For $y=\sqrt{3 x+9}$, find $d y$, given $x=2$ and $\Delta x=0.27$
\[
d y=
\]
(Round to the nearest thousandth as needed.)

Step 1 ：We are given the function \(y = \sqrt{3x + 9}\), and we are asked to find the differential of y, \(dy\), given \(x = 2\) and \(\Delta x = 0.27\).

Step 2 ：The differential \(dy\) is given by the derivative of y with respect to x, \(\frac{dy}{dx}\), times the change in x, \(\Delta x\). So, we first need to find the derivative of y with respect to x.

Step 3 ：The derivative of \(y = \sqrt{3x + 9}\) with respect to x is \(\frac{3}{2\sqrt{3x + 9}}\).

Step 4 ：Substituting \(x = 2\) into the derivative, we get \(\frac{\sqrt{15}}{10}\).

Step 5 ：Finally, we multiply the derivative at \(x = 2\) by \(\Delta x = 0.27\) to find the differential of y, \(dy\).

Step 6 ：\(dy = \frac{\sqrt{15}}{10} \times 0.27 = 0.105\).

Step 7 ：Final Answer: The differential of y, \(dy\), given \(x = 2\) and \(\Delta x = 0.27\) for the function \(y = \sqrt{3x + 9}\) is \(\boxed{0.105}\).