Find the rate of change $$ \frac{d y}{d t} $$ at $$ y=4 $$, if $$ y=\sqrt{x+5} $$ and $$ \frac{d x}{d t}=3 $$

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Differentiate the equation  
$$
y = \sqrt{x+5}
$$  
with respect to $$ t $$ using the chain rule. We have  
$$
\frac{dy}{dt} = \frac{d}{dt} \sqrt{x+5} = \frac{1}{2\sqrt{x+5}} \cdot \frac{dx}{dt}.
$$

Since $$ y = \sqrt{x+5} $$, at $$ y = 4 $$ we have  
$$
4 = \sqrt{x+5}.
$$

Square both sides to solve for $$ x $$:  
$$
16 = x + 5 \quad \Longrightarrow \quad x = 11.
$$

Now substitute $$ x = 11 $$ and $$ \frac{dx}{dt} = 3 $$ into the derivative:  
$$
\frac{dy}{dt} = \frac{1}{2\sqrt{11+5}} \cdot 3 = \frac{1}{2\sqrt{16}} \cdot 3.
$$

Since $$\sqrt{16} = 4$$, this simplifies to  
$$
\frac{dy}{dt} = \frac{3}{2 \times 4} = \frac{3}{8}.
$$
The rate of change $$ \frac{dy}{dt} $$ at $$ y=4 $$ is $$ \frac{3}{8} $$.

Find the rate of change at , if and

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