Question

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

• A. $\frac{x^2}{1-x^4}$
• B. $\frac{2x^2}{1-x^4}$
• C. $\frac{x^2}{2(1-x^4)}$
• D. None of these

Answer 1

Answer: (A)

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

$\frac{dy}{dx}=\frac{1}{(\frac{1+x}{1-x}{)}^{\frac{1}{4}}}\times$ $\frac{1}{4\times (\frac{1+x}{1-x}{)}^{\frac{3}{4}}}\times$ $\frac{(1-x)\left(1\right)-(1+x)(-1)}{(1-x{)}^2}-$ $\frac{1}{2}\frac{1}{x^2+1}$

$\frac{1-x}{4(1+x)}\times \frac{2}{(1-x{)}^2}$ $-\frac{1}{2(x^2+1)}$ $=\frac{1}{2(1-x^2)}-$ $\frac{1}{2(x^2+1)}$ =$\frac{x^2}{1-x^4}$