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

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We are given that

$$
y=\sqrt{x+5}\quad\text{and}\quad \frac{dx}{dt}=3,
$$

and we need to find $$ \frac{dy}{dt} $$ when $$ y=4 $$.

**Step 1. Differentiate $$ y $$ with respect to $$ x $$:**

$$
y=(x+5)^{\frac{1}{2}}
$$

Differentiating with respect to $$ x $$ gives

$$
\frac{dy}{dx}=\frac{1}{2}(x+5)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x+5}}.
$$

**Step 2. Use the chain rule to compute $$ \frac{dy}{dt} $$:**

$$
\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}=\frac{1}{2\sqrt{x+5}}\cdot3.
$$

Thus,

$$
\frac{dy}{dt}=\frac{3}{2\sqrt{x+5}}.
$$

**Step 3. Determine $$ x $$ when $$ y=4 $$:**

Given $$ y=\sqrt{x+5} $$, set

$$
4=\sqrt{x+5}.
$$

Squaring both sides,

$$
16=x+5 \quad\Rightarrow\quad x=11.
$$

**Step 4. Substitute $$ x=11 $$ back into the expression for $$ \frac{dy}{dt} $$:**

$$
\frac{dy}{dt}=\frac{3}{2\sqrt{11+5}}=\frac{3}{2\sqrt{16}}=\frac{3}{2\cdot 4}=\frac{3}{8}.
$$

Thus, the rate of change at $$ y=4 $$ is

$$
\frac{dy}{dt}=\frac{3}{8}.
$$
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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