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

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Answer 1 · 2015-09-13T18:22:47

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

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 $f'(1)=1/2-pi/2$

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