what is the second derivative of cos^2?

Question

pleez help

Answer 1

y = cos^2 (x) = (cos(x)) ^2
so use general power rule

y' = 2 cos(x) (-sin(x)) = - 2 cos(x) sin(x)
now use product rule

y'' = (-2 cos(x)) (cos(x)) + sin(x) * -2 (-sin(x))
= -2 cos^2 (x) + 2 sin^2 (x) =
-2 (cos^2 (x) - sin^2 (x)) = -2 cos(2x)

Answer 2

That's a bit of a problem ... what you have is a floating function with no variable.

I'm going to assume you meant cos^2(x) though.

f(x) = cos^2(x)

Solving this for the first derivative involves the chain rule.

f'(x) = 2cos(x) [-sin(x)]. Simplifying, we get
f'(x) = -2sin(x)cos(x)

Solving for the second derivative involves the product rule. Remember that whenever we take the derivative, we can ignore the constant (in this case -2) and take the derivative of everything else.

Product rule verbally: "the derivative of the first times the second plus the derivative of the second times the first"

f''(x) = -2 [cos(x)cos(x) + (-sin(x)sin(x))
f''(x) = -2 [cos^2(x) - sin^2(x)]

This is actually a famous identity; the double angle formula. We can condense those difference of trig squares to cos2x.

f''(x) = -2cos(2x)

Answer 3

Use the chain rule. First derivative is -2 cos x sin x. This equals -sin 2x. The derivative of this is -2 cos 2x.

Answer 4

- 2 cos x

Answer 5

You are missing a variable. I am going to assume you meant cos² x.

d(cos² x)/dx = 2(cos x)(-sin x) = -2(sin x)(cos x)
d²(cos² x)/dx² = d[-2(sin x)(cos x)]/dx
= -2cos² x + 2sin² x = -2(cos² x - sin² x) = -2cos(2x)

Answer 6

-2sin^2 + 2cos^2 = 2(cos^2 - sin^2) = -2cos(2*theta)

Answer 7

2cosx(-sinx)
=-2sinxcosx
=-sin2x