Let y = sin x sin 2x sin 3x sin 4x
Take log both sides:
⇒ log y = log (sin x sin 2x sin 3x sin 4x)
⇒ log y = log (sin x ) + log (sin 2x) + log (sin 3x) + log (sin 4x)
\(\Big\{\)log (ab) = log a + log b; log\(\big(\cfrac{a}{b}\big)\) = log a - log b\(\Big\}\)
Differentiating with respect to x:
\(\Bigg\{\) Using chain rule, \(\cfrac{d(u +a)}{d\text x}=\cfrac{du}{d\text x}+\cfrac{da}{d\text x}\) where a and u are any variables \(\Bigg\}\)
\(\Bigg\{\) Using chain rule, \(\cfrac{d(au)}{d\text x}\) = a\(\cfrac{du}{d\text x}\) where a is any constant and u is any variable \(\Bigg\}\)
