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(2x2 − 3) Sin X - Mathematics

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प्रश्न

(2x2 − 3) sin 

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उत्तर

\[\text{ Let } u = 2 x^2 - 3; v = \sin x\]
\[\text{ Then }, u' = 4x; v' = \cos x\]
\[\text{ Using theproduct rule }:\]
\[\frac{d}{dx}\left( uv \right) = uv' + vu'\]
\[\frac{d}{dx}\left[ \left( 2 x^2 - 3 \right) \sin x \right] = \left( 2 x^2 - 3 \right) \cos x + 4x \sin x \]
\[\]

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पाठ 30: Derivatives - Exercise 30.4 [पृष्ठ ३९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 30 Derivatives
Exercise 30.4 | Q 21 | पृष्ठ ३९

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