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

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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,
$$
f'''(x)=\frac{81}{8}(3x+2)^{-\frac{5}{2}}.
$$

$$
\boxed{f'''(x)=\frac{81}{8}(3x+2)^{-\frac{5}{2}}}
$$
$$ f'''(x) = \frac{81}{8}(3x+2)^{-\frac{5}{2}} $$

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

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