Question:
Mark $(\sqrt{)}$ against the correct answer in the following:
If $y=\cot ^{-1}\left(\frac{1-x}{1+x}\right)$ then $\frac{d y}{d x}=?$
A. $\frac{-1}{\left(1+x^{2}\right)}$
B. $\frac{1}{\left(1+x^{2}\right)}$
C. $\frac{-1}{\left(1+x^{2}\right)^{3 / 2}}$
D. none of these
Solution:
