Question

If $y=\frac{1}{\sqrt{(x^2+a^2)}+\sqrt{(x^2+b^2)}},$ find $\frac{dy}{dx}.$

• A. $=\frac{x}{(a^2-b^2)}[\frac{1}{\sqrt{(x^2+a^2)}}-\frac{1}{\sqrt{(x^2+b^2)}}.]$
• B. $=\frac{x}{(a^2-b^2)}[\frac{1}{\sqrt{(x^2-a^2)}}-\frac{1}{\sqrt{(x^2-b^2)}}.]$
• C. $=\frac{x}{(a^2+b^2)}[\frac{1}{\sqrt{(x^2+a^2)}}-\frac{1}{\sqrt{(x^2+b^2)}}.]$
• D. $=\frac{x}{(a^2+b^2)}[\frac{1}{\sqrt{(x^2-a^2)}}-\frac{1}{\sqrt{(x^2-b^2)}}.]$

Answer 1

The correct option is A $=\frac{x}{(a^2-b^2)}[\frac{1}{\sqrt{(x^2+a^2)}}-\frac{1}{\sqrt{(x^2+b^2)}}.]$

Given, $y=\frac{1}{\sqrt{(x^2+a^2)}+\sqrt{(x^2+b^2)}},$

$y=\frac{1}{\sqrt{(x^2+a^2)}+\sqrt{(x^2+b^2)}}\times \frac{\sqrt{(x^2+a^2)}-\sqrt{(x^2+b^2)}}{\sqrt{(x^2+a^2)}-\sqrt{(x^2+b^2)}}$

$=\frac{\sqrt{(x^2+a^2)}-\sqrt{(x^2+b^2)}}{(x^2+a^2)-(x^2+b^2)}$

$=\frac{1}{(a^2-b^2)}[\sqrt{(x^2+a^2)}-\sqrt{(x^2+b^2)}]$

Now using $\frac{d}{dx}\sqrt{u}=\frac{1}{2\sqrt{u}}\frac{du}{dx}$

$\therefore \frac{dy}{dx}=\frac{1}{(a^2-b^2)}[\frac{1}{2\sqrt{(x^2+a^2)}}.2x-\frac{1}{2\sqrt{(x^2+b^2)}}.2x]$

$=\frac{x}{(a^2-b^2)}[\frac{1}{\sqrt{(x^2+a^2)}}-\frac{1}{\sqrt{(x^2+b^2)}}.]$