# How do you differentiate arcsin(csc(x^3)) ) using the chain rule?

Feb 17, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} \csc \left({x}^{3}\right) \cot \left({x}^{3}\right)}{\sqrt{1 - {\left(\csc \left({x}^{3}\right)\right)}^{2}}}$

#### Explanation:

Given:
$y = \arcsin \left(\csc \left({x}^{3}\right)\right)$
$\arcsin \left(\csc \left({x}^{3}\right)\right) = {\sin}^{-} 1 \left(\csc \left({x}^{3}\right)\right)$

Let $t = \csc \left({x}^{3}\right)$
$y = {\sin}^{-} 1 t$
(dy)/dt=1/sqrt(1-t^2)=1/sqrt(1-(csc(x^3))^2

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

$t = \csc \left({x}^{3}\right)$

Let $u = {x}^{3}$

$t = \csc u$

$\frac{\mathrm{dt}}{\mathrm{du}} = \csc u \cot u = \csc \left({x}^{3}\right) \cot \left({x}^{3}\right)$

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

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \left(\csc \left({x}^{3}\right) \cot \left({x}^{3}\right)\right) \left(3 {x}^{2}\right) = 3 {x}^{2} \csc \left({x}^{3}\right) \cot \left({x}^{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sqrt{1 - {\left(\csc \left({x}^{3}\right)\right)}^{2}}} \left(3 {x}^{2} \csc \left({x}^{3}\right) \cot \left({x}^{3}\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} \csc \left({x}^{3}\right) \cot \left({x}^{3}\right)}{\sqrt{1 - {\left(\csc \left({x}^{3}\right)\right)}^{2}}}$