Question

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

Answer 1

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

Explanation:

If `f(x)=ln(g(x))`
Then `f'(x)=(g'(x))/(g(x))`

`g(x)=(1-x^2)^(-1/2)=(h(x))^n`
`g'(x)=n* h'(x)* h(x)^(n-1)`
`h'(x)=-2x`
`g'(x)=-1/2*-2x*(1-x^2)^(-3/2)=x(1-x^2)^(-3/2)`

`f'(x)=(x(1-x^2)^(-3/2))/((1-x^2)^(-1/2))=(x(1-x^2)^(1/2))/((1-x^2)^(3/2))=x/((1-x^2)^(2/2))=x/((1-x^2))=x(1-x^2)^(-1)`