Question

If $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}$, then $f\text{'}\left(x\right)$ is equal to

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

Answer 1

The correct option is A

$\frac{\mathrm{x}}{{\mathrm{a}}^2-{\mathrm{b}}^2}\left[\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}-\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}\right]$

Step1: Given a function

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}+\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}$

Step 2: Multiplying numerator and denominator by, $\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}-\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}$

$\begin{array}{rcl}\therefore \mathrm{f}\left(\mathrm{x}\right)& =& \frac{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}-\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}{\left\{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)-\left({\mathrm{x}}^2+{\mathrm{b}}^2\right)\right\}}\\ & =& \frac{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}-\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}{{\mathrm{x}}^2+{\mathrm{a}}^2-{\mathrm{x}}^2-{\mathrm{b}}^2}\\ & =& \frac{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}-\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}{{\mathrm{a}}^2-{\mathrm{b}}^2}\\ & =& \frac{1}{{\mathrm{a}}^2-{\mathrm{b}}^2}\left(\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}-\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}\right)\\ \mathrm{Now},\mathrm{using}\frac{\mathrm{d}}{\mathrm{dx}}\sqrt{\mathrm{u}}& =& \frac{1}{2\sqrt{\mathrm{u}}}\frac{\mathrm{du}}{\mathrm{dx}}\\ & =& \frac{1}{{\mathrm{a}}^2-{\mathrm{b}}^2}\left[\frac{2\mathrm{x}}{2\sqrt{\left({\mathrm{x}}^2+{\mathrm{a}}^2\right)}}-\frac{2\mathrm{x}}{2\sqrt{({\mathrm{x}}^2+{\mathrm{b}}^2)}}\right]\\ & =& \frac{\mathrm{x}}{{\mathrm{a}}^2-{\mathrm{b}}^2}\left[\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}-\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{b}}^2}}\right]\end{array}$

Hence, the correct option is an option (A).