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Question

Differentiate d5(exsin2x)dx

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Solution

let y=5exsin2x
apply log on both sides
logy=(exsin2x)log5
now differentiate on both sides with respect to x
ddx(logy)=log5ddxexsin2x
y!y=log5(exsin2x+exsin2x)
therefore dydx=(exsin2x+exsin2x)5exsin2xlog5

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