Find dy/dx, when y = 3^(3x) sin 4x 2^x

Question

Find \(\cfrac{dy}{d\text x}\), when

y = 3^(3x) sin 4x 2^x

Answer 1

Let y = e^3x sin 4x 2^x

Take log both sides:

⇒ log y = log (e^3x sin 4x 2^x )

⇒ log y = log (e^3x ) + log (sin 4x) + log (2^x )

\(\Big\{\) 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 x^a = 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 = e^3x sin 4x 2^x

⇒ \(\cfrac{dy}{d\mathrm x} \) = e^3x sin 4x 2^x{3 + 4 cot 4x + log 2}