Question

How do you differentiate `f(x) = 1/sqrt(arctan(2x^3)` using the chain rule?

Answer 1

`f'(g(x))g'(x) = 1/2(arctan(2x^3))*(6x^2)/(1+(2x^3)^2)`

Explanation:

first, take the derivative of the denominator using chain rule:
`f=sqrt(x)`
`f'=1/2x^(-1/2)`
`g=arctan(2x^3)`
`g'=(6x^2)/(1+(2x^3)^2)`

`f'(g(x))g'(x) = 1/2(arctan(2x^3))^(-1/2)*(6x^2)/(1+(2x^3)^2)`

that's it! if you want to simplify it further you can but my calc teacher only requires this.

note: recognize that the derivative of arctan has the chain rule built into it already if that makes sense, as `d/dx[arctan(u)]=(u')/(1+u^2)`