मराठी

(X Sin X + Cos X) (X Cos X − Sin X)

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प्रश्न

(x sin x + cos x) (x cos x − sin x

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उत्तर

\[u = x \sin x + \cos x; v = x \cos x - \sin x\]
\[u' = x \cos x + \sin x - \sin x = x \cos x ; v' = - x \sin x + \cos x - \cos x = - x \sin x\]
\[ \]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = uv' + vu'\]
\[\frac{d}{dx}\left[ \left( x \sin x + \cos x \right)\left( x \cos x - \sin x \right) \right] = \left( x \sin x + \cos x \right)\left( - x \sin x \right) + \left( x \cos x - \sin x \right)\left( x \cos x \right)\]
\[ = - x^2 \sin^2 x - x \cos x \sin x + x^2 \cos^2 x - x \cos x \sin x \]
\[ = x^2 \left( \cos^2 x - \sin^2 x \right) - x\left( 2 \sin x \cos x \right)\]
\[ = x^2 \cos \left( 2x \right) - x\left( \sin \left( 2x \right) \right)\]
\[ = x \left[ x \cos \left( 2x \right) - \sin \left( 2x \right) \right]\]

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पाठ 30: Derivatives - Exercise 30.4 [पृष्ठ ३९]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 30 Derivatives
Exercise 30.4 | Q 11 | पृष्ठ ३९

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