5.Find 𝜕𝑧
𝜕𝑥 ,
𝜕𝑧
𝜕𝑦 using partial derivatives for
x𝑒
𝑦 + 𝑦𝑒
𝑧 + log 𝑥 − 2 − 3 log 2 = 0 𝑎𝑡 (1, log 2, log 3)

Given the equation:

\[
x e^y + y e^z + \log x - 2 - 3 \log 2 = 0
\]

We want to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the point \((x, y, z) = (1, \log 2, \log 3)\).

### 1. Finding \(\frac{\partial z}{\partial x}\):

First, implicitly differentiate the given equation with respect to \(x\), treating \(z\) as a function of \(x\) (i.e., \(z = z(x)\)):

\[
\frac{\partial}{\partial x}\left( x e^y + y e^z + \log x - 2 - 3 \log 2 \right) = 0
\]

Differentiating term by term:

\[
\frac{\partial}{\partial x} \left( x e^y \right) + \frac{\partial}{\partial x} \left( y e^z \right) + \frac{\partial}{\partial x} (\log x) = 0
\]

\[
e^y + y e^z \frac{\partial z}{\partial x} + \frac{1}{x} = 0
\]

Now, solve for \(\frac{\partial z}{\partial x}\):

\[
y e^z \frac{\partial z}{\partial x} = -\left( e^y + \frac{1}{x} \right)
\]

\[
\frac{\partial z}{\partial x} = -\frac{e^y + \frac{1}{x}}{y e^z}
\]

### 2. Finding \(\frac{\partial z}{\partial y}\):

Next, implicitly differentiate the given equation with respect to \(y\), treating \(z\) as a function of \(y\) (i.e., \(z = z(y)\)):

\[
\frac{\partial}{\partial y}\left( x e^y + y e^z + \log x - 2 - 3 \log 2 \right) = 0
\]

Differentiating term by term:

\[
\frac{\partial}{\partial y} \left( x e^y \right) + \frac{\partial}{\partial y} \left( y e^z \right) + \frac{\partial}{\partial y} (\log x) = 0
\]

\[
x e^y + e^z + y e^z \frac{\partial z}{\partial y} = 0
\]

Now, solve for \(\frac{\partial z}{\partial y}\):

\[
y e^z \frac{\partial z}{\partial y} = -\left( x e^y + e^z \right)
\]

\[
\frac{\partial z}{\partial y} = -\frac{x e^y + e^z}{y e^z}
\]

### 3. Evaluate at the Point \((x, y, z) = (1, \log 2, \log 3)\):

Substitute \(x = 1\), \(y = \log 2\), and \(z = \log 3\) into the partial derivatives:

#### \(\frac{\partial z}{\partial x}\) at \((1, \log 2, \log 3)\):

\[
\frac{\partial z}{\partial x} = -\frac{e^{\log 2} + \frac{1}{1}}{\log 2 \cdot e^{\log 3}}
\]

\[
\frac{\partial z}{\partial x} = -\frac{2 + 1}{\log 2 \cdot 3} = -\frac{3}{3 \log 2} = -\frac{1}{\log 2}
\]

#### \(\frac{\partial z}{\partial y}\) at \((1, \log 2, \log 3)\):

\[
\frac{\partial z}{\partial y} = -\frac{1 \cdot e^{\log 2} + e^{\log 3}}{\log 2 \cdot e^{\log 3}}
\]

\[
\frac{\partial z}{\partial y} = -\frac{2 + 3}{\log 2 \cdot 3} = -\frac{5}{3 \log 2}
\]

### Final Answers:
- \(\frac{\partial z}{\partial x} = -\frac{1}{\log 2}\)
- \(\frac{\partial z}{\partial y} = -\frac{5}{3 \log 2}\)

Would you like a more detailed explanation of any step?

Here are 5 related questions:
1. How do you interpret partial derivatives geometrically?
2. Can partial derivatives be used in optimization problems? How?
3. How do mixed partial derivatives work, and are they always equal?
4. How do you compute higher-order partial derivatives?
5. How do you solve implicit differentiation problems involving more variables?

**Tip:** Always double-check the point of evaluation in partial derivatives to ensure accuracy.

Learn how to find partial derivatives 𝜕𝑧/𝜕𝑥, 𝜕𝑧/𝜕𝑦 using implicit differentiation with detailed step-by-step solutions. This problem involves advanced calculus concepts, including the chain rule and implicit differentiation formula. Suitable for advanced college-level students.

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