Solve this

Let $y=\sin x \sin 2 x \sin 3 x \sin 4 x$

Take log both sides:

$\Rightarrow \log y=\log (\sin x \sin 2 x \sin 3 x \sin 4 x)$

$\Rightarrow \log y=\log (\sin x)+\log (\sin 2 x)+\log (\sin 3 x)+\log (\sin 4 x)$

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

Differentiating with respect to $x$ :

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

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

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

$\left\{\frac{\mathrm{d}(\log \mathrm{u})}{\mathrm{dx}}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

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

$\left\{\frac{\mathrm{d}(\sin \mathrm{u})}{\mathrm{dx}}=\cos \mathrm{u} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

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

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{au})}{\mathrm{dx}}=\mathrm{a} \frac{\mathrm{du}}{\mathrm{dx}}$ where $\mathrm{a}$ is any constant and $\mathrm{u}$ is any variable $\}$

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

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

Put the value of $y=\sin x \sin 2 x \sin 3 x \sin 4 x$ :

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