\( \left. \begin{array} { c } { f ( x ) = ( x + 2 ) \sqrt { 4 x - x ^ { 2 } } } \\ { f ^ { \prime } ( x ) = j } \end{array} \right. \)

To find the derivative \( f'(x) \) of the function \[ f(x) = (x + 2) \sqrt{4x - x^2}, \] we'll use the **product rule** and the **chain rule** from calculus. ### Step 1: Apply the Product Rule The product rule states that if you have a function \( f(x) = u(x) \cdot v(x) \), then the derivative \( f'(x) \) is: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x). \] Here, let: - \( u(x) = x + 2 \) - \( v(x) = \sqrt{4x - x^2} \) First, compute the derivatives \( u'(x) \) and \( v'(x) \): - \( u'(x) = \frac{d}{dx}(x + 2) = 1 \) - To find \( v'(x) \), we'll use the chain rule. ### Step 2: Apply the Chain Rule to \( v(x) \) The chain rule states that if \( v(x) = \sqrt{g(x)} \), then: \[ v'(x) = \frac{1}{2\sqrt{g(x)}} \cdot g'(x). \] Here, \( g(x) = 4x - x^2 \), so: - \( g'(x) = 4 - 2x \) - Thus, \[ v'(x) = \frac{1}{2\sqrt{4x - x^2}} \cdot (4 - 2x) = \frac{4 - 2x}{2\sqrt{4x - x^2}}. \] ### Step 3: Combine Using the Product Rule Now, substitute \( u'(x) \), \( u(x) \), and \( v'(x) \) into the product rule formula: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 1 \cdot \sqrt{4x - x^2} + (x + 2) \cdot \frac{4 - 2x}{2\sqrt{4x - x^2}}. \] ### Step 4: Simplify the Expression Combine the terms over a common denominator: \[ f'(x) = \frac{2(4x - x^2) + (x + 2)(4 - 2x)}{2\sqrt{4x - x^2}}. \] Calculate the numerator: \[ \begin{align*} 2(4x - x^2) &= 8x - 2x^2, \\ (x + 2)(4 - 2x) &= 4x + 8 - 2x^2 - 4x = 8 - 2x^2. \end{align*} \] So, \[ 8x - 2x^2 + 8 - 2x^2 = -4x^2 + 8x + 8. \] Therefore, \[ f'(x) = \frac{-4x^2 + 8x + 8}{2\sqrt{4x - x^2}}. \] ### Step 5: Simplify Further (Optional) Factor out a common factor in the numerator: \[ -4x^2 + 8x + 8 = -2(2x^2 - 4x - 4) = -2(x^2 - 2x - 2). \] Thus, the derivative simplifies to: \[ f'(x) = \frac{-2(x^2 - 2x - 2)}{\sqrt{4x - x^2}}. \] ### Final Answer The derivative of \( f(x) = (x + 2)\sqrt{4x - x^2} \) is: \[ f'(x) = \frac{-2x^2 + 4x + 4}{\sqrt{4x - x^2}}. \]