If \( y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right) \), then \( \frac{d y}{d x} \) is equal to

\[
\begin{array}{l}
y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right) \\
\frac{d y}{d x}=\frac{1}{\frac{1-x^{2}}{1+x^{2}}}\left[\frac{d\left(1-x^{2}\right)}{d x} \cdot\left(1+x^{2}\right)-\frac{d\left(1+x^{2}\right)}{d x} \cdot\left(1-x^{2}\right)\right]
\end{array}
\]
(Using quotient rule)
\[
\begin{array}{l}
\frac{d y}{d x}=\frac{1+x^{2}}{1-x^{2}}\left[\frac{-2 x\left(1+x^{2}\right)-2 x\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right] \\
\frac{d y}{d x}=\frac{1}{1-x^{2}}\left[\frac{-2 x\left(1+x^{2}+1-x^{2}\right)}{\left(1+x^{2}\right)}\right] \\
\frac{d y}{d x}=\left[\frac{-4 x}{1^{2}-\left(x^{2}\right)^{2}}\right] \\
\therefore \frac{d y}{d x}=\left[\frac{-4 x}{1-x^{4}}\right]
\end{array}
\]