$$ =\left(\frac{1 / 2}{x+9}\right)^{-1 / 2}\left(\frac{9}{(x+9)^{2}}\right) $$ Therefore, we have the following result. $$ y^{\prime}=\square $$

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We start with
$$
y'=\left(\frac{1/2}{x+9}\right)^{-1/2}\frac{9}{(x+9)^2}.
$$

**Step 1. Rewrite the expression inside the parentheses.**

Notice that
$$
\frac{1/2}{x+9}=\frac{1}{2(x+9)}.
$$

**Step 2. Simplify the negative exponent.**

Recall that for any positive quantity $$ A $$,
$$
A^{-1/2}=\frac{1}{\sqrt{A}}.
$$
Thus,
$$
\left(\frac{1}{2(x+9)}\right)^{-1/2}=\left(\frac{1}{2(x+9)}\right)^{-1/2}=\sqrt{2(x+9)}.
$$

**Step 3. Substitute back into the expression.**

Now, the derivative becomes
$$
y'=\sqrt{2(x+9)}\cdot\frac{9}{(x+9)^2}.
$$

**Step 4. Write $$\sqrt{2(x+9)}$$ as $$\sqrt{2}\sqrt{x+9}$$.**

$$
y'=\frac{9\sqrt{2}\sqrt{x+9}}{(x+9)^2}.
$$

**Step 5. Simplify the fraction.**

Recognize that
$$
\frac{\sqrt{x+9}}{(x+9)^2}=\frac{1}{(x+9)^{3/2}},
$$
because
$$
(x+9)^2=\left(\sqrt{x+9}\right)^4.
$$

So, we can write
$$
y'=\frac{9\sqrt{2}}{(x+9)^{3/2}}.
$$

**Final Result:**

$$
y'=\frac{9\sqrt{2}}{(x+9)^{3/2}}.
$$
$$ y' = \frac{9\sqrt{2}}{(x+9)^{3/2}} $$

\[ =\left(\frac{1 / 2}{x+9}\right)^{-1 / 2}\left(\frac{9}{(x+9)^{2}}\right) \] Therefore, we have the following result. \( y^{\prime}=\square \)

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