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Let \(f\) be a differentiable function of one variable, and let \(z=f\left(x^{2}+y^{2}\right) .\) Show that $$ y \frac{\partial z}{\partial x}-x \frac{\partial z}{\partial y}=0 $$

Short Answer

Expert verified
The expression \( y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} \) simplifies to zero.

Step by step solution

01

Understand the problem

We need to show that the expression \( y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} \) simplifies to zero. Given that \( z = f(x^2 + y^2) \), we will use partial differentiation and chain rule to achieve this.
02

Set up related variables

Given \( u = x^2 + y^2 \) where \( z = f(u) \), note that \( u \) is implicitly a function of \( x \) and \( y \). This relation will be essential when applying the chain rule for partial derivatives.
03

Differentiate partially with respect to x

Use the chain rule: \( \frac{\partial z}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x} \). Since \( u = x^2 + y^2 \), then \( \frac{\partial u}{\partial x} = 2x \), so \( \frac{\partial z}{\partial x} = \frac{df}{du} \cdot 2x \).
04

Differentiate partially with respect to y

Similarly, use the chain rule: \( \frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y} \). Since \( u = x^2 + y^2 \), then \( \frac{\partial u}{\partial y} = 2y \), so \( \frac{\partial z}{\partial y} = \frac{df}{du} \cdot 2y \).
05

Substitute into main expression

Substitute the expressions for \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) back into the main equation: \( y(2x \frac{df}{du}) - x(2y \frac{df}{du}) \).
06

Simplify the expression

Both terms in the expression contain \( \frac{df}{du} \). Factor it out: \( \frac{df}{du}(2xy - 2xy) = \frac{df}{du} \cdot 0 = 0 \). So, the expression simplifies to zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule in Partial Differentiation
The chain rule is a powerful tool in calculus, especially when dealing with composite functions. In the context of partial differentiation, it involves taking the derivative of a function with respect to multiple variables indirectly. To understand how the chain rule aids in partial differentiation, consider a function composed of functions of different variables.
For example, if you have a function like \( z = f(u) \), where \( u = x^2 + y^2 \) is another function involving \( x \) and \( y \), the chain rule helps you determine how changes in \( x \) or \( y \) influence \( z \), through \( u \).
  • The first step involves identifying the inner and outer functions. Here, \( u = x^2 + y^2 \) represents the inner function and \( f(u) \) the outer function.
  • Derive \( u \) with respect to each variable to capture how \( u \) varies. Thus, \( \frac{\partial u}{\partial x} = 2x \) and \( \frac{\partial u}{\partial y} = 2y \).
  • Next, apply the chain rule: \( \frac{\partial z}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x} \), and \( \frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y} \).
This systematic approach allows us to maintain clarity even with complex, multivariable functions.
Understanding Differentiable Functions
Differentiable functions are the cornerstone of calculus, allowing us to consider how a function behaves as its inputs change. A function is said to be differentiable if it can be represented in a linear, smooth manner at any point within its domain. This means it has a well-defined derivative.
  • This concept ensures continuity. Any function that’s differentiable must also be continuous, although the converse might not always hold.
  • In practical terms, for a function of one variable like \( f \), differentiability implies that the graph of \( f \) has no abrupt corners or jumps.
  • When considering a function of multiple variables like \( z = f(x^2 + y^2) \), differentiability means the set transformations or mappings happen smoothly as \( x \) and \( y \) vary.
In summary, differentiable functions provide a structured, predictable mapping of variables, crucial for tasks like partial differentiation.
Partial Derivatives and Their Application
Partial derivatives extend the concept of a derivative to functions of more than one variable. Instead of looking at the rate of change in one direction, partial derivatives tell us about the rate of change of a function with respect to each variable independently, holding other variables constant. Let’s break it down using the function \( z = f(x^2 + y^2) \) as an example:
  • To find the partial derivative of \( z \) with respect to \( x \), we differentiate \( z \) considering \( y \) constant. This gives us \( \frac{\partial z}{\partial x} = \frac{df}{du} \cdot 2x \).
  • Similarly, the partial derivative with respect to \( y \) is calculated by holding \( x \) constant, resulting in \( \frac{\partial z}{\partial y} = \frac{df}{du} \cdot 2y \).
Partial derivatives are crucial, especially in optimization and solving differential equations, as they allow us to understand how a function behaves with each independent variable.

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