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Magazine Chapter 13: Problem 53
Let \(z=f(x-y, y-x) .\) Show that \(\partial z / \partial x+\partial z / \partial y=0\)
Short Answer
Expert verified
We proved \( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \).
Step by step solution
01
Recognize Function Composition
The function given is \( z = f(x-y, y-x) \). This means that \( z \) is a function of two variables, \( u = x-y \) and \( v = y-x \). We will express partial derivatives of \( z \) using the chain rule for these new variables.
02
Compute Partial Derivative of z with Respect to x
Using the chain rule, we have:\[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \]Here, \( \frac{\partial u}{\partial x} = 1 \) and \( \frac{\partial v}{\partial x} = -1 \). Thus, the equation becomes:\[ \frac{\partial z}{\partial x} = f_u \cdot 1 + f_v \cdot (-1) = f_u - f_v \]
03
Compute Partial Derivative of z with Respect to y
Similarly, using the chain rule:\[ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} \]Here, \( \frac{\partial u}{\partial y} = -1 \) and \( \frac{\partial v}{\partial y} = 1 \). Therefore, we get:\[ \frac{\partial z}{\partial y} = f_u \cdot (-1) + f_v \cdot 1 = -f_u + f_v \]
04
Sum the Partial Derivatives
Now, adding the results from Step 2 and Step 3, we have:\[ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = (f_u - f_v) + (-f_u + f_v) \]Simplifying gives:\[ f_u - f_v - f_u + f_v = 0 \]
05
Conclude the Proof
Since the sum of the partial derivatives is zero, we have shown that:\[ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. When you have a function like \( z = f(u,v) \), where \( u \) and \( v \) are themselves functions of other variables like \( x \) and \( y \), the partial derivative describes how \( z \) changes as you vary one of these variables while keeping the others constant.
- The partial derivative of \( z \) with respect to \( x \) is calculated by noting how both intermediate variables, \( u \) and \( v \), depend on \( x \).
- This involves using the chain rule, which helps to "chain" the effect of \( x \) via \( u \) and \( v \) on \( z \).
- Understanding these derivatives gives insights into the influence each variable has on the function's output.
Function Composition
Function composition is a technique where one function is applied to the result of another. In our exercise, \( z = f(x-y, y-x) \) expresses this concept because the variables \( x \) and \( y \) are used to create intermediary variables \( u \) and \( v \), which are then passed into function \( f \).
because it simplifies complicated calculations by focusing Attention on each function鈥檚 contribution separately.
In our example, calculating \( \frac{\partial z}{\partial x} \) involved finding how
\( u \) and \( v \) are individually impacted by changes in \( x \).
- This composition allows complex functions to be broken down into simpler parts that are easier to differentiate and analyze.
- The chain rule is crucial here as it provides the method to connect changes in \( x \) or \( y \) through their effects on \( u \) and \( v \).
because it simplifies complicated calculations by focusing Attention on each function鈥檚 contribution separately.
In our example, calculating \( \frac{\partial z}{\partial x} \) involved finding how
\( u \) and \( v \) are individually impacted by changes in \( x \).
Mathematics Proofs
Mathematics proofs serve as logical arguments verifying That mathematical statements are true. The exercise uses a proof-like approach to show \( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \).
develops a deeper comprehension of mathematical concepts. This specific proof illustrates the consistency
and interdependence of our variables \( x \) and \( y \) within the given function \( f \).
- We first calculated each partial derivative separately using the chain rule and analyzed their dependency.
- Then, by adding these derivatives, we simplified the expression to demonstrate the desired result.
- Proofs in mathematics often follow this pattern: setting up claims, manipulating expressions, and showing endings that confirm the hypothesis.
develops a deeper comprehension of mathematical concepts. This specific proof illustrates the consistency
and interdependence of our variables \( x \) and \( y \) within the given function \( f \).
