Question

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

(a) $\frac{4x^3}{1-x^4}$

(b) $-\frac{4x}{1-x^4}$

(c) $\frac{1}{4-x^4}$

(d) $-\frac{4x^3}{1-x^4}$

Answer 1

(b) $-\frac{4x}{1-x^4}$
$$\begin{gathered} \mathrm{We}\mathrm{have},y=\log \left(\frac{1-x^2}{1+x^2}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{{\displaystyle \frac{1-x^2}{1+x^2}}}\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1+x^2}{1-x^2}\left[\frac{\left(1+x^2\right)\left(-2x\right)-\left(1-x^2\right)\left(2x\right)}{{\left(1+x^2\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1-x^2}\left[\frac{-2x-2x^3-2x+2x^3}{\left(1+x^2\right)}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{-4x}{1-x^4} \end{gathered}$$