y = Sin(2x)Cos(x). Find dy/dx.

Assume base differentiation knowledge: Sin(x) = Cos(x), Cos(x) = -Sin(x)The question combines the chain and product rule. To begin, start by splitting the equation: Sin(2x)Cos(x) = Sin(2x) x Cos(x)The product rule formula is dy/dx = u(dv/dx) + v(du/dx), where in this case u = Sin(2x) and v = Cos(x).Firstly, work out du/dx: This is done using the chain rule. (The chain rule formula: y = f(g(x)), dy/dx = f'(g(x))g'(x))Use f() = Sin(), g(x) = 2x. f'() = Cos(), and g'(x) = 2. Combining these, you get Cos(2x)(2) = 2Cos(2x).dv/dx is slightly simpler as it does not involve the chain rule. Cos(x) = -Sin(x).Combining these for final values (u, v, du/dx, dv/dx) in the product rule formula gives:(Sin(2x))(-Sin(x)) + (Cos(x))(2Cos(2x)), which simplifies to 2Cos(2x)Cos(x) - Sin(2x)Sin(x).