Question

If `f(x) =cot(-x/3)` and `g(x) = sqrt(x^2+1`, what is `f'(g(x))`?

Answer 1

`f'(g(x))=1/3csc^2(-sqrt(x^2+1)/3)`

Explanation:

As `f(x)=cot(-x/3)`

`f'(x)=d/(d(-x/3))(cot(-x/3))xxd/(dx)(-x/3)`

= `-csc^2(-x/3)xx(-1/3)=1/3csc^2(-x/3)`

As `g(x)=sqrt(x^2+1)`

`f'(g(x))=1/3csc^2(-sqrt(x^2+1)/3)`