Zeigen Sie: Die Funktion \(z=a \cdot \mathrm{e}^{x / y}\) erfullt die Gleichung \(x z_{x}+y z_{y}=0\) ( \(a:\) Konstante).

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one of its input variables changes, while the others are held constant.
For the given function \(z = a \, \text{e}^{x / y}\), we find the partial derivatives with respect to both \(x\) and \(y\).
An important rule to remember is that the derivative of \( \text{e}^{u} \) with respect to \(u\) remains \( \text{e}^{u} \).
When differentiating \( \text{e}^{x / y} \), the chain rule is applied. The partial derivative with respect to \(x\) leads to multiplying by \( 1 / y \), and the partial derivative with respect to \(y\) results in multiplication by \(-x / y^2\). These calculations are crucial for solving the provided exercise.

The exponential function \( \text{e}^{u} \) is ubiquitous in mathematics, especially in calculus.
It grows rapidly and has unique properties like \( \frac{d}{du} \text{e}^{u} = \text{e}^{u} \).
In the context of partial differential equations, the exponential function often arises. For our function \(z = a \, \text{e}^{x / y}\), the exponential term \( \text{e}^{x / y} \) describes how \(z\) scales with \(x\) and \(y\).
Understanding its behavior is key to performing the differentiations in the given exercise.

Differential calculus focuses on how functions change and includes techniques for finding rates of change and slopes of curves.
In this exercise, we apply differential calculus to find partial derivatives which tell us how \(z\) changes with each variable individually.

• We start by differentiating with respect to \(x\), leading to \( z_{x} = a \, \text{e}^{x / y} \, (1 / y) \).
• Then, with respect to \(y\), yielding \( z_{y} = a \, \text{e}^{x / y} \, (-x / y^2) \).

These derivative forms are then plugged into the original equation \( x z_{x} + y z_{y} = 0 \), simplifying to prove the original statement.

A mathematical proof is a logical argument that demonstrates the truth of a given statement using previously established facts and principles.
Here, we use partial derivatives and properties of exponential functions to show that \( x z_{x} + y z_{y} = 0 \).
This involves:

• Substituting the partial derivatives \( z_{x} \) and \( z_{y} \) into the equation.
• Simplifying the resulting expression.
• Showing that each term cancels out to zero.

This step-by-step verification strengthens our confidence in the solution's accuracy and showcases the power of differential calculus.