Question

Find $\frac{dy}{dx}$

$y=e^{3x}\sin 4x·2^x$

Answer 1

$\mathrm{We}\mathrm{have},y=e^{3x}\times \sin 4x\times 2^x...\left(i\right)$
Taking log on both sides,
$$\begin{gathered} \log y=\log e^{3x}+\log \sin 4x+\log 2^x \\ \Rightarrow \log y=3x\log e+\mathrm{logsin}4x+x\log 2 \\ \Rightarrow \log y=3x+\log \sin 4x+x\log 2 \end{gathered}$$
Differentiating with respect to x,
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(\log \sin 4x\right)+\frac{d}{dx}\left(x\log 2\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3+\frac{1}{\sin 4x}\frac{d}{dx}\left(\sin 4x\right)+\log 2\left(1\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3+\frac{1}{\sin 4x}\left(\cos 4x\right)\frac{d}{dx}\left(4x\right)+\log 2 \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3+\cot 4x\left(4\right)+\log 2 \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3+4\cot 4x+\log 2 \\ \Rightarrow \frac{dy}{dx}=y\left[3+4\cot 4x+\log 2\right] \\ \Rightarrow \frac{dy}{dx}=e^{3x}\sin 4x{2}^x\left[3+4\cot 4x+\log 2\right]\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\right] \end{gathered}$$