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Mathematics 13 Online
OpenStudy (anonymous):

how do you differentiate 3sin^2(x)cos(x)

OpenStudy (cwrw238):

use product and chain rules

OpenStudy (anonymous):

would it be 6sinxcosx-3sinx?

OpenStudy (tkhunny):

Seems unlikely. Who told you that? Please show YOUR work. Think a little harder about the Chain Rule.

OpenStudy (dan815):

d/dx[f(g(x))*h(x)] = f'(g(x))*g'(x)*h(x)+h'(x)*f(g(x))

OpenStudy (callisto):

If you don't want to use product rule, you may consider simplify the expression a bit. \[3\sin^2(x)\cos(x) = \frac{2\times\sin(x)\times \cos(x)\times 3\cos(x)}{2}=\frac{3\sin(2x)\sin(x)}{2}\] Then use \(\sin(a)\sin(b) = \frac{1}{2}(\cos(a-b)-\cos(a+b))\) to further simplify it. It should be pretty easy to integrate.

OpenStudy (dan815):

Lets start with this : What is the derivative o 3sin^2(x)

OpenStudy (anonymous):

i got it thanks 6sinxcos^2x-3sin^3x

OpenStudy (anonymous):

gotta go to school now!

OpenStudy (cwrw238):

thats correct

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