If y=log1+x1-x1/4-12Tan-1(x), then dydx=

A
$\frac{x}{1 - x^{2}}$
B
$\frac{x^{2}}{1 - x^{4}}$
C
$\frac{x}{1 + x^{4}}$
D
$\frac{x}{1 - x^{4}}$

Solution:

$Given y = \log \{{(\frac{1 + x}{1 - x})}^{\frac{1}{4}}\} - \frac{1}{2} \tan^{- 1} x \Rightarrow y = \frac{1}{4} \log \{\frac{1 + x}{1 - x})\} - \frac{1}{2} \tan^{- 1} x (∵ \log (a^{m}) = mloga) y = \frac{1}{4} \log (1 + x) - \frac{1}{4} \log (1 - x) - \frac{1}{2} {Tan}^{- 1} x (∵ \log (\frac{a}{b}) = loga - logb)$

$d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s \frac{dy}{dx} = \frac{1}{4 (1 + x)} + \frac{1}{4 (1 - x)} - \frac{1}{2 (1 + x^{2})} = \frac{1 - x + 1 + x}{4 (1 + x) (1 - x)} - \frac{1}{2 (1 + x^{2})} = \frac{1}{2 (1 - x^{2})} - \frac{1}{2 (1 + x^{2})} = \frac{x^{2}}{1 - x^{4}}$

If $y = \log \{{(\frac{1 + x}{1 - x})}^{1 / 4}\} - \frac{1}{2} {Tan}^{- 1} (x), then \frac{dy}{dx} =$

Solution:

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