Question:
Mark $(\sqrt{ })$ against the correct answer in the following:
If $y=\tan ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right)$ then $\frac{d y}{d x}=?$
A. $\frac{2 x}{\left(1+x^{4}\right)}$
B. $\frac{-2 x}{\left(1+x^{4}\right)}$
C. $\frac{\mathrm{x}}{\left(1+\mathrm{x}^{4}\right)}$
D. none of these
Solution:
