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Mathematics 40 Online
OpenStudy (anonymous):

Differentiate \[xsin^2x\]

OpenStudy (anonymous):

product rule: [x*sin^2x]' = [x]'*sin^2x + x*[sin^2x]' =1*sin^2x + x*(2*sinx*cosx) = sin^2x + 2xsinxcosx or using double angle formula = sin^2x + xsin(2x)

OpenStudy (anonymous):

How do you differentiate sin^2x. That's what I had trouble with.

OpenStudy (anonymous):

|dw:1331889589089:dw| but a faster way would be the chain rule...

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