Assume that $$ F(x, y, z(x, y))=0 $$ implicitly defines $$ z $$ as a differentiable function of $$ x $$ and $$ y $$. The partial derivatives of $$ z $$ are $$ \frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} $$ and $$ \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}} $$. Evaluate $$ \frac{\partial z}{\partial x} $$ and $$ \frac{\partial z}{\partial y} $$ for $$ x y z+2 x+3 y-5 z=0 $$. $$ \frac{\partial z}{\partial x}=\square $$

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We are given the implicit equation

$$
xy\,z + 2x + 3y - 5z = 0.
$$

Define

$$
F(x,y,z)=xy\,z + 2x + 3y - 5z.
$$

The formulas for the partial derivatives are

$$
\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}, \quad \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}},
$$

where $$ F_{x}=\frac{\partial F}{\partial x} $$, $$ F_{y}=\frac{\partial F}{\partial y} $$, and $$ F_{z}=\frac{\partial F}{\partial z} $$.

**Step 1. Compute $$ F_x $$:**

Differentiate $$ F(x,y,z) $$ with respect to $$ x $$:

$$
F_x = \frac{\partial}{\partial x} (xy\,z) + \frac{\partial}{\partial x} (2x) + \frac{\partial}{\partial x} (3y) - \frac{\partial}{\partial x} (5z).
$$

- For $$ xy\,z $$: since $$ y $$ and $$ z $$ are treated as functions or constants with respect to $$ x $$, we get $$ y\,z $$.
- For $$ 2x $$: the derivative is $$ 2 $$.
- The term $$ 3y $$ is independent of $$ x $$ so its derivative is $$ 0 $$.
- The term $$ -5z $$ gives $$ 0 $$ if we consider $$ z $$ as a variable that is implicitly defined (its derivative with respect to $$ x $$ is not needed in this partial differentiation because $$ F_x $$ holds $$ z $$ as an independent variable).

Thus,

$$
F_x = y\,z + 2.
$$

**Step 2. Compute $$ F_y $$:**

Differentiate $$ F(x,y,z) $$ with respect to $$ y $$:

$$
F_y = \frac{\partial}{\partial y} (xy\,z) + \frac{\partial}{\partial y} (2x) + \frac{\partial}{\partial y} (3y) - \frac{\partial}{\partial y} (5z).
$$

- For $$ xy\,z $$: the derivative is $$ x\,z $$.
- For $$ 2x $$: it does not depend on $$ y $$ so the derivative is $$ 0 $$.
- For $$ 3y $$: the derivative is $$ 3 $$.
- For $$ -5z $$: again, treat $$ z $$ as an independent variable for partial differentiation so the derivative is $$ 0 $$.

Thus,

$$
F_y = x\,z + 3.
$$

**Step 3. Compute $$ F_z $$:**

Differentiate $$ F(x,y,z) $$ with respect to $$ z $$:

$$
F_z = \frac{\partial}{\partial z} (xy\,z) + \frac{\partial}{\partial z} (2x) + \frac{\partial}{\partial z} (3y) - \frac{\partial}{\partial z} (5z).
$$

- For $$ xy\,z $$: the derivative is $$ xy $$.
- $$ 2x $$ and $$ 3y $$ are independent of $$ z $$ so their derivatives are $$ 0 $$.
- For $$ -5z $$: the derivative is $$-5$$.

Thus,

$$
F_z = xy - 5.
$$

**Step 4. Evaluate $$ \frac{\partial z}{\partial x} $$:**

Using

$$
\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}},
$$

substitute the computed values:

$$
\frac{\partial z}{\partial x} = -\frac{y\,z + 2}{xy - 5}.
$$

Thus, the answer is

$$
\frac{\partial z}{\partial x} = -\frac{y\,z + 2}{xy - 5}.
$$
$$ \frac{\partial z}{\partial x} = -\frac{y\,z + 2}{xy - 5} $$

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