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