Problem 47 - Calculus Early Transcendentals

Question

\(45-48\) Assume that all the given functions are differentiable. If \(z=f(x-y),\) show that \(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=0\)

Step-by-Step Solution

Answer

The partial derivatives sum to zero: \( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \).

Step 1: Understanding the Function z

The function given is in the form of a composite function: \( z = f(x - y) \). This means that \( z \) is a function of a single variable \( u = x - y \), where \( u \) is a function of \( x \) and \( y \).

Step 2: Express the Partial Derivatives

To show \( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \), we need to express the partial derivatives of \( z \) with respect to both \( x \) and \( y \).

Step 3: Use the Chain Rule for Partial Derivatives

The chain rule helps us find the partial derivatives. First, for \( \frac{\partial z}{\partial x} \), we use the derivative of \( f \) with respect to \( u \):\[ \frac{\partial z}{\partial x} = \frac{df}{du} \cdot \frac{\partial (x-y)}{\partial x} = f'(u) \cdot 1 = f'(x-y) \]Similarly, for \( \frac{\partial z}{\partial y} \):\[ \frac{\partial z}{\partial y} = \frac{df}{du} \cdot \frac{\partial (x-y)}{\partial y} = f'(u) \cdot (-1) = -f'(x-y) \]

Step 4: Add the Partial Derivatives

Finally, add the two partial derivatives:\[ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = f'(x-y) + (-f'(x-y)) = 0 \]

Step 5: Conclusion: Verify the Result

Thus, we have shown that \( \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = 0 \), as required.

Key Concepts

Chain Rule

The chain rule is a fundamental tool in differential calculus that allows us to calculate the derivative of a composite function. In simpler terms, the chain rule helps us understand how changes in one variable affect another through a composite relationship. When dealing with functions of multiple variables, like in partial derivatives, the chain rule becomes particularly useful. In this exercise, we have a function \( z = f(x-y) \), which is a composite function because \( x-y \) is itself a function of \( x \) and \( y \).In partial differentiation, the chain rule can be expressed as follows:

• For \( \frac{\partial z}{\partial x} \), we differentiate the function \( f \) with respect to \( u \) (where \( u = x-y \)) and multiply by the derivative of \( u \) with respect to \( x \).
• For \( \frac{\partial z}{\partial y} \), we again differentiate \( f \) with respect to \( u \), but this time multiply by the derivative of \( u \) with respect to \( y \).

By applying the chain rule, we break the problem down into manageable steps, simplifying complex differentiation tasks. Here, it allows us to find partial derivatives of the composite function \( f(x-y) \) with respect to \( x \) and \( y \).

Differential Calculus

Differential calculus is the branch of mathematics that deals with the study of rates at which quantities change. It provides us with tools to find derivatives, which are fundamentally rate-of-change measures. In the context of functions with more than one variable, like those involving partial derivatives, differential calculus lets us look at how a particular variable affects the function while keeping others constant.In our exercise, we deal with a single expression \( z = f(x-y) \). We need to compute the partial derivatives of \( z \) with respect to \( x \) and \( y \). Differential calculus gives us the framework to do this by focusing on the rate of change of the function with respect to these variables:

• \( \frac{\partial z}{\partial x} \) measures how \( z \) changes as \( x \) changes, while \( y \) remains constant.
• \( \frac{\partial z}{\partial y} \) measures the change in \( z \) due to changes in \( y \) with \( x \) held constant.

The ultimate goal in this problem is to show that the sum of these partial derivatives equals zero, embodying an important concept in differential calculus concerned with balanced changes in composite systems.

Composite Functions

Composite functions are functions that are formed by combining two or more functions. In the context of our exercise, the function \( z = f(x-y) \) is a composite function where \( f \) is applied to the expression \( x-y \). Understanding composite functions is crucial because they allow us to tackle complex functional relationships by decomposing them into simpler parts.For instance, when you have a function like \( f(g(x)) \), you treat \( g(x) \) as an intermediary step in the process of evaluating \( f \). In our problem:

• \( u = x-y \) is considered the inner function.
• \( z = f(u) \) is the outer function.

The power of composite functions lies in their ability to represent intricate relationships in mathematics, allowing us to manage derivations and computations more effectively. The exercise leverages this by showing that even if each piece (or component function) behaves differently, their combined effect can obey identifiable and sometimes simpler rules, like seeing "zero" change through their sum of derivatives.