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What is derivative of the function y=2x*ln(2x)*sin(2x)?

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carrotpie | Student | eNoter

Posted May 22, 2011 at 1:49 PM via web

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What is derivative of the function y=2x*ln(2x)*sin(2x)?

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giorgiana1976 | College Teacher | Valedictorian

Posted May 22, 2011 at 2:02 PM (Answer #1)

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We'll have to apply product rule and chain rule to determine the 1st derivative of the given function:

y' = (2x)'*ln(2x)*sin(2x) + (2x)*[ln(2x)]'*sin(2x) + (2x)*ln(2x)*[sin(2x)]'

y' = 2*ln(2x)*sin(2x) + [2x*sin(2x)]/x + (2x)*ln(2x)*[2*cos(2x)]

y' = [ln (4x^2)]*sin(2x) + 2*sin(2x) + 4x*ln(2x)*cos(2x)

y' = 2*(sin 2x)*{ln[e*(2x)]} + 4x*ln(2x)*cos(2x)

The first derivative of the given function is:

y' = 2*(sin 2x)*{ln[e*(2x)]} + 4x*ln(2x)*cos(2x)

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