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10 X Sin X

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प्रश्न

\[\frac{{10}^x}{\sin x}\] 

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उत्तर

\[\text{ Let } u = {10}^x ; v = \sin x\]
\[\text{ Then }, u' = {10}^x \log 10; v' = \cos x\]
\[\text{ Using the quotient rule }:\]
\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2}\]
\[\frac{d}{dx}\left( \frac{{10}^x}{\sin x} \right) = \frac{\sin x {10}^x \log 10 - {10}^x \cos x}{\sin^2 x}\]
\[ = \frac{\sin x {10}^x \log 10}{\sin^2 x} - \frac{{10}^x \cos x}{\sin^2 x}\]
\[ = {10}^x \log 10 \cos ec x - {10}^x cosec x \cot x\]
\[ = {10}^x cosec x\left( \log 10 - \cot x \right)\]

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अध्याय 30: Derivatives - Exercise 30.5 [पृष्ठ ४४]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.5 | Q 17 | पृष्ठ ४४

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