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