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.

Question

UpStudy AI · Accepted Answer

Given that

$$
y = \sqrt{x+1},
$$

we differentiate both sides with respect to $$ t $$ using the chain rule. First, write the function as

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

Differentiating with respect to $$ t $$, we have

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

This can be rewritten as

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

We are given $$ \frac{dx}{dt} = 9 $$ and need to find $$ \frac{dy}{dt} $$ when $$ y = 1 $$.

Since

$$
y = \sqrt{x+1},
$$

setting $$ y = 1 $$ gives

$$
1 = \sqrt{x+1}.
$$

Squaring both sides,

$$
1 = x+1,
$$

which simplifies to

$$
x = 0.
$$

Substitute $$ x=0 $$ and $$ \frac{dx}{dt}=9 $$ into the derivative expression:

$$
\frac{dy}{dt} = \frac{1}{2\sqrt{0+1}} \times 9 = \frac{1}{2} \times 9 = \frac{9}{2}.
$$

Thus, the exact rate of change is

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

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