# How do you find the derivative f(x)=arctan(x/alpha)?

Jan 6, 2017

$f ' \left(x\right) = \frac{\alpha}{{x}^{2} + {\alpha}^{2}}$

#### Explanation:

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.

$y = \arctan \left(\frac{x}{\alpha}\right) \iff \tan y = \frac{x}{\alpha}$

Differentiate Implicitly:

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\alpha}$ ..... [1]

Using the $\tan \text{/} \sec$ identity;

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

Substituting into [1]

$\left({\left(\frac{x}{\alpha}\right)}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\alpha}$
$\therefore \left(\frac{{x}^{2} + {\alpha}^{2}}{\alpha} ^ 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\alpha}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\alpha} \cdot {\alpha}^{2} / \left({x}^{2} + {\alpha}^{2}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\alpha}{{x}^{2} + {\alpha}^{2}}$