If \( f(x)=2 x \sin x \cos x \), find \( f^{\prime}(x)=\square \)

To find the derivative of the function \( f(x) = 2x \sin x \cos x \), we can use the product rule and the chain rule. First, we can simplify \( f(x) \) using the double angle identity for sine: \[ \sin(2x) = 2 \sin x \cos x \] Thus, we can rewrite \( f(x) \): \[ f(x) = x \sin(2x) \] Now, we will differentiate \( f(x) \) using the product rule. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then: \[ (uv)' = u'v + uv' \] Let \( u(x) = x \) and \( v(x) = \sin(2x) \). We need to find \( u' \) and \( v' \): 1. \( u'(x) = 1 \) 2. To find \( v'(x) \), we use the chain rule: \[ v(x) = \sin(2x) \implies v'(x) = \cos(2x) \cdot (2) = 2 \cos(2x) \] Now we can apply the product rule: \[ f'(x) = u'v + uv' = (1)(\sin(2x)) + (x)(2 \cos(2x)) \] This simplifies to: \[ f'(x) = \sin(2x) + 2x \cos(2x) \] Thus, the derivative \( f'(x) \) is: \[ \boxed{\sin(2x) + 2x \cos(2x)} \]