Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(f(x, y)=(2 x-y)^{4}\)

Step-by-Step Solution

Answer

The partial derivatives are \( \frac{\partial f}{\partial x} = 8(2x-y)^3 \) and \( \frac{\partial f}{\partial y} = -4(2x-y)^3 \).

Step 1: Understand Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while keeping the other variables constant. In this problem, we need to find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

Step 2: Find \( \frac{\partial f}{\partial x} \)

Treat \( y \) as a constant and differentiate \( f(x, y)=(2 x-y)^{4} \) with respect to \( x \). Using the chain rule, the derivative of \( (2x - y)^4 \) is \( 4(2x-y)^3 \cdot \frac{d}{dx}(2x-y) \). Since \( \frac{d}{dx}(2x-y) = 2 \), it follows that:\[\frac{\partial f}{\partial x} = 4(2x-y)^3 \cdot 2 = 8(2x-y)^3\]

Step 3: Find \( \frac{\partial f}{\partial y} \)

Treat \( x \) as a constant and differentiate \( f(x, y)=(2 x-y)^{4} \) with respect to \( y \). Using the chain rule, the derivative of \( (2x - y)^4 \) is \( 4(2x-y)^3 \cdot \frac{d}{dy}(2x-y) \). Since \( \frac{d}{dy}(2x-y) = -1 \), it follows that:\[\frac{\partial f}{\partial y} = 4(2x-y)^3 \cdot (-1) = -4(2x-y)^3\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. In simple terms, it is a method used to differentiate a function that is nested within another function. The key idea is to work from the outermost function inward, repeatedly applying the differentiation rules.
When applying the chain rule, if you have a composite function like \( f(g(x)) \), the derivative is determined by the formula:

• \( \frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)

This can be extended to functions of multiple variables, which is the case in our problem, where the function is \( f(x, y) = (2x - y)^4 \). Here, it's crucial to understand that although the expression looks complex, the rule simplifies the differentiation process considerably by breaking it into more manageable parts.
For instance, when calculating the partial derivative with respect to \( x \), keep \( y \) constant. The derivative first treats the outer function \( (2x - y)^4 \), and then it applies the derivative to the inside function \( 2x - y \). This method is systematically helpful when working through problems that involve intricate nested functions.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus beyond single-variable functions to functions with two or more variables, such as \( f(x, y) \). In this realm, partial derivatives become essential because they allow us to examine the individual influences that each variable has on the function's behavior.
One fundamental tool in multivariable calculus is the concept of treating all variables, except the one you are differentiating with respect to, as constants. This process helps to simplify the function's behavior when isolating the impact of one specific variable, like in our exercise with \( f(x, y) = (2x - y)^4 \).
Understanding how to take partial derivatives isn't just about knowing where each variable affects the function but also recognizing how changes in each variable affect the surface defined by the function. This insight is essential when analyzing real-world problems where multiple independent factors come into play, such as physics or economics. Mastering these skills equips students to tackle an array of complex, multidimensional challenges.

Differentiation

Differentiation, a core operation in calculus, is the process of finding a function that describes the rate at which one quantity changes with respect to another. In the context of partial derivatives as seen in the problem, differentiation plays a crucial role in understanding the intricate dependency of a function on its various variables.
Partial differentiation specifically deals with functions of several variables. By taking the derivative of a function with respect to one variable at a time, while holding others constant, insights into the behavior of multivariable functions are gained. For the function \( f(x, y) = (2x - y)^4 \), differentiation involves applying the rules we know, like the power rule and chain rule, to each variable. This often involves simplifying the problem by breaking it down into these smaller, more manageable calculations.
Differentiation allows us to model and solve real-world problems where varying factors need to be calculated. It gives us tools to understand how slight changes in one part of an equation can affect the whole, making it a powerful concept not only in mathematics but in any discipline requiring analytical rigor.