Differentiate f(x) = 2sin(x^2)*cos(x^2).

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

We'll use the product rule and the
chain rule:

(u*v)' = u'*v +
u*v'

We'll put u(x) = 2sin(x^2) => u'(x) = 2(cos
x^2)*(x^2)'

u'(x) = 4x*cos
x^2

v(x) = cos(x^2) => v'(x) =
[-sin(x^2)]*(x^2)'

v'(x) =
-2x*sin(x^2)

We'll substitute f,g,f',g' in the expression
of product:

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

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

We'll factorize by
4x:

f'(x) = 4x[(cos x^2)^2 - (sin
x^2)^2]

f'(x) = 4xcos
(2x^2)