D-dxloge 1-x-1-x 1-4-1-2tan-1x-

The correct option is A x^2/1 x^4. Let y=1/4 [ log ( 1+x ) log ( 1 x ) ] 1/2tan^ 1x. ⇒dy/dx=1/4 [ 1/1+x 1/1 x ( 1 ) ] 1/2.1/1+x^2 =1/ ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

d/dxloge 1+x/...

Question

$\frac{d}{d x} ({log}_{e} {(\frac{1 + x}{1 - x})}^{1 / 4} - \frac{1}{2} {tan}^{- 1} x .)$

\frac{x^{2}}{1 - x^{4}} .

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\frac{x^{3}}{1 - x^{4}} .

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\frac{x^{4}}{1 - x^{4}} .

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- \frac{x^{2}}{1 - x^{4}} .

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Solution

The correct option is A $\frac{x^{2}}{1 - x^{4}} .$
Let $y = \frac{1}{4} [log (1 + x) - log (1 - x)] - \frac{1}{2} {tan}^{- 1} x .$
$\Rightarrow \frac{d y}{d x} = \frac{1}{4} [\frac{1}{1 + x} - \frac{1}{1 - x} (- 1)] - \frac{1}{2} . \frac{1}{1 + x^{2}}$
$= \frac{1}{2 (1 - x^{2})} - \frac{1}{2 (1 + x^{2})} = \frac{x^{2}}{1 - x^{4}} .$

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ddx(loge(1+x1−x)1/4−12tan−1x.)

The correct option is A x21−x4.Let y=14[log(1+x)−log(1−x)]−12tan−1x.⇒dydx=14[11+x−11−x(−1)]−12.11+x2=12(1−x2)−12(1+x2)=x21−x4.

$\frac{d}{d x} ({log}_{e} {(\frac{1 + x}{1 - x})}^{1 / 4} - \frac{1}{2} {tan}^{- 1} x .)$