# How do you differentiate given f (x) = 3 arcsin (x^4)?

Nov 13, 2016

$f ' \left(x\right) = \frac{12 {x}^{3}}{\sqrt{1 - {x}^{8}}}$

#### Explanation:

Let # y = f(x) = 3arcsin(x^4) , Then

$\frac{y}{3} = \arcsin \left({x}^{4}\right)$
$\sin \left(\frac{y}{3}\right) = {x}^{4}$ ..... [1]

We can now differentiate implicitly to get:

$\cos \left(\frac{y}{3}\right) \left(\frac{1}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3}$
$\therefore \cos \left(\frac{y}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{3}$ ..... [2]

Using the fundamental trig identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we can write:

${\sin}^{2} \left(\frac{y}{3}\right) + {\cos}^{2} \left(\frac{y}{3}\right) = 1$
${\left({x}^{4}\right)}^{2} + {\cos}^{2} \left(\frac{y}{3}\right) = 1$ (from [1])
${\cos}^{2} \left(\frac{y}{3}\right) = 1 - {x}^{8}$
$\cos \left(\frac{y}{3}\right) = \sqrt{1 - {x}^{8}}$

Substituting into [2] we get:

$\sqrt{1 - {x}^{8}} \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{12 {x}^{3}}{\sqrt{1 - {x}^{8}}}$