Question

Find $\frac{dy}{dx}$

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

Answer 1

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