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

Mar 31, 2018

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

#### Explanation:

$f \left(x\right) = 3 \arcsin \left({x}^{4}\right)$

Let
$y = f \left(x\right)$

$3 \arcsin \left({x}^{4}\right) = 3 {\sin}^{-} 1 \left({x}^{4}\right)$

$y = 3 {\sin}^{-} 1 \left({x}^{4}\right)$

Let $u = {x}^{4}$

$y = 3 {\sin}^{-} 1 u$

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

$y = 3 {\sin}^{-} 1 u$

$\frac{\mathrm{dy}}{\mathrm{du}} = 3 \times \frac{1}{\sqrt{1 - {u}^{2}}}$

$u = {x}^{4}$

${u}^{2} = {\left({x}^{4}\right)}^{2}$
${u}^{2} = {x}^{8}$

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

$u = {x}^{4}$

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

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

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

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