Question

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

Answer 1

`f(x)=1/(sqrt(sin^2(2-x^2)))=(sin^2(2-x^2))^(-1/2)=((sin(2-x^2))^2)^(-1/2)=(sin(2-x^2))^-1`

In order to differentiate this, we need to use the chain rule twice on `sin(f(x))^n`.

`d/dx[sin(f(x))]^n=n[sin(f(x))]^(n-1)cos(f(x)) f'(x)`

`f'(x)=-1(sin^2(2-x^2))^(-2)cos(2-x^2)-2x=`

`=(2xcos(2-x^2))/(sin^2(2-x^2))^2`