Posted by **Julie** on Thursday, April 12, 2012 at 12:33am.

differentiate the function (

a) y = cos³x

b) y = Sin²xCos3x

c) y = Sin (x³)

- Calculus grade 12 -
**Reiny**, Thursday, April 12, 2012 at 7:40am
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

- Calculus grade 12 -
**Julie**, Thursday, April 12, 2012 at 12:47pm
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)

- Calculus grade 12 -
**Reiny**, Thursday, April 12, 2012 at 1:05pm
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)

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