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Differentiate Sin 2 { Log ( 2 X + 3 ) } ? - Mathematics

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प्रश्न

Differentiate \[\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}\] ?

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उत्तर

\[\text{Let } y = \sin^2 \left[ \log\left( 2x + 3 \right) \right]\]

\[\Rightarrow \frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left\{ \log\left( 2x + 3 \right) \right\} \right]\]

\[ = 2 \sin\left\{ \log\left( 2x + 3 \right) \right\}\frac{d}{dx}\sin\left\{ \log\left( 2x + 3 \right) \right\} \left[ \text{Using chain rule} \right]\]

\[ = 2\sin\left\{ \log\left( 2x + 3 \right) \right\} \cos\left\{ \log\left( 2x + 3 \right) \right\}\frac{d}{dx}\log\left( 2x + 3 \right)\]

\[ = \sin\left\{ 2\log\left( 2x + 3 \right) \right\} \times \frac{1}{\left( 2x + 3 \right)}\frac{d}{dx}\left( 2x + 3 \right) \left[ \because 2\sin A \cos A = \sin2A \right]\]

\[ = \sin\left\{ 2\log\left( 2x + 3 \right) \right\}\left( \frac{2}{\left( 2x + 3 \right)} \right)\]

\[So, \frac{d}{dx}\left[ \sin^2 \left\{ \log\left( 2x + 3 \right) \right\} \right] = \sin\left\{ 2 \log\left( 2x + 3 \right) \right\}\left( \frac{2}{\left( 2x + 3 \right)} \right)\]

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अध्याय 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.02 | Q 43 | पृष्ठ ३७

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