Find the differential $$ d y $$ of the given function. $$ y=\frac{x+3}{3 x-4} $$ $$ d y=\square $$

Question

UpStudy AI · Accepted Answer

We are given
$$
y = \frac{x+3}{3x-4}.
$$
To find $$ dy $$, we first compute the derivative $$ y' $$ using the quotient rule.

Let
$$
u = x+3 \quad \text{and} \quad v = 3x-4.
$$
Then,
$$
u' = 1 \quad \text{and} \quad v' = 3.
$$

The quotient rule states that
$$
y' = \frac{u'v - v'u}{v^2}.
$$

Substitute the expressions:
$$
y' = \frac{1\cdot (3x-4) - 3\cdot (x+3)}{(3x-4)^2}.
$$

Simplify the numerator:
$$
1\cdot (3x-4) = 3x-4,
$$
$$
3\cdot (x+3) = 3x + 9.
$$
So,
$$
3x-4 - (3x+9) = 3x-4-3x-9 = -13.
$$

Thus,
$$
y' = \frac{-13}{(3x-4)^2}.
$$

Since $$ dy = y' \, dx $$, we have
$$
dy = \frac{-13}{(3x-4)^2} \, dx.
$$

Therefore,
$$
\boxed{dy = \frac{-13}{(3x-4)^2} \, dx.}
$$
$$ dy = \frac{-13}{(3x-4)^2} \, dx $$

Find the differential of the given function.