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Mathematics 17 Online
OpenStudy (anonymous):

derivative of f(x)= cos(x) sin(3x)

OpenStudy (pawanyadav):

Derivative of function of type (UV) Is (U'V+V'U) Where U and V are two different function.

OpenStudy (pawanyadav):

Here U=cos x V=sin 3x Can you tell what are U' and V'

OpenStudy (anonymous):

U' = -sinx V' = cos3x

OpenStudy (mathmale):

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.

OpenStudy (anonymous):

V'= 3xcos(3x)

OpenStudy (unklerhaukus):

not quite, what is the derivative of 3x?

OpenStudy (anonymous):

I did it a different way and messed up. Whoops. it should be 3cos(3x)

OpenStudy (unklerhaukus):

that's better

OpenStudy (unklerhaukus):

so you have U = cos x , U' = -sin x V = sin 3x , V' = 3cos 3x plug these into (UV)' = (U'V+V'U)

OpenStudy (anonymous):

(UV)' = -sin^2(3x^2) + 3cos^2(3x^2)

OpenStudy (pawanyadav):

@gingycreamsicle is this correct Can (sin a)(sin b)=sin^2 (ab) ???

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