Question

$\frac{d}{dx}\left\{\log \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right\}=$

• A. $\frac{1}{x\sqrt{1-x^2}}$
• B. $-\frac{1}{\sqrt{1-x^2}}$
• C. $-\frac{1}{x\sqrt{1-x^2}}$
• D. none of these

Answer 1

The correct option is C $-\frac{1}{x\sqrt{1-x^2}}$

let $y=\log \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)$

let $t=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$

$=\frac{(\sqrt{1+x}+\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$

$\therefore t=\frac{1+\sqrt{1-x^2}}{x}$

$\frac{dt}{dx}=\frac{x\left(\frac{-2x}{2\sqrt{1-x^2}}\right)-x(1+\sqrt{1-x^2})}{x^2}$

$=\frac{-x^2-\sqrt{1-x^2}-(1-x^2)}{x^2\sqrt{1-x^2}}$

$y=\log t$

differentiating on both sides wrt to $x$

$\frac{dy}{dx}=\frac{1}{t}\frac{dt}{dx}$

$=\left(\frac{x}{1+\sqrt{1-x^2}}\right)\left(\frac{-(1+\sqrt{1-x^2})}{x^2\sqrt{1-x^2}}\right)$

$\therefore \frac{dy}{dx}=\frac{-1}{x\sqrt{1-x^2}}$