Asked by Anonymous
                How do I take the derivative of 3(sin(x))^2 cos(x)? Are these right--> 3(sin(x))^2 *-sin(x) + cos(x)*6sin(x)*cos(x)=(-3sin(x))^3 +6sin(x)(cos(x))^2?
            
            
        Answers
                    Answered by
            Reiny
            
    you are correct
y = [3(sin(x))^2][ cos(x)] <--- chain rule embedded in product rule
y' = 3(sin(x))^2 (-sinx) + (cosx)(6 sinx cosx) = -3sin^3 x + 6sinx cos^2 x
{ note sin^3 x = (sinx)^3 }
simplifying a bit
= -3sinx(sin^2 x - 2cos^2 x)
or
= -3sinx(1 - 3cos^2 x)
or
    
y = [3(sin(x))^2][ cos(x)] <--- chain rule embedded in product rule
y' = 3(sin(x))^2 (-sinx) + (cosx)(6 sinx cosx) = -3sin^3 x + 6sinx cos^2 x
{ note sin^3 x = (sinx)^3 }
simplifying a bit
= -3sinx(sin^2 x - 2cos^2 x)
or
= -3sinx(1 - 3cos^2 x)
or
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