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Mathematics 9 Online
OpenStudy (anonymous):

derivative of sin^2x

OpenStudy (anonymous):

you have to use the power rule and chain rule

OpenStudy (anonymous):

sin^2x=(1-cos2x)/2

OpenStudy (anonymous):

actually it should be 2sinxcosx

OpenStudy (anonymous):

2sinxcosx=sin2x but in this case, \[\sin ^{2}x\]

OpenStudy (anonymous):

sin2x would be 2cos2x

OpenStudy (anonymous):

no, sin2x=2sinxcosx

OpenStudy (anonymous):

http://www.derivative-calculator.net/ you go ahead and plug those in. Just to show you. you're not applying chain rule correctly.

OpenStudy (anonymous):

oh, big misunderstanding! I mean sin2x=2sinxcosx not derivative of sin2x equals 2sinxcosx

OpenStudy (anonymous):

ok

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