e rate of change $$ \frac{d y}{d t} $$ at $$ y=4 $$, if $$ y=\sqrt{x+5} $$ and $$ \frac{d x}{d t}= $$

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To find the rate of change $$ \frac{d y}{d t} $$ at $$ y=4 $$ given that $$ y=\sqrt{x+5} $$ and $$ \frac{d x}{d t} $$ is provided, we will follow these steps:

1. **Find the value of $$ x $$ when $$ y=4 $$**.
2. **Differentiate $$ y $$ with respect to $$ t $$** using the chain rule.
3. **Substitute the values to find $$ \frac{d y}{d t} $$**.

### Step 1: Find $$ x $$ when $$ y=4 $$

We start with the equation:
$$
y = \sqrt{x + 5}
$$
Setting $$ y = 4 $$:
$$
4 = \sqrt{x + 5}
$$
Now, we will square both sides to solve for $$ x $$:
$$
16 = x + 5
$$
Thus,
$$
x = 16 - 5 = 11
$$

### Step 2: Differentiate $$ y $$ with respect to $$ t $$

Using the chain rule:
$$
\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}
$$
First, we need to find $$ \frac{d y}{d x} $$:
$$
y = (x + 5)^{1/2}
$$
Differentiating with respect to $$ x $$:
$$
\frac{d y}{d x} = \frac{1}{2}(x + 5)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x + 5}}
$$

### Step 3: Substitute $$ x = 11 $$ into $$ \frac{d y}{d x} $$

Now we substitute $$ x = 11 $$:
$$
\frac{d y}{d x} = \frac{1}{2\sqrt{11 + 5}} = \frac{1}{2\sqrt{16}} = \frac{1}{2 \cdot 4} = \frac{1}{8}
$$

### Final Calculation

Now we can express $$ \frac{d y}{d t} $$:
$$
\frac{d y}{d t} = \frac{1}{8} \cdot \frac{d x}{d t}
$$

If you provide the value of $$ \frac{d x}{d t} $$, we can calculate $$ \frac{d y}{d t} $$.
To find $$ \frac{d y}{d t} $$ when $$ y = 4 $$ and $$ y = \sqrt{x + 5} $$, first find $$ x = 11 $$. Then, differentiate $$ y $$ with respect to $$ t $$ using the chain rule: $$ \frac{d y}{d t} = \frac{1}{8} \cdot \frac{d x}{d t} $$ So, $$ \frac{d y}{d t} = \frac{1}{8} \times \frac{d x}{d t} $$.

e rate of change at , if and

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