Question

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

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

Answer 1

The correct option is A

$\frac{x^2}{\left(1-x^4\right)}$

Find the $\frac{dy}{dx}$:

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

Using the property of logarithmic i.e.; $\log a^b=b\log a$.

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

Now differentiate $y$ with respect to $x$ using quotient rule i.e.; $\frac{d\left({\displaystyle \frac{u}{v}}\right)}{dx}=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \left[\frac{1}{4}\frac{(1-\mathrm{x})}{(1+\mathrm{x})}\times \frac{\left[\right(1-\mathrm{x})+(1+\mathrm{x}\left)\right]}{(1-\mathrm{x}{)}^2}\right]-\frac{1}{2}\times \frac{1}{\left(1+{\mathrm{x}}^2\right)}\\ & =& \left[\frac{1}{4}\frac{(1-\mathrm{x})}{(1+\mathrm{x})}\times \frac{2}{(1-\mathrm{x}{)}^2}\right]-\frac{1}{2}\times \frac{1}{\left(1+{\mathrm{x}}^2\right)}\\ & =& \left[\frac{1}{2}\frac{1}{(1+\mathrm{x})\left(1-\mathrm{x}\right)}\right]-\frac{1}{2}\times \frac{1}{\left(1+{\mathrm{x}}^2\right)}\\ & =& \left[\frac{1}{2}\frac{1}{\left(1-{\mathrm{x}}^2\right)}\right]-\frac{1}{2}\times \frac{1}{\left(1+{\mathrm{x}}^2\right)}\\ & =& \frac{1}{2}\left[\frac{1+x^2-1+x^2}{\left(1-x^2\right)\left(1+x^2\right)}\right]\\ & =& \frac{x^2}{1-x^4}\end{array}$

Hence, option A is correct.