Question

What is the derivative of `f(x) = arcsin(2x^3 - 1)`?

Answer 1

`(df)/(dx)=(3x^2)/(xsqrt(x(1-x^3))`

Explanation:

We can use here chain formula. As differential of `arcsinx` is `1/sqrt(1-x^2)`

derivative of `f(x)=arcsin(2x^3-1)` is

`(df)/(dx)=1/sqrt(1-(2x^3-1)^2)*d/(dx)(2x^3-1)`

= `1/sqrt(1-4x^6+4x^3-1)*6x^2`

= `(3x^2)/(xsqrt(x(1-x^3))`

Answer 2

`(3x^2)/sqrt(x^3(1-x^3)`

Explanation:

We use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=2x^3-1,:.(du)/dx=6x^2`.

Then `y=arcsinu,=>dy/(du)=1/sqrt(1-u^2)`.

Combining together, we get,

`dy/dx=1/sqrt(1-u^2)*6x^2`

`=(6x^2)/sqrt(1-u^2)`

Final step is to substitute back `u=2x^3-1`, and we get,

`=(6x^2)/sqrt(1-(2x^3-1)^2)`

`=(3x^2)/sqrt(x^3(1-x^3)`