Problem 49 - Calculus Early Transcendentals

Question

Find $$\partial z / \partial x \text { and } \partial z / \partial y$$ (a) \(z$$=f(x)+g(y)\) (b) \(z=f(x+y)\)

Step-by-Step Solution

Answer

(a) \( \partial z / \partial x = f'(x), \partial z / \partial y = g'(y); \) (b) \( \partial z / \partial x = f'(x+y), \partial z / \partial y = f'(x+y).\)

Step 1: Understanding the Problem

We are given two functions: (a) \( z = f(x) + g(y) \) and (b) \( z = f(x+y) \). We need to find the partial derivatives \( \partial z / \partial x \) and \( \partial z / \partial y \) for each function separately.

Step 2: Finding Partial Derivatives for (a)

Given \( z = f(x) + g(y) \), notice that \( z \) is the sum of two separate functions of \( x \) and \( y \). **For \( \partial z / \partial x \):** - Treat \( g(y) \) as a constant. - Differentiate \( f(x) \) with respect to \( x \): \( \partial z / \partial x = f'(x) \).**For \( \partial z / \partial y \):**- Treat \( f(x) \) as a constant. - Differentiate \( g(y) \) with respect to \( y \): \( \partial z / \partial y = g'(y) \).

Step 3: Finding Partial Derivatives for (b)

Given \( z = f(x+y) \), use the chain rule for differentiation.**For \( \partial z / \partial x \):**- Let \( u = x+y \), so \( z = f(u) \).- Differentiate \( z \) with respect to \( x \): \( \partial z / \partial x = f'(u) \cdot \partial u / \partial x = f'(x+y) \cdot 1 = f'(x+y) \).**For \( \partial z / \partial y \):**- Let \( u = x+y \), so \( z = f(u) \).- Differentiate \( z \) with respect to \( y \): \( \partial z / \partial y = f'(u) \cdot \partial u / \partial y = f'(x+y) \cdot 1 = f'(x+y) \).

Step 4: Summarizing the Results

From (a):- \( \partial z / \partial x = f'(x) \)- \( \partial z / \partial y = g'(y) \)From (b):- \( \partial z / \partial x = f'(x+y) \)- \( \partial z / \partial y = f'(x+y) \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus, especially when dealing with functions where one variable depends on another. If you have a composite function, say, \( z = f(g(x)) \), the chain rule helps in differentiating it by providing a structured process. Suppose you want to differentiate \( z \) with respect to \( x \). Here, you'd treat \( g(x) \) as an intermediate function, \( u \), and apply the chain rule:

• Find \( \frac{du}{dx} \) (differentiation of \( g(x) \) with respect to \( x \)).
• Then, differentiate \( f(u) \) with respect to \( u \), that is \( \frac{df}{du} \).
• Finally, apply the chain rule: \( \frac{dz}{dx} = \frac{df}{du} \times \frac{du}{dx} \).

This rule is particularly useful when differentiating a function such as \( z = f(x+y) \), where both \( x \) and \( y \) are inputs to \( f \). By treating \( x+y \) as a single variable \( u \), partial derivatives can be effortlessly calculated using the chain rule.

Function Differentiation

Function differentiation is all about finding the rate at which one quantity changes with respect to another. In simpler terms, it's determining how small changes in an input variable affect the function’s output. When working with a function like \( z = f(x) + g(y) \), the task is to find the derivatives to understand how \( z \) changes when either \( x \) or \( y \) changes.

• For a single-variable function, say \( y = h(x) \), the derivative \( \frac{dy}{dx} \) gives the slope, or the rate of change of \( y \) concerning \( x \).
• For multivariable functions, like \( z = f(x) + g(y) \), you often find partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). This process involves differentiating each function part independently.

Differentiation tells us how outputted variables respond to their inputs, and it’s foundational in calculus because it allows us to model real-world changes and predictions.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables. It’s crucial in analyzing scenarios where functions depend on two or more independent variables. For example, \( z = f(x, y) \) is a common form where you need to find out how changes in both \( x \) and \( y \) affect \( z \).

• Partial derivatives are your main tools here. They reveal how \( z \) changes concerning one variable at a time, holding the others constant.
• With partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), you gain insight into the behavior and rate of change of \( z \) in its respective directions.
• The handling of functions like \( z = f(x+y) \) shows the importance of understanding both the interaction between variables and the chain rule to simplify the process.

This field of calculus not only aids in theoretical analysis but is applied in numerous fields such as physics, engineering, and even economics to solve complex problems involving multiple influencing factors.