# What is the derivative of  arcsin(x^4)?

Jun 2, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3}}{\sqrt{1 - {x}^{8}}}$

#### Explanation:

Let $u = {x}^{4}$ then $\frac{\mathrm{du}}{\mathrm{dx}} = 4 {x}^{3}$

Let $y = \arcsin \left(u\right)$

Take the arcsin of both sides

$\sin \left(y\right) = u$

Differentiate both sides with respect to $u$

$\frac{\mathrm{dy}}{\mathrm{du}} \cos \left(y\right) = 1$

Divide both sides by $\cos \left(y\right)$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\cos} \left(y\right)$

recall that $S {\in}^{2} \left(y\right) + {\cos}^{2} \left(y\right) = 1$

So $\cos \left(y\right) = \sqrt{1 - {\sin}^{2} \left(y\right)}$
NOTE: Take only the positive root

From above we said that $\sin \left(y\right) = u$ so we can write

$\cos \left(y\right) = \sqrt{1 - {u}^{2}}$

So $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}}$

Recall that $u = {x}^{4}$

So $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {\left({x}^{4}\right)}^{2}}}$

Now using the chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {x}^{8}}} \left(4 {x}^{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 {x}^{3}}{\sqrt{1 - {x}^{8}}}$