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Question
Differentiate \[\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}\] ?
Solution
\[\text{We have,} \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}\]
\[\text{ By rationalising we get }, \]
\[\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}} \times \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}\]
\[ = \frac{\left( \sqrt{x^2 + 1} + \sqrt{x^2 - 1} \right)^2}{\left( \sqrt{x^2 + 1} \right)^2 - \left( \sqrt{x^2 - 1} \right)^2}\]
\[ = \frac{\left( \sqrt{x^2 + 1} \right)^2 + \left( \sqrt{x^2 - 1} \right)^2 + 2\left( \sqrt{x^2 + 1} \right)\left( \sqrt{x^2 - 1} \right)}{x^2 + 1 - x^2 + 1}\]
\[ = \frac{x^2 + 1 + x^2 - 1 + 2\sqrt{x^4 - 1}}{2}\]
\[ = \frac{2 x^2 + 2\sqrt{x^4 - 1}}{2}\]
\[ = x^2 + \sqrt{x^4 - 1}\]
\[\text{Now, Let } y = x^2 + \sqrt{x^4 - 1}\]
Differentiate it with respect to x we get,
\[\frac{dy}{dx} = \frac{d}{dx}\left( x^2 + \sqrt{x^4 - 1} \right)\]
\[ = 2x + \frac{1}{2\sqrt{x^4 - 1}} \times \frac{d}{dx}\left( x^4 - 1 \right)\]
\[ = 2x + \frac{1}{2\sqrt{x^4 - 1}} \times \left( 4 x^3 \right)\]
\[ = 2x + \frac{2 x^3}{\sqrt{x^4 - 1}}\]
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