How do you differentiate $y=(sin^-1(5x^2))^3$ ?

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Answer 1 · 2016-11-06T19:02:33

$\fracdydx=\frac{30x(arcsin(5x^2))^2}{\sqrt{1-5x^2}}$

The derivative of $\arcsin(u)$ (or $sin^-1(u)$ ) is $\frac{1}{\sqrt{1-u^2}}\frac{du}dx$

So by using the chain rule,
$\fracdydx=3(arcsin(5x^2))^2\cdot\frac1{\sqrt{1-5x^2}}\cdot10x$

Which can be simplified to:
$\fracdydx=\frac{30x(arcsin(5x^2))^2}{\sqrt{1-5x^2}}$

How do you differentiate y=(sin^-1(5x^2))^3y=(sin^-1(5x^2))^3?