Question

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

Answer 1

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

Explanation:

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}}`