Question

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

The correct option is B $\frac{x^2}{1-x^4}$

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