What is the derivative of arctan(x)^(1/2)?

1 Answer
Nov 3, 2016

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

Explanation:

I usually make no attempt to remember the derivatives of inverse trig function and would have to look up the result.

Therefore, Inverse trig functions are more easily differentiated by rewriting in terms of the non-inverse trig function and implicit differentiating.

Let $y = \arctan {\left(x\right)}^{\frac{1}{2}}$

$\therefore y = \sqrt{\arctan x}$
$\therefore {y}^{2} = \arctan x$
$\therefore \tan \left({y}^{2}\right) = x$

If we (implicity) differentiate we get:
$\left({\sec}^{2} \left({y}^{2}\right)\right) \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$2 y {\sec}^{2} \left({y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Using the trig identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$ we can see
${\sec}^{2} \left({y}^{2}\right) = {\tan}^{2} \left({y}^{2}\right) + 1$
$\therefore {\sec}^{2} \left({y}^{2}\right) = {x}^{2} + 1$

And so we have:
$2 y {\sec}^{2} \left({y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1 \implies 2 y \left({x}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\therefore 2 \sqrt{\arctan x} \left({x}^{2} + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left({x}^{2} + 1\right) \sqrt{\arctan x}}$