Question

If $y=\log {\left[\frac{\sqrt{(1+x)+\sqrt{(1-x)}}}{\sqrt{(1+x)-\sqrt{(1-x)}}}\right]}^{1/2}$ then show that $\frac{dy}{dx}=\frac{1}{2x\sqrt{(1-x^2)}}$

Answer 1

$y=\log {\left[\frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{\sqrt{(1+x)}-\sqrt{(1-x)}}\right]}^{1/2}$

Put $x=\sin 2\theta$

$y=\log {\left[\frac{\sqrt{(1+\sin 2\theta )}+\sqrt{(1-\sin 2\theta )}}{\sqrt{(1+\sin 2\theta )}-\sqrt{(1-\sin 2\theta )}}\right]}^{1/2}$

$\because \sqrt{1\pm \sin 2\theta }=\cos \theta \pm \sin \theta$

$y=\log {\left[\frac{(\cos \theta +\sin \theta )+(\cos \theta -\sin \theta )}{(\cos \theta +\sin \theta )-(\cos \theta -\sin \theta )}\right]}^{1/2}$

$\therefore y=\frac{1}{2}\log \left(\frac{2\cos \theta }{2\sin \theta }\right)$

$y=\frac{1}{2}\log \cot \theta$

$\therefore \frac{dy}{d\theta }=\frac{1}{2}\frac{1}{\cot \theta }.(-cose{c}^2\theta )=-\frac{1}{2\cos \theta \sin \theta }=-\frac{1}{\sin 2\theta }$

$\therefore \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}$

$=-\frac{1}{\sin 2\theta }.\frac{1}{2\cos 2\theta }$ ($\because x=\sin 2\theta \Rightarrow \frac{dx}{d\theta }=2\cos 2\theta$)

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