Find the indicated derivative. If \( f(x)=\sqrt{3 x+2} \), find \( f^{\prime \prime \prime}(x) \) \( f^{\prime \prime \prime}(x)=\square \)

We start with the function \[ f(x)=\sqrt{3x+2}=(3x+2)^{\frac{1}{2}}. \] **Step 1: Compute \( f'(x) \)** Using the chain rule: \[ f'(x)=\frac{d}{dx}\left((3x+2)^{\frac{1}{2}}\right)=\frac{1}{2}(3x+2)^{-\frac{1}{2}}\cdot 3. \] Thus, \[ f'(x)=\frac{3}{2}(3x+2)^{-\frac{1}{2}}. \] **Step 2: Compute \( f''(x) \)** Differentiate \( f'(x) \): \[ f'(x)=\frac{3}{2}(3x+2)^{-\frac{1}{2}}. \] Using the chain rule again: \[ f''(x)=\frac{3}{2}\cdot\left(-\frac{1}{2}\right)(3x+2)^{-\frac{3}{2}}\cdot3. \] Simplify the constants: \[ \frac{3}{2}\cdot\left(-\frac{1}{2}\right)\cdot3 = -\frac{9}{4}, \] so that \[ f''(x)=-\frac{9}{4}(3x+2)^{-\frac{3}{2}}. \] **Step 3: Compute \( f'''(x) \)** Differentiate \( f''(x) \): \[ f''(x)=-\frac{9}{4}(3x+2)^{-\frac{3}{2}}. \] Differentiate using the chain rule where the derivative of \((3x+2)^{n}\) is \(n(3x+2)^{n-1}\cdot3\). Here, \(n=-\frac{3}{2}\): \[ f'''(x)=-\frac{9}{4}\cdot\left(-\frac{3}{2}\right)(3x+2)^{-\frac{5}{2}}\cdot3. \] Now, combine the constants: \[ -\frac{9}{4}\cdot\left(-\frac{3}{2}\right)=\frac{27}{8}, \] and multiplying by 3 gives: \[ \frac{27}{8}\cdot3=\frac{81}{8}. \] Thus, \[ f'''(x)=\frac{81}{8}(3x+2)^{-\frac{5}{2}}. \] \[ \boxed{f'''(x)=\frac{81}{8}(3x+2)^{-\frac{5}{2}}} \]