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derivative of f(x)= cos(x) sin(3x)
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Derivative of function of type (UV) Is (U'V+V'U) Where U and V are two different function.
Here U=cos x V=sin 3x Can you tell what are U' and V'
U' = -sinx V' = cos3x
Finding the derivative of V=sin 3x requires use of the Chain Rule. Focus carefully on that 3x. It's true that the derivative of v=sin x is v' cos x.
V'= 3xcos(3x)
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not quite, what is the derivative of 3x?
I did it a different way and messed up. Whoops. it should be 3cos(3x)
that's better
so you have U = cos x , U' = -sin x V = sin 3x , V' = 3cos 3x plug these into (UV)' = (U'V+V'U)
(UV)' = -sin^2(3x^2) + 3cos^2(3x^2)
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@gingycreamsicle is this correct Can (sin a)(sin b)=sin^2 (ab) ???
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