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Question
Find \[\frac{dy}{dx}\] \[y = e^{3x} \sin 4x \cdot 2^x\] ?
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Solution
\[\text{ We have, y } = e^{3x} \times \sin4x \times 2^x . . . \left( i \right)\]
Taking log on both sides,
\[\log y = \log e^{3x} + \log\sin4x + \log 2^x \]
\[ \Rightarrow \log y = 3x \log e + \log\sin4x + x \log2 \]
\[ \Rightarrow \log y = 3x + \log\sin4x + x \log2\]
Differentiating with respect to x,
\[\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left( 3x \right) + \frac{d}{dx}\left( \log \sin4x \right) + \frac{d}{dx}\left( x \log2 \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + \frac{1}{\sin4x}\frac{d}{dx}\left( \sin4x \right) + \log2\left( 1 \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + \frac{1}{\sin4x}\left( \cos4x \right)\frac{d}{dx}\left( 4x \right) + \log2\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + \cot4x\left( 4 \right) + \log2\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3 + 4\cot4x + \log2\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ 3 + 4\cot4x + \log2 \right]\]
\[ \Rightarrow \frac{dy}{dx} = e^{3x} \sin4x 2^x \left[ 3 + 4\cot4x + \log2 \right] \left[ \text{ Using equation} \left( i \right) \right]\]
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