101Ryan101
what's the derivative of cosxsinx



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lalaly
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use the product rule

anonymous
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product rule
\[(fg)'=f'g+g'f\] use
\[f(x)=\sin(x), f'(x)=\cos(x), g(x)=\cos(x), g'(x)=\sin(x)\]

lalaly
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f(x)=cos x
g(x)=sin x
product rule:
d/dx ( f(x)g(x) ) = f'(x)g(x) + f(x) g'(x)

101Ryan101
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dy/dx of cosxsinx = (sinx)(sinx) + cosxcosx = cos^2x  sin^2x = ??

lalaly
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yupp thats right....and cos^2xsin^2x = cos(2x)

101Ryan101
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= (1+cos2x)/2  (1cos2x)/2 = 2(cos2x)2 = yeaaaaeee!

lalaly
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hehe