\( \begin{aligned} \text { If } f(x) & =3 x(\sin x+\cos x), \text { find } \\ f^{\prime}(x) & =\square \\ f^{\prime}(1) & =\square\end{aligned} \)

To find the derivative \( f'(x) \) of the function \( f(x) = 3x(\sin x + \cos x) \), we will use the product rule. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \] In our case, we can let: - \( u(x) = 3x \) - \( v(x) = \sin x + \cos x \) Now we need to find \( u'(x) \) and \( v'(x) \): 1. **Finding \( u'(x) \)**: \[ u'(x) = 3 \] 2. **Finding \( v'(x) \)**: \[ v'(x) = \cos x - \sin x \] Now we can apply the product rule: \[ f'(x) = u'v + uv' \] Substituting \( u \), \( u' \), \( v \), and \( v' \): \[ f'(x) = (3)(\sin x + \cos x) + (3x)(\cos x - \sin x) \] Now we can simplify this expression: \[ f'(x) = 3(\sin x + \cos x) + 3x(\cos x - \sin x) \] Distributing \( 3x \): \[ f'(x) = 3\sin x + 3\cos x + 3x\cos x - 3x\sin x \] Combining like terms, we have: \[ f'(x) = 3\sin x + 3\cos x + 3x\cos x - 3x\sin x \] Thus, the derivative is: \[ f'(x) = 3(\sin x + \cos x) + 3x(\cos x - \sin x) \] Next, we need to evaluate \( f'(1) \): Substituting \( x = 1 \): \[ f'(1) = 3(\sin 1 + \cos 1) + 3(1)(\cos 1 - \sin 1) \] This simplifies to: \[ f'(1) = 3(\sin 1 + \cos 1) + 3(\cos 1 - \sin 1) \] Combining the terms: \[ f'(1) = 3\sin 1 + 3\cos 1 + 3\cos 1 - 3\sin 1 \] This simplifies to: \[ f'(1) = 3\cos 1 + 3\cos 1 = 6\cos 1 \] Thus, the final answers are: \[ f'(x) = 3(\sin x + \cos x) + 3x(\cos x - \sin x) \] \[ f'(1) = 6\cos 1 \]