Find the first three derivatives of the function f(x)= 2cos x sin 2x?

Answer 1

Egad. For the first derivative, let u = cos(x) and v = sin(2x).

The first derivative is

2 * u * dv/dx + 2 * v * du/dx

dv/dx = 2cos(2x)
du/dx = -sin(x)

Putting it all together gives

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

Answer 2

First, we'll rewrite this a bit. Use the double-angle identity,

sin 2x = 2 sin x cos x.

Then f(x) = 4 sin x cos^2 x.

Now differentiate, using the product rule

(uv)' = v u' + u v'

to get

f'(x) = 4 (sin x cos^2 x)'
= 4(cos^2 x(sin x)' + sin x(cos^2 x)')
= 4 cos^3 x - 8 sin^2 x cos x
= 4 cos^3 x - 8(1 - cos^2 x)cos x
= 4 cos^3 x - 8 cos x + 8 cos^3 x
= 12 cos^3 x - 8 cos x.

Then,
f''(x) = -36 cos^2 x sin x + 8 sin x
and
f'''(x) = -36(cos^2 x sin x)' + 8 cos x
= -36(sin x(-2 cos x sin x) + cos^3 x) + 8 cos x
= 72 sin^2 x cos x - 36 cos^3 x + 8 cos x
= 72(1-cos^2 x)cos x - 36 cos^3 x + 8 cos x
= 80 cos x - 108 cos^3 x