A) \[\frac{{{x}^{2}}}{1-{{x}^{4}}}\]
B) \[\frac{2{{x}^{2}}}{1-{{x}^{4}}}\]
C) \[\frac{{{x}^{2}}}{2\,\,(1-{{x}^{4}})}\]
D) None of these
Correct Answer: A
Solution :
\[y=\log {{\left( \frac{1+x}{1-x} \right)}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x\] Differentiating w.r.t. x of y, we get \[\frac{dy}{dx}={{\left( \frac{1-x}{1+x} \right)}^{1/4}}\frac{1}{4}{{\left( \frac{1+x}{1-x} \right)}^{-3/4}}\left[ \frac{(1-x)+(1+x)}{{{(1-x)}^{2}}} \right]-\frac{1}{2}.\frac{1}{1+{{x}^{2}}}\] \[=\frac{1}{2}\left( \frac{1-x}{1+x} \right)\frac{1}{{{(1-x)}^{2}}}-\frac{1}{2(1+{{x}^{2}})}\] \[=\frac{1}{2}.\frac{1}{(1-{{x}^{2}})}-\frac{1}{2}\frac{1}{(1+{{x}^{2}})}=\frac{{{x}^{2}}}{1-{{x}^{4}}}\].
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