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प्रश्न
Prove that \[\frac{d}{dx} \left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}\] ?
उत्तर
\[\frac{d}{dx}\left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\} = \sqrt{a^2 - x^2}\]
\[\text{ LHS } = \frac{d}{dx}\left\{ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right\}\]
\[ = \frac{d}{dx}\left( \frac{x}{2}\sqrt{a^2 - x^2} \right) + \frac{d}{dx}\left( \frac{a^2}{2} \sin^{- 1} \frac{x}{a} \right)\]
\[ = \frac{1}{2}\left[ x\frac{d}{dx}\sqrt{a^2 - x^2} + \sqrt{a^2 - x^2}\frac{d}{dx}\left( x \right) \right] + \frac{a^2}{2} \times \frac{1}{\sqrt{1 - \left( \frac{x}{a} \right)^2}} \times \frac{d}{dx}\left( \frac{x}{a} \right) \]
\[ = \frac{1}{2}\left[ x \times \frac{1}{2\sqrt{a^2 - x^2}}\frac{d}{dx}\left( a^2 - x^2 \right) + \sqrt{a^2 - x^2} \right] + \left[ \frac{a^2}{2} \right] \times \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \times \left( \frac{1}{a} \right)\]
\[ = \frac{1}{2}\left[ \frac{x\left( - 2x \right)}{2\sqrt{a^2 - x^2}} + \sqrt{a^2 - x^2} \right] + \left( \frac{a^2}{2} \right)\frac{a}{\sqrt{a^2 - x^2}} \times \left( \frac{1}{a} \right)\]
\[ = \frac{1}{2}\left[ \frac{- 2 x^2 + 2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{1}{2}\left[ \frac{2\left( a^2 - 2 x^2 \right)}{2\sqrt{a^2 - x^2}} \right] + \frac{a^2}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{a^2 - 2 x^2}{2\sqrt{a^2 - x^2}} + \frac{a^2}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{a^2 - 2 x^2 + a^2}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{2 a^2 - 2 x^2}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{2\left( a^2 - x^2 \right)}{2\sqrt{a^2 - x^2}}\]
\[ = \frac{\left( a^2 - x^2 \right)}{\sqrt{a^2 - x^2}}\]
\[ = \sqrt{a^2 - x^2} = RHS\]
\[\text{ Hence proved }\]
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