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)

To solve this problem, we need to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for the given implicit function:

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

### Step 1: Find \(\frac{\partial z}{\partial x}\)

We differentiate the equation implicitly with respect to \(x\):

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

Differentiating each term:

1. The derivative of \(x e^y\) with respect to \(x\) is \(e^y\).
2. The derivative of \(y e^z\) with respect to \(x\) is \(y e^z \frac{\partial z}{\partial x}\).
3. The derivative of \(\log x\) with respect to \(x\) is \(\frac{1}{x}\).
4. The derivative of constants like \(-2\) and \(-3 \log 2\) is 0.

So, we have:

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

Solving for \(\frac{\partial z}{\partial x}\):

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

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

### Step 2: Find \(\frac{\partial z}{\partial y}\)

Now, differentiate the equation implicitly with respect to \(y\):

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

Differentiating each term:

1. The derivative of \(x e^y\) with respect to \(y\) is \(x e^y\).
2. The derivative of \(y e^z\) with respect to \(y\) is \(e^z + y e^z \frac{\partial z}{\partial y}\).
3. The derivative of \(\log x\) with respect to \(y\) is 0 (since \(\log x\) is independent of \(y\)).

So, we have:

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

Solving for \(\frac{\partial z}{\partial y}\):

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

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

### Step 3: Evaluate at the given point \((x, y, z) = (1, \log 2, \log 3)\)

Substitute \(x = 1\), \(y = \log 2\), and \(z = \log 3\) into the expressions for \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

For \(\frac{\partial z}{\partial x}\):

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

For \(\frac{\partial z}{\partial y}\):

\[
\frac{\partial z}{\partial y} = \frac{-1 \cdot e^{\log 2} - e^{\log 3}}{\log 2 \cdot e^{\log 3}} = \frac{-2 - 3}{\log 2 \cdot 3} = \frac{-5}{3 \log 2}
\]

### Final Answer:
\[
\frac{\partial z}{\partial x} = \frac{-1}{\log 2}, \quad \frac{\partial z}{\partial y} = \frac{-5}{3 \log 2}
\]

Would you like more details on any of the steps, or do you have any other questions?

Here are 5 related questions for further exploration:
1. How do you find the partial derivative of a multivariable function?
2. What is implicit differentiation, and how is it used in finding partial derivatives?
3. How do you apply the chain rule in partial differentiation?
4. How can you interpret the physical meaning of \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) in this context?
5. What are the applications of partial derivatives in fields like economics or engineering?

**Tip:** When performing implicit differentiation, remember to treat other variables as constants when differentiating with respect to a particular variable.

Learn how to find partial derivatives 𝜕𝑧/𝜕𝑥 and 𝜕𝑧/𝜕𝑦 using implicit differentiation for the multivariable function x𝑒^𝑦 + 𝑦𝑒^𝑧 + log 𝑥 − 2 − 3 log 2 = 0 at the point (1, log 2, log 3). This problem involves advanced concepts in multivariable calculus.

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