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प्रश्न
Differentiate \[\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)\] ?
उत्तर
\[\text{Let } y = \sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right) \right\}\]
\[ = \frac{1}{\sqrt{1 - \left( \frac{x}{\sqrt{x^2 + a^2}} \right)^2}} \times \frac{d}{dx}\left( \frac{x}{\sqrt{x^2 + a^2}} \right) \left[ \text{Using chain rule and quotient rule} \right]\]
\[ = \frac{1}{\sqrt{1 - \left( \frac{x}{\sqrt{x^2 + a^2}} \right)^2}} \times \left[ \frac{\left( x^2 + a^2 \right)^\frac{1}{2} \frac{d}{dx}\left( x \right) - x\frac{d}{dx} \left( x^2 + a^2 \right)^\frac{1}{2}}{\left[ \left( x^2 + a^2 \right)^\frac{1}{2} \right]^2} \right]\]
\[ = \frac{\sqrt{x^2 + a^2}}{\sqrt{x^2 + a^2 - x^2}}\left[ \frac{\sqrt{x^2 + a^2} - \frac{x}{2\sqrt{x^2 + a^2}}\frac{d}{dx}\left( x^2 + a^2 \right)}{\left( x^2 + a^2 \right)} \right]\]
\[ = \frac{\sqrt{x^2 + a^2}}{a\left( x^2 + a^2 \right)}\left[ \sqrt{x^2 + a^2} - \frac{x}{2\sqrt{x^2 + a^2}} \times 2x \right]\]
\[ = \frac{\sqrt{x^2 + a^2}}{a\left( x^2 + a^2 \right)}\left[ \frac{x^2 + a^2 - x^2}{\sqrt{x^2 + a^2}} \right]\]
\[ = \frac{a^2}{a\left( x^2 + a^2 \right)}\]
\[ = \frac{a}{\left( x^2 + a^2 \right)}\]
\[So, \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right) \right\} = \frac{a}{\left( x^2 + a^2 \right)}\]
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