Let y = e3x sin 4x 2x
Take log both sides:
⇒ log y = log (e3x sin 4x 2x )
⇒ log y = log (e3x ) + log (sin 4x) + log (2x )
\(\Big\{\) log (ab) = log a + log b; log \(\big(\cfrac{a}{b}\big)\)= log a - log b \(\Big\}\)
⇒ log y = 3x log e+ log (sin 4x) + x log 2{log xa = a log x}
⇒ log y = 3x + log (sin 4x) + x log 2 {log e = 1}
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, \(\cfrac{d(u +a)}{d\text x}=\cfrac{du}{d\text x}+\cfrac{da}{d\text x}\) where a and u are any variables; \(\cfrac{d(log\,u)}{d\text x}=\cfrac{1}{u}\cfrac{du}{d\text x}\) \(\Bigg\}\)
Put the value of y = e3x sin 4x 2x
⇒ \(\cfrac{dy}{d\mathrm x}
\) = e3x sin 4x 2x{3 + 4 cot 4x + log 2}