We have to differentiate f(x) = 2sin(x^2)*cos(x^2).

Now we know that 2 *sin x* cos x = sin 2x

Therefore 2*sin(x^2)*cos(x^2)

=> sin (2*x^2)

To find the derivative we use the chain rule.

=> f'(x) = (sin2x^2)'

=> f'(x) = (2x^2)'*cos(2x^2)

=> f'(x) = (4x)*cos(2x^2)

=> f'(x) = 4x*cos(2x^2)

**Therefore the required derivative is 4x*cos(2x^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)**