# What is the derivative of arcsin^3(5x)?

Sep 9, 2015

The final answer is $f ' \left(x\right) = \frac{15 \arcsin {\left(5 x\right)}^{2}}{\sqrt{1 - 25 {x}^{2}}}$

#### Explanation:

Let $f \left(x\right) = y = {\arcsin}^{3} \left(5 x\right) = {\left(\arcsin \left(5 x\right)\right)}^{3}$.
We need to use the chain rule here.

Recall that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dy}}{\mathrm{du}}$.
First let $u = \arcsin \left(5 x\right)$ and $y = {u}^{3}$.

To solve for $\frac{\mathrm{dy}}{\mathrm{du}}$ where $y = {u}^{3}$, we can use the power rule:
$\frac{\mathrm{dy}}{\mathrm{du}} = 3 {u}^{2}$

To solve for $\frac{\mathrm{du}}{\mathrm{dx}}$ where $u = \arcsin \left(5 x\right)$, we can use the inverse sin derivative rule, combined with the chain rule again.
Let $u ' = 5 x$, so $u = \arcsin \left(5 x\right) = \arcsin \left(u '\right)$.

By the inverse sin derivative rule:
$\frac{\mathrm{du}}{\mathrm{du} '} = \frac{1}{\sqrt{1 - u {'}^{2}}}$

By the power rule:
$\frac{\mathrm{du} '}{\mathrm{dx}} = 5$

Then multiplying together and substituting in for $u '$, we have:
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{du} '} \cdot \frac{\mathrm{du} '}{\mathrm{dx}}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - {\left(5 x\right)}^{2}}}$
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - 25 {x}^{2}}}$

Finally, we have:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} \cdot \frac{\mathrm{dy}}{\mathrm{du}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - 25 {x}^{2}}} \cdot 3 {u}^{2}$

Plugging back in $u$ and simplifying:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5}{\sqrt{1 - 25 {x}^{2}}} \cdot 3 \arcsin {\left(5 x\right)}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{15 \arcsin {\left(5 x\right)}^{2}}{\sqrt{1 - 25 {x}^{2}}}$