# Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

$y=e^{3 x} \sin 4 x 2^{x}$

Solution:

Let $y=e^{3 x} \sin 4 x 2^{x}$

Take log both sides:

$\Rightarrow \log y=\log \left(e^{3 x} \sin 4 x 2^{x}\right)$

$\Rightarrow \log y=\log \left(e^{3 x}\right)+\log (\sin 4 x)+\log \left(2^{x}\right)$

$\left\{\log (a b)=\log a+\log b ; \log \left(\frac{a}{b}\right)=\log a-\log b\right\}$

$\Rightarrow \log y=3 x \log e+\log (\sin 4 x)+x \log 2\left\{\log x^{a}=a \log x\right\}$

$\Rightarrow \log y=3 x+\log (\sin 4 x)+x \log 2\{\log e=1\}$

Differentiating with respect to $\mathrm{x}$ :

$\Rightarrow \frac{\mathrm{d}(\log y)}{\mathrm{dx}}=\frac{\mathrm{d}(3 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(\log \sin 4 \mathrm{x})}{\mathrm{dx}}+\frac{\mathrm{d}(\mathrm{x} \log 2)}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where $\mathrm{a}$ and $\mathrm{u}$ are any variables $\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=3 \frac{d x}{d x}+\frac{1}{\sin 4 x} \frac{d(\sin 4 x)}{d x}+\log 2 \frac{d x}{d x}$

$\left\{\begin{array}{c}\text { Using chain rule, } \frac{\mathrm{d}(\mathrm{au})}{\mathrm{dx}}=\mathrm{a} \frac{\mathrm{du}}{\mathrm{dx}} \text { where } \mathrm{a} \text { is any constant and } \mathrm{u} \text { is any variable ; } \\ \frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\end{array}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=3+\frac{\cos 4 x}{\sin 4 x} \frac{d(4 x)}{d x}+\log 2$

$\left\{\frac{d(\sin u)}{d x}=\cos u \frac{d u}{d x}\right\}$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=3+\cot 4 x \times 4 \frac{d x}{d x}+\log 2$

$\Rightarrow \frac{d y}{d x}=y\{3+4 \cot 4 x+\log 2\}$

Put the value of $y=e^{3 x} \sin 4 x 2^{x}$ :

$\Rightarrow \frac{d y}{d x}=e^{3 x} \sin 4 x 2^{x}\{3+4 \cot 4 x+\log 2\}$