what is the 2nd derivative of f when f(x)= cos2x- sin ^2x?

Question

is it f''(x)= -6 cos2x

Answer 1

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

f''(x) = -4cos(2x) - 2cos(2x) = -6 cos(2x)

you are right.

Answer 2

1st derivative= -2sin2x - 2cosxsinx
2nd derivative= -4cos2x + 2(sin^2)x -2(cos^2)x
simplifies to= -4(2(cos^2)x -1) + 2(sin^2)x-2(cos^2)x
=-10(cos^2)x+4+2(1-(cos^2)x)
=-12(cos^2)x+6
=-6(2(cos^2)x-1)
=-6(cos2x)
yes, you are correct haha, I just wanted to practice my calculus

Answer 3

f(x) = cos2x - sin^2x
f ' (x) = -sin2x * (2) - 2sin x *(cos x)
f ' (x) = - 2sin2x - sin2x (sin 2x = 2 sin x cos x ; trig identity)
f ' (x) = -3sin2x
f " (x) = -3cos2x * 2 = -6cos2x

Answer 4

I assume you mean

f(x) = cos(2x) - sin²x

f'(x) = -2sin(2x) - 2(sinx)(cosx) = -3sin(2x)

f''(x) = -6cos(2x)

You are correct.