how does thederivative of -2xcos(x^2)= -4xcos(x^2) -2sin(x^2) and not -4x^2cos(x^2) -2sin(x^2)?
Huh? If you let
u = x^2, du = 2x dx, so
∫ -2x cos(x^2) dx = ∫ -cosu du = -sinu = -sin(x^2)
I suspect a typo.
if y = (-2x)(cos(x^2))
then the derivative dy/dx = (-2x)(-2x)(sin(x^2)) + (-2)cos(x^2) , by the product rule
= (4x^2)sin(x^2) - 2cos(x^2)
Neither of your choices matches that, so like oobleck, I suspect a typo.
oops. I thought it said anti-derivative. Mental typo!
To find the derivative of the function -2xcos(x^2), we can use the product rule of differentiation. The product rule states that if we have a function u(x) multiplied by v(x), the derivative of the product is given by:
(d(uv))/(dx) = u'v + uv'
Let's apply the product rule to the function -2xcos(x^2):
u(x) = -2x
v(x) = cos(x^2)
First, we need to find the derivative of u(x) and v(x):
u'(x) = -2 (derivative of -2x with respect to x)
v'(x) = -sin(x^2) * 2x (chain rule: derivative of cos(x^2) with respect to x is -sin(x^2) times the derivative of x^2, which is 2x)
Now, we can apply the product rule:
(d(-2xcos(x^2)))/(dx) = u'v + uv'
= (-2) * cos(x^2) + (-2x) * (-sin(x^2) * 2x)
Simplifying this expression, we get:
= -2cos(x^2) + 4x^2sin(x^2)
So, the derivative of -2xcos(x^2) is -2cos(x^2) + 4x^2sin(x^2), not -4x^2cos(x^2) - 2sin(x^2).