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Differentiate √ a 2 − X 2 a 2 + X 2 ? - Mathematics

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प्रश्न

Differentiate \[\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}\] ?

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उत्तर

\[\text{ Let } y = \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}\]

\[ \Rightarrow y = \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^\frac{1}{2} \]

\[\text{Differentiate it with respect to x we get}, \]

\[\frac{d y}{d x} = \frac{d}{dx} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^\frac{1}{2} \]

\[ = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{a^2 - x^2}{a^2 + x^2} \right) \left[ \text{Using chain rule} \right]\]

\[ = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( a^2 + x^2 \right)\frac{d}{dx}\left( a^2 - x^2 \right) - \left( a^2 - x^2 \right)\frac{d}{dx}\left( a^2 + x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\} \]

\[ = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x\left( a^2 + x^2 \right) - 2x\left( a^2 - x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\}\]

\[ = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x a^2 - 2 x^3 - 2x a^2 + 2 x^3}{\left( a^2 + x^2 \right)^2} \right\}\]

\[ = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x a^2}{\left( a^2 + x^2 \right)^2} \right\}\]

\[ = \frac{- 2x a^2}{\sqrt{a^2 - x^2} \left( a^2 + x^2 \right)^\frac{3}{2}}\]

\[So, \frac{d}{dx}\left( \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \right) = \frac{- 2 a^2 x}{\sqrt{a^2 - x^2} \left( a^2 + x^2 \right)^\frac{3}{2}}\]

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अध्याय 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.02 | Q 14 | पृष्ठ ३७

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