Calculus grade 12
posted by Julie on .
differentiate the function (
a) y = cos³x
b) y = Sin²xCos3x
c) y = Sin (x³)

a)
chain rule
dy/dx = 3(cos^2 x)(sinx)
= 3sinx cos^2 x
b) product rule
dy/dx = (sin^2 x)(3sin (3x) ) + (cos (3x))(2sinx(cosx))
= ....
c) you try it, let me know what you got 
Hey Reiny, I think b) is wrong.
I got:
b)y' = 2SinxCos3x + CosxCos3x * (Sinx)
y' = 2SinxCos3x  SinxCosxCos3x
... and I am not sure how to do c) 
don't know how you got your answer, I will stick to mine
derivative of sin^2 x or (sinx)^2 is
2sinx cosx
the derivative of cos 3x = (sin 3x)(3) = 3sin 3x
so dy/dx = (sin^2 x)(3sin (3x)) + cos(3x) (2sinxcosx)
like I had
all you have to do is trying to simplify it a bit.
3rd one:
y = sin (x^3)
dy/dx = cos (x^3) (3x^2)
= 3x^2 cos(x^3)