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)
To differentiate each of these functions, we will use the chain rule and the product rule.
a) y = cos³x
To differentiate this function, we apply the chain rule. The derivative of cos(x) is -sin(x), so we have:
dy/dx = 3 * cos²x * (-sin(x))
= -3sin(x) * cos²x
b) y = sin²x * cos(3x)
To differentiate this function, we use the product rule. The derivative of the first function (sin²x) with respect to x is 2sin(x)cos(x). The derivative of the second function (cos(3x)) with respect to x is -3sin(3x). Using the product rule formula (d(uv)/dx = u * dv/dx + v * du/dx), we have:
dy/dx = (2sin(x)cos(x)) * cos(3x) + sin²x * (-3sin(3x))
= 2sin(x)cos²(x)cos(3x) - 3sin²(3x)sin²(x)
c) y = sin(x³)
To differentiate this function, we apply the chain rule. The derivative of sin(x) is cos(x), so we have:
dy/dx = 3x² * cos(x³)
These are the derivatives of the given functions.