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

1 Answer

Konstantinos Michailidis

Sep 13, 2015

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#

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