# What is the derivative of this function y=sin^-1(3x^5+1)^3?

Jan 1, 2018

So, y = ${\sin}^{-} 1 {\left(3 {x}^{5} + 1\right)}^{3}$
$\frac{\mathrm{dy}}{\mathrm{dx}}$= $\frac{15 {x}^{4}}{\sqrt{1 - {\left(3 {x}^{5} + 1\right)}^{6}}}$

#### Explanation:

You have to use chain rule and a substitution(to make is simpler). You don't have to use substitution but I did. You should know that the derivative of arcsin x equals 1/(sqrt(1-x^2) .
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$ From this you can see the du will cancel and $\frac{\mathrm{dy}}{\mathrm{dx}}$ is obtained.
So, y = ${\sin}^{-} 1 {\left(3 {x}^{5} + 1\right)}^{3}$
use u = $3 {x}^{5} + 1$
$\frac{\mathrm{du}}{\mathrm{dx}}$ = $15 {x}^{4}$

Substitute u in:
y= ${\sin}^{-} 1 {\left(u\right)}^{3}$
$\frac{\mathrm{dy}}{\mathrm{du}}$= 1/sqrt(1-(u^3)^2
= $\frac{1}{\sqrt{1 - {u}^{6}}}$
Now to obtain $\frac{\mathrm{dy}}{\mathrm{dx}}$ we need to multiply $\frac{\mathrm{dy}}{\mathrm{du}}$ by $\frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{15 {x}^{4}}{\sqrt{1 - {u}^{6}}}$
Substitute $3 {x}^{5} + 1$ back in to get rid of u
$\frac{\mathrm{dy}}{\mathrm{dx}}$= $\frac{15 {x}^{4}}{\sqrt{1 - {\left(3 {x}^{5} + 1\right)}^{6}}}$