For f(x)=arctan(1−x1+x)f(x) = \arctan(\frac{1-x}{1+x})f(x)=arctan(1+x1−x), what is f′(x)f'(x)f′(x)?
1/(1+x2)1/(1+x^2)1/(1+x2)
−1/(1+x2)-1/(1+x^2)−1/(1+x2)
1/(1−x2)1/(1-x^2)1/(1−x2)
−1-1−1