Find the rate of change $$ \frac{d y}{d t} $$ at $$ y=1 $$, if $$ y=\sqrt{x+1} $$ and $$ \frac{d x}{d t}=9 $$. Write the exact answer. Do not round.

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We are given the relationship
$$
y = \sqrt{x+1}
$$
and need to find $$\frac{dy}{dt}$$ when $$y=1$$ given that $$\frac{dx}{dt} = 9$$.

**Step 1.** Differentiate $$y = \sqrt{x+1}$$ with respect to $$x$$ to find $$\frac{dy}{dx}$$.

Rewriting the equation, we have
$$
y = (x+1)^{1/2}.
$$
Differentiating with respect to $$x$$:
$$
\frac{dy}{dx} = \frac{1}{2}(x+1)^{-1/2} = \frac{1}{2\sqrt{x+1}}.
$$

**Step 2.** Use the chain rule to relate $$\frac{dy}{dt}$$ and $$\frac{dx}{dt}$$:
$$
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}.
$$

Substitute the expression for $$\frac{dy}{dx}$$ and the given value for $$\frac{dx}{dt}$$:
$$
\frac{dy}{dt} = \frac{1}{2\sqrt{x+1}} \cdot 9.
$$

**Step 3.** Determine $$x$$ when $$y = 1$$.

Since $$y = \sqrt{x+1}$$, set $$y = 1$$:
$$
1 = \sqrt{x+1}.
$$
Square both sides:
$$
1 = x+1 \quad \Longrightarrow \quad x = 0.
$$

**Step 4.** Substitute $$x = 0$$ into the expression for $$\frac{dy}{dt}$$:
$$
\frac{dy}{dt} = \frac{1}{2\sqrt{0+1}} \cdot 9 = \frac{1}{2\sqrt{1}} \cdot 9 = \frac{9}{2}.
$$

The exact rate of change when $$y=1$$ is 
$$
\frac{dy}{dt} = \frac{9}{2}.
$$
The rate of change $$ \frac{dy}{dt} $$ when $$ y = 1 $$ is $$ \frac{9}{2} $$.

Find the rate of change at , if and . Write the exact answer. Do not round.

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