1-tan(arctan(x))tan(arctan(1/x))

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

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

$\tan \left(y\right) = \frac{{x}^{2} + 1}{0}$

This is an undefined value: however, we know that the tangent of $y$ is undefined, and the tangent function is undefined at $\frac{\pi}{2}$ and $- \frac{\pi}{2}$.

So, we know that

$y = \frac{\pi}{2} \text{ }$ or $\text{ } y = - \frac{\pi}{2}$

Thus,

$\arctan \left(x\right) + \arctan \left(\frac{1}{x}\right) = \pm \frac{\pi}{2}$

And the derivative of $\frac{\pi}{2}$ or $- \frac{\pi}{2}$, which are both constant, is just $0$.