Question

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

Answer 1

So, y = `sin^-1(3x^5+1)^3`
`(dy)/(dx)`= `(15x^4)/sqrt(1-(3x^5+1)^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)` .
`dy/(dx)=dy/(du)*(du)/(dx)` From this you can see the du will cancel and `dy/(dx)` is obtained.
So, y = `sin^-1(3x^5+1)^3`
use u = `3x^5 +1`
`(du)/(dx)` = `15x^4`

Substitute u in:
y= `sin^-1 (u)^3`
`(dy)/(du)`= `1/sqrt(1-(u^3)^2`
= `1/sqrt(1-u^6)`
Now to obtain `(dy)/(dx)` we need to multiply `(dy)/(du)` by `(du)/(dx)`

`(dy)/(dx)` = `(15x^4)/sqrt(1-u^6)`
Substitute `3x^5 +1` back in to get rid of u
`(dy)/(dx)`= `(15x^4)/sqrt(1-(3x^5+1)^6)`