Question

Find  \[\frac{dy}{dx}\] \[y = \sin x \sin 2x \sin 3x \sin 4x\] ?

 

Answer 1

\[\text{ We have, y } = \sin x \sin2x \sin3x \sin4x . . . \left( i \right)\]

Taking log on both sides

\[\log y = \log\left( \sin x \sin2x \sin3x \sin4x \right)\]
\[ \Rightarrow \log y = \log\sin x + \log\sin2x + \log\sin3x + \log\sin4x\]

Differentiating with respect to x using chain rule,

\[\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left( \log\sin x \right) + \frac{d}{dx}\left( \log\sin2x \right) + \frac{d}{dx}\left( \log\sin3x \right) + \frac{d}{dx}\left( \log\sin4x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{\sin x}\frac{d}{dx}\left( \sin x \right) + \frac{1}{\sin2x}\frac{d}{dx}\left( \sin2x \right) + \frac{1}{\sin3x}\frac{d}{dx}\left( \sin3x \right) + \frac{1}{\sin4x}\frac{d}{dx}\left( \sin4x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{\sin x}\left( \cos x \right) + \frac{1}{\sin2x}\left( \cos2x \right)\frac{d}{dx}\left( 2x \right) + \frac{1}{\sin3x}\left( \cos3x \right)\frac{d}{dx}\left( 3x \right) + \frac{1}{\sin4x}\left( \cos4x \right)\frac{d}{dx}\left( 4x \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \left[ \cot x + \cot2x\left( 2 \right) + \cot3x\left( 3 \right) + \cot4x\left( 4 \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \cot x + 2\cot2x + 3\cot3x + 4\cot4x \right]\]
\[ \Rightarrow \frac{dy}{dx} = \left( \sin x \sin2x \sin3x \sin4x \right)\left[ \cot x + 2\cot2x + 3\cot3x + 4\cot4x \right] \left[ \text{Using equation } \left( i \right) \right]\]