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If $y=log\bigg(\large \frac{1-x^2}{1+x^2}\normalsize \bigg),then\;\large \frac{dy}{dx}$ is equal to

\[ \begin{array}{1 1}(A)\;\frac{4x^3}{1-x^4} & (B)\;\frac{-4x}{1-x^4}\\(C)\;\frac{1}{4-x^4} &(D)\;\frac{-4x^3}{1-x^4}\end{array}\]

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Toolbox:
  • Chain Rule :If $z=f(y)$ and $y=g(x)$,then $\large\frac{dz}{dx}=\frac{dz}{dy}-\frac{dy}{dx}$
$y=\log\bigg|\large\frac{1-x^2}{1+x^2}\bigg|$
$\large\frac{dy}{dx}=\large\frac{1}{\Large\frac{(1-x^2)}{(1+x^2)}}.\frac{(1+x^2)(-2x)-(1-x^2)(2x)}{(1+x^2)^2}$
$\Rightarrow \large\frac{(1+x^2)}{(1-x^2)}\times \frac{-2x-2x^3-2x+2x^3}{(1+x^2)^2}$
On simplifying we get,
$\Rightarrow \large\frac{-4x}{(1-x^4)}$
Hence the correct option is $B$
answered Jul 4, 2013 by sreemathi.v
 
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