How do you find f'(1) if $f (x) = x^{2} \cdot {tan}^{- 1} x$ $f(x)= x^2*tan^-1 x$ ?

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Answer 1 · 2015-05-18T16:57:23

It would be $\frac{π}{2} + \frac{1}{2}$ $pi/2+1/2$

f' (x)= $2 x {tan}^{- 1} x + \frac{x^{2}}{1 + x^{2}}$ $2x tan^-1 x + x^2 /(1+x^2)$

f' (1)= $2 {tan}^{- 1} 1 + \frac{1}{2}$ $2tan^-1 1+ 1/2$

= $2 \frac{π}{4} + \frac{1}{2}$ $2pi/4 +1/2$

= $\frac{π}{2} + \frac{1}{2}$ $pi/2 +1/2$

How do you find f'(1) if f(x)=x2⋅tan−1xf(x)= x^2*tan^-1 x?