To solve this differential, firstly note that 2cos(x)sin(x) = sin(2x) (by the Double-Angle Sine Identity), this makes computing the differential a lot easier. To differentiate sin(2x) we need to use the chain rule so let, t = 2x then y = sin(t). Differentiating t = 2x with respect to x gives, dt/dx = 2. Differentiating y = sin(t) with respect to t gives, dy/dt = cos(t) Then by the chain rule, dy/dx = dy/dt * dt/dx. So dy/dx = cos(t) * 2 = 2cos(t). Write t in terms of x, we know from our definition of t that t = 2x. Therefore, dy/dx = 2cos(2x) So the differential of 2cos(x)sin(x) with respect to x is 2cos(2x). (Note: To check the answer try computing 2cos(x)sin(x) using the product rule.)