Question

If `f(x)= x^2*tan^-1 x` then how do you find f'(1)?

Answer 1

It is `f'(1)=1/2-pi/2`

Explanation:

Since `f(x)=x^2*tan^-1(x)` the derivative is

`f'(x)=2x*tan^-1(x)+x^2*1/(1+x^2)`

hence `f'(1)=2*tan^(-1)(-1)+1/2`

But `tan^(-1)(-1)=tan^(-1)(tan(-pi/4))=-pi/4`

So `f'(1)=1/2-pi/2`