Chapter 13: Problem 50
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 simplifies to 0 because \(2(yx - xy) = 0\).
Step by step solution
01
Define the Composite Function
We have the function \(z = f(u)\) where \(u = x^2 + y^2\). Since \(z\) is a composition of \(f\) and \(u\), \(z\) is differentiable because \(f\) is differentiable and \(x^2 + y^2\) is a differentiable function.
02
Apply the Chain Rule to Partial Derivatives
To find \(\frac{\partial z}{\partial x}\), use the chain rule: \(\frac{\partial z}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x}\). Similarly, for \(\frac{\partial z}{\partial y}\), we have \(\frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y}\).
03
Compute \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\)
Since \(u = x^2 + y^2\), compute \(\frac{\partial u}{\partial x} = 2x\) and \(\frac{\partial u}{\partial y} = 2y\).
04
Substitute Derivatives into Partial Derivatives
Substitute the derivatives of \(u\) into the chain rule expressions: \(\frac{\partial z}{\partial x} = \frac{df}{du} \cdot 2x\) and \(\frac{\partial z}{\partial y} = \frac{df}{du} \cdot 2y\).
05
Simplify the Expression
Calculate the left-hand side of the equation: \(y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = y (\frac{df}{du} \cdot 2x) - x (\frac{df}{du} \cdot 2y)\).
06
Factor and Simplify
Factor out the common term \(\frac{df}{du} \cdot 2\): \(y \cdot 2x - x \cdot 2y = 2(yx - xy) = 0\). Both terms \(yx\) and \(xy\) cancel each other out, leaving 0.
07
Conclude the Derivation
Therefore, the expression \(y \frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = 0\) holds true and is shown to be 0 through direct simplification.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In calculus, partial derivatives represent how a function changes as one of its independent variables is varied, while all other variables are held constant. When you have a function with several variables, like a multivariable function, each variable can affect the output independently.
In our exercise, the function we are dealing with is noted as \( z = f(x^2 + y^2) \) where \( u = x^2 + y^2 \). To find how this function changes with respect to each variable, we compute the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
The chain rule assists in finding these partial derivatives. The expression becomes:
In our exercise, the function we are dealing with is noted as \( z = f(x^2 + y^2) \) where \( u = x^2 + y^2 \). To find how this function changes with respect to each variable, we compute the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
The chain rule assists in finding these partial derivatives. The expression becomes:
- \( \frac{\partial z}{\partial x} = \frac{df}{du} \cdot \frac{\partial u}{\partial x} \)
- \( \frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial u}{\partial y} \)
Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. Simply put, this means you can find a tangent line at any point on the curve of the function, which represents the function's slope at that point. Being differentiable implies continuity, but it's a bit more specific.
In our exercise, it is given that \( f \) is a differentiable function of one variable. This characteristic ensures smoothness in the curve of \( f \), allowing us to compute derivatives without encountering points of discontinuity or sharp turns. When considering \( z = f(x^2 + y^2) \), this property extends to our composite function, meaning we can work with its derivatives reliably.
In our exercise, it is given that \( f \) is a differentiable function of one variable. This characteristic ensures smoothness in the curve of \( f \), allowing us to compute derivatives without encountering points of discontinuity or sharp turns. When considering \( z = f(x^2 + y^2) \), this property extends to our composite function, meaning we can work with its derivatives reliably.
- Differentiability means the ability to apply derivative techniques smoothly.
- This property is crucial for computations involving composition of functions.
Composite Function
Composite functions are functions within functions. They can be more complex, but the idea is straightforward: you have one function inside another. In many problems, including our exercise, dealing with composite functions is common because it allows a single dependent variable to depend on multiple independent ones simultaneously.
In this scenario, \( z = f(u) \) and \( u = x^2 + y^2 \). The composition ensures the behavior of \( z \) is influenced by both \( x \) and \( y \). A composite function is crucial when you want to express affects brought on by multiple variables succinctly.
In this scenario, \( z = f(u) \) and \( u = x^2 + y^2 \). The composition ensures the behavior of \( z \) is influenced by both \( x \) and \( y \). A composite function is crucial when you want to express affects brought on by multiple variables succinctly.
- Lets \( z \) depend on more than direct input variables — expanding applicability.
- Simplifies influence of complex relationships between variables when viewing output.