Question

If $y= \log (\frac {1-x^2}{1+x^2})$ then $\frac {dy}{dx}$ is equal to _______

• A. $\frac {-4x}{1-x^4}$
• B. $\frac {4x^3}{1-x^4}$
• C. $\frac {1}{4-x^4}$
• D. $\frac {-4x^{3}}{1-x^4}$

Answer 1

Answer: (A)

Given, $y =\log \left(\frac{1-x^{2}}{1+x^{2}}\right) $
$=\log \left(1-x^{2}\right)-\log \left(1+x^{2}\right) $
$\therefore \frac{d y}{d x} =\frac{1}{1-x^{2}}(-2 x)-\frac{1}{1+x^{2}}(2 x)$
$=-2 x\left[\frac{1}{1-x^{2}}+\frac{1}{1+x^{2}}\right] $
$=-2 x\left[\frac{1+x^{2}+1-x^{2}}{1-x^{4}}\right]=\frac{-4 x}{1-x^{4}}$