Question

Differentiate $\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ w.r.t $\sqrt{1-x^4}$

• A. $\frac{1-\sqrt{1-x^4}}{x^6}$
• B. $\frac{1+\sqrt{1-x^4}}{x^6}$
• C. $\frac{1-\sqrt{1-x^4}}{x^4}$
• D. $\frac{1+\sqrt{1-x^4}}{x^4}$

Answer 1

The correct option is B $\frac{1+\sqrt{1-x^4}}{x^6}$

Let $v=\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ and $u=\sqrt{1-x^4}$

Put $x^2=\cos 2\theta$, then

$v=\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}=\frac{(\cos \theta +\sin \theta )}{(\cos \theta -\sin \theta )}\times \frac{(\cos \theta +\sin \theta )}{(\cos \theta +\sin \theta )}$

$=\frac{1+\sin 2\theta }{\cos 2\theta }=\sec 2\theta +\tan 2\theta$

And $u=\sqrt{1-{\cos }^{2}2\theta }=\sin 2\theta$

Therefore $\frac{dv}{d\theta }=2\sec 2\theta \tan 2\theta +2{\sec }^{2}2\theta =2\sec 2\theta (\tan 2\theta +\sec 2\theta )$

And $\frac{du}{d\theta }=2\cos 2\theta$

Hence $\frac{dv}{du}=\frac{2\sec 2\theta (\tan 2\theta +\sec 2\theta )}{2\cos 2\theta }=\frac{(\frac{\sin 2\theta }{\cos 2\theta }+\frac{1}{\cos 2\theta })}{{\cos }^{2}2\theta }$

$=\frac{(1+\sin 2\theta )}{{\cos }^{3}2\theta }=\frac{1+\sqrt{1-x^4}}{x^6}$