# Question #25a9e

Mar 11, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{5}}{\sqrt{x} \left(2 + 10 x\right)}$

Here's how to do it.

$y = \arctan \left(\sqrt{5 x}\right)$

Because the equation is in the form of $f \left(g \left(x\right)\right)$, we should use the chain rule.

$f \left(x\right) = \arctan \left(g \left(x\right)\right)$
$g \left(x\right) = \sqrt{5 x}$

The chain rule:
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

Our case:
$f ' \left(g \left(x\right)\right) = \frac{1}{1 + {\left(\sqrt{5 x}\right)}^{2}} = \frac{1}{1 + 5 x}$
$g ' \left(x\right) = \left(\frac{\sqrt{5}}{2}\right) {\left(x\right)}^{- \frac{1}{2}}$

Plugging these values back into the chain rule gives us the following answer:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{5}}{\sqrt{x} \left(2 + 10 x\right)}$