Question

Differentiate:
2^x.e^3x.sin4x​

Answer 1

Answer:

`logy=xlog2+3x+log sin 4x`

`rArr(1)/(y).(dy)/(dx)=(log2)+3+(4cos4x)/(sin4x).`

Answer 2

Answer:

We have,y = 2x e3x . sin 4xTaking logarithm both sides, we getlog y = log[2x e3x . sin 4x]⇒log y = log 2x + log e3x + log sin 4x⇒log y = x log 2 + 3x + log sin 4xDifferentiating both sides with respect to x, we get   1ydydx = log 2 + 3 + 1sin 4x × 4cos 4x⇒1ydydx =  log 2 + 3 + 4 cot 4x⇒dydx = y[log 2 + 3 + 4 cot 4x]⇒dydx = 2x e3x . sin 4x [log 2 + 3 + 4 cot 4x]

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