could somebody check this derivative problem?

Question

Find the first and second derivative of this function.

f(x)= (4x²)(Cos²[3x])

I got f '(x)= 8x Cos²[3x]- 12x² Sin[6x] and
f ''(x)= 8 Cos²[3x] - 48x Sin[6x]-72x² Cos[6x], but I am not sure if either is correct.

Answer 1

f'(x) = 8xcos^2(3x) - 24x^2sin(3x)cos(3x)
= 8xcos(3x)[cos(3x) - 3xsin(3x)]

Same as your f'(x), just a different form.

f"(x) = 8cos^2(3x) - 96sin(3x)cos(3x) - 48xsin(3x)cos(3x) - 72x^2cos^2(3x) + 72x^2sin^2(3x)
= 72x^2 +8cos(3x)[cos(3x) - 48(2 + x)sin(3x)]

I think I did it all right. Easy enough to check.

Answer 2

Looks good to me. I can't remember if cos x dx = -sin x or sin x dx = -cos x, though.

Answer 3

y´=8xcos^2(3x)-4x^2*2cos(3x)*sin(3x)*3
y´=8x cos^2(3x)-12 x^2 sin(6x)

y¨ =8cos^2(3x)-8x*2cos(3x)sin(3x)*3-72x^2cos(6x)
y´´=8cos^2(3x)-8xsin(6x) -72x^2 cos(6x)

Answer 4

my math program gives
f'(x)=8xcos(3x)²-24x² sin(3x) cos(3x)

f"(x)=8(1-18x²) cos(3x)²-96x sin(3x) cos (3x) + 72x²

Answer 5

Yes both are correct

Answer 6

They are actually wrong, remember that (f(x)*g(x))'=f'(x)*g(x)+
f(x)*g'(x)
Check this:

f'(x)=(4x^2)'*(Cos[3x])^2+
(4x^2)*[(Cos[3x])^2]'

f'(x)=8x(Cos[3x])^2-
4x^2*2Cos[3x]*(Cos[3x])'

f'(x)=8x(Cos[3x])^2-
24x^2*Cos[3x]*Sin[3x]

and you will get for the second derivative:

f''(x)=8(Cos[3x])^2-
-72*x^2*(Cos[3x])^2-
-96*x*Cos[3x]*Sin[3x]+
+72*x^2*(Sin[3x])^2

Answer 7

They both look correct.