Question

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

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

Answer 1

Answer: (B)

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

$\frac{dy}{dx}$ = $\frac{1}{2}$ [$\frac{1}{1-x^2}$ - $\frac{1}{1+x^2}$]

=$\frac{x^2}{1-x^4}$

So, the correct option is 'B'.