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प्रश्न
Differentiate \[\frac{3 x^2 \sin x}{\sqrt{7 - x^2}}\] ?
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उत्तर
\[\text{Let } y = \frac{3 x^2 \sin x}{\sqrt{7 - x^2}}\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{3 x^2 sinx}{\left( 7 - x^2 \right)^\frac{1}{2}} \right\}\]
\[ = \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times \frac{d}{dx}\left( 3 x^2 \sin x \right) - \left( 3 x^2 \sin x \right)\frac{d}{dx} \left( 7 - x^2 \right)^\frac{1}{2}}{\left[ \left( 7 - x^2 \right)^\frac{1}{2} \right]^2} \left[ \text{Using quotient rule, chain rule and product rule} \right]\]
\[ = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left[ x^2 \frac{d}{dx}\left( \sin x \right) + \sin x\frac{d}{dx}\left( x^2 \right) \right] - 3 x^2 \sin x \times \frac{1}{2}\left( 7 - x^2 \right) \times \frac{d}{dx}\left( 7 - x^2 \right)}{\left( 7 - x^2 \right)} \right]\]
\[ = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} 3\left( x^2 \cos x + 2x \sin x \right) - 3 x^2 \sin x \times \frac{1}{2} \left( 7 - x^2 \right)^\frac{- 1}{2} \left( - 2x \right)}{\left( 7 - x^2 \right)} \right]\]
\[ = \left[ \frac{\left( 7 - x^2 \right)^\frac{1}{2} \times 3\left( x^2 \cos x + 2x \sin x \right)}{\left( 7 - x^2 \right)} + \frac{3 x^3 \sin x \left( 7 - x^2 \right)^\frac{- 1}{2}}{\left( 7 - x^2 \right)} \right]\]
\[ = \left[ \frac{6x \sin x + 3 x^2 \cos x}{\sqrt{\left( 7 - x^2 \right)}} + \frac{3 x^3 \sin x}{\left( 7 - x^2 \right)^\frac{3}{2}} \right]\]
\[So, \frac{d}{dx}\left( \frac{3 x^2 \sin x}{\sqrt{7 - x^2}} \right) = \left[ \frac{6x \sin x + 3 x^2 \cos x}{\sqrt{\left( 7 - x^2 \right)}} + \frac{3 x^3 \sin x}{\left( 7 - x^2 \right)^\frac{3}{2}} \right]\]
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