Choose the correct alternative in the following:
If $y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$, then $\frac{d y}{d x}=$
A. $\frac{4 \mathrm{x}^{3}}{1-\mathrm{x}^{4}}$
B. $-\frac{4 x}{1-x^{4}}$
C. $\frac{1}{4-x^{4}}$
D. $\frac{4 x^{3}}{1-x^{4}}$
$y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\frac{d y}{d x}=\frac{1}{\frac{1-x^{2}}{1+x^{2}}}\left[\frac{\frac{d\left(1-x^{2}\right)}{d x} \cdot\left(1+x^{2}\right) \frac{d\left(1+x^{2}\right)}{d x} \cdot\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$ (Using quotient rule)
$\frac{d y}{d x}=\frac{1+x^{2}}{1-x^{2}}\left[\frac{-2 x\left(1+x^{2}\right)-2 x\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right]$
$\frac{d y}{d x}=\frac{1}{1-x^{2}}\left[\frac{-2 x\left(1+x^{2}+1-x^{2}\right)}{\left(1+x^{2}\right)}\right]$
$\frac{d y}{d x}=\left[\frac{-4 x}{1^{2}-\left(x^{2}\right)^{2}}\right]$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{-4 \mathrm{x}}{1-\mathrm{x}^{4}}\right]$
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