If f(x) = frac{1}{sqrt{x^2 + a^2} + sqrt{x^2 | vedclass

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(A) Given $f(x) = \frac{1}{\sqrt{x^2 + a^2} + \sqrt{x^2 + b^2}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{x^

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

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(A) Given $f(x) = \frac{1}{\sqrt{x^2 + a^2} + \sqrt{x^2 + b^2}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2})$:$f(x) = \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{(\sqrt{x^2 + a^2} + \sqrt{x^2 + b^2})(\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2})}$$f(x) = \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{(x^2 + a^2) - (x^2 + b^2)}$$f(x) = \frac{1}{a^2 - b^2} [\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}]$Now,differentiating with respect to $x$:$f'(x) = \frac{1}{a^2 - b^2} \left[ \frac{d}{dx}(\sqrt{x^2 + a^2}) - \frac{d}{dx}(\sqrt{x^2 + b^2}) \right]$$f'(x) = \frac{1}{a^2 - b^2} \left[ \frac{1}{2\sqrt{x^2 + a^2}} \cdot 2x - \frac{1}{2\sqrt{x^2 + b^2}} \cdot 2x \right]$$f'(x) = \frac{x}{a^2 - b^2} \left[ \frac{1}{\sqrt{x^2 + a^2}} - \frac{1}{\sqrt{x^2 + b^2}} \right]$.

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

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