Anonymous

h(x)=sin(2x)cos(2x) => h(x)=sin(4x)/2 => h '(x)=(1/2){d[sin(4x)]/dx}d(4x)/dx => h '(x)=(1/2)cos(4x)(4) => h '(x)=2cos(4x)

Anonymous

h (x) = f (x) g (x) h ` (x) = f `(x) g(x) + g`(x) f (x) h ` (x) = 2 cos (2x) cos (2x) - 2 sin (2x) sin (2x) h ` (x) = 2 [ cos²(2x) - sin²(2x) ] g ` (x) = 2 cos 4x

Anonymous

h = (1/2)sin(4x) dh/dx = 2cos(4x) That was the more compact solution but these identities will give equivalent forms. cos(4x) = cos^2(2x) - sin^2(2x) cos(4x) = 2cos^2(2x) - 1 cos(4x) = 1 - 2sin^2(2x)