x^2cosx-2xsinx-2cosx
To simplify the expression x^2cosx - 2xsinx - 2cosx, we can use some algebraic manipulation and trigonometric identities.
1. First, let's group the terms with cosx together and the terms with sinx together:
x^2cosx - 2cosx - 2xsinx
2. Next, we can factor out cosx from the first two terms:
cosx(x^2 - 2) - 2xsinx
3. Now, let's focus on the second term. We can rewrite -2xsinx as -2sinx * x.
cosx(x^2 - 2) - 2sinx * x
4. Since sinx * x is the same as xsinx, we can rewrite the expression as:
cosx(x^2 - 2) - 2xsinx
5. We now have a common factor of (x^2 - 2), so we can factor that out:
(x^2 - 2) * (cosx - 2sinx)
So the simplified expression is (x^2 - 2) * (cosx - 2sinx).
To get to this answer, we grouped the terms, factored out common factors, and rearranged the expression using trigonometric identities.
yup!
sloppiness on my part
d/dx sinx = cosx
I assume that you want the derivative, since you titles your post that way.
if y = x^2cosx-2xsinx-2cosx
dy/dx = (x^2)(-sinx) + 2x cosx - 2x(-cosx) - 2sinx + 2sinx
= -x^2( sinx) + 4x( cosx)
etc if needed