Q. 234.3( 3 Votes )

# Find , when

y = e^{3x} sin 4x 2^{x}

Answer :

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})

⇒ log y = 3x log e+ log (sin 4x) + x log 2 {log x^{a} = alog x}

⇒ log y = 3x + log (sin 4x) + x log 2 {log e = 1}

Differentiating with respect to x:

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