Question

differentiate the function (

a) y = cos³x

b) y = Sin²xCos3x

c) y = Sin (x³)

Answer 1

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

Answer 2

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)

Answer 3

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)