https://r2.erweima.ai/mathbot/image/1724597889897ggcb5jk5.jpg

Let's solve the problem given in the image. The problem states:

Given \( z = xy + xe^{\frac{y}{x}} \), prove that:
\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = xy + z
\]

### Step 1: Find the partial derivatives of \( z \) with respect to \( x \) and \( y \).

First, let's differentiate \( z \) with respect to \( x \):

\[
z = xy + xe^{\frac{y}{x}}
\]

**Partial derivative with respect to \( x \):**

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

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

For the first term:

\[
\frac{\partial}{\partial x} (xy) = y
\]

For the second term, use the product rule:

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

Now, differentiate \( e^{\frac{y}{x}} \) with respect to \( x \):

\[
\frac{\partial}{\partial x} \left( e^{\frac{y}{x}} \right) = e^{\frac{y}{x}} \cdot \frac{\partial}{\partial x} \left( \frac{y}{x} \right) = e^{\frac{y}{x}} \cdot \left( -\frac{y}{x^2} \right)
\]

Thus:

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

So the total derivative with respect to \( x \) is:

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

### Partial derivative with respect to \( y \):

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

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

For the first term:

\[
\frac{\partial}{\partial y} (xy) = x
\]

For the second term:

\[
\frac{\partial}{\partial y} \left( xe^{\frac{y}{x}} \right) = x \cdot \frac{\partial}{\partial y} \left( e^{\frac{y}{x}} \right)
\]

Differentiate \( e^{\frac{y}{x}} \) with respect to \( y \):

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

So:

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

### Step 2: Substitute the partial derivatives into the expression.

We want to prove:

\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = xy + z
\]

Substitute the values:

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

Distribute the terms:

\[
xy + x e^{\frac{y}{x}} - y e^{\frac{y}{x}} + xy + y e^{\frac{y}{x}}
\]

Simplify:

\[
xy + xy + x e^{\frac{y}{x}} = 2xy + x e^{\frac{y}{x}}
\]

Notice that:

\[
z = xy + xe^{\frac{y}{x}}
\]

So the expression simplifies to:

\[
2xy + x e^{\frac{y}{x}} - xy = xy + z
\]

This matches the right-hand side of the equation, proving the given expression is correct.

### Conclusion:
We have shown that:
\[
x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = xy + z
\]

Would you like more details on any step, or have any other questions?

### Related Questions:
1. What is the general approach to solving partial differential equations?
2. How do you apply the product rule in partial differentiation?
3. What are the common applications of this type of proof in physics or engineering?
4. Can this method be extended to functions of more than two variables?
5. How do higher-order partial derivatives factor into this type of proof?

### Tip:
When working with partial derivatives, always remember to keep track of the variables you are holding constant during differentiation.

Learn how to prove the derivative of z with respect to x and y for a given function z = xy + xe^(y/x). This problem covers partial differentiation, the product rule, and functions of two variables. Suitable for advanced undergraduate students and those studying differential calculus.

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