Problem 2
Question
Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\frac{2 x}{y}-\frac{3}{x y^{2}}\)
Step-by-Step Solution
Verified Answer
The partial derivatives are \( \frac{\partial f}{\partial x} = \frac{2}{y} + \frac{3}{x^2y^2} \) and \( \frac{\partial f}{\partial y} = -\frac{2x}{y^2} + \frac{6}{xy^3} \).
1Step 1: Identify the Function Components
The function given is \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \). We need to find the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). The function components are \( \frac{2x}{y} \) and \( -\frac{3}{xy^2} \).
2Step 2: Differentiate with Respect to x
For \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. The derivative of \( \frac{2x}{y} \) with respect to \( x \) is \( \frac{d}{dx}\left(\frac{2x}{y}\right) = \frac{2}{y} \). The derivative of \( -\frac{3}{xy^2} \) with respect to \( x \) is \( \frac{d}{dx}\left(-\frac{3}{xy^2}\right) = \frac{3}{x^2y^2} \). Thus, \( \frac{\partial f}{\partial x} = \frac{2}{y} + \frac{3}{x^2y^2} \).
3Step 3: Differentiate with Respect to y
For \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant. The derivative of \( \frac{2x}{y} \) with respect to \( y \) is \( \frac{d}{dy}\left(\frac{2x}{y}\right) = -\frac{2x}{y^2} \). The derivative of \( -\frac{3}{xy^2} \) with respect to \( y \) is \( \frac{d}{dy}\left(-\frac{3}{xy^2}\right) = \frac{6}{xy^3} \). Thus, \( \frac{\partial f}{\partial y} = -\frac{2x}{y^2} + \frac{6}{xy^3} \).
Key Concepts
Understanding Partial DerivativesExploring Multivariable CalculusThe Process of Function Differentiation
Understanding Partial Derivatives
In multivariable calculus, when we have functions that depend on more than one variable, we often want to see how the function changes with respect to one of these variables, while keeping others constant. This is where partial derivatives come in. They help us understand the rate of change of a multivariable function with respect to a single variable. In our given exercise, we compute the partial derivatives of the function \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \).
To find \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant and differentiate the function with respect to \( x \). Similarly, for \( \frac{\partial f}{\partial y} \), \( x \) is treated as a constant. This separation allows us to isolate the effect of one variable at a time, making it easier to analyze the behavior of the function in a multidimensional space.
To find \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant and differentiate the function with respect to \( x \). Similarly, for \( \frac{\partial f}{\partial y} \), \( x \) is treated as a constant. This separation allows us to isolate the effect of one variable at a time, making it easier to analyze the behavior of the function in a multidimensional space.
Exploring Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of more than one variable. It's integral in many fields like physics, engineering, and economics.
In our example, the function \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \) is a function of two variables. This means it takes an input pair \((x, y)\) and outputs a single value.
To understand such functions graphically, we imagine a surface in 3D space. The height of this surface at any point \( (x, y) \) depends on the value of the function.
By using partial derivatives, multivariable calculus provides tools for optimizing functions and modeling real-world phenomena that involve multiple changing factors.
In our example, the function \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \) is a function of two variables. This means it takes an input pair \((x, y)\) and outputs a single value.
To understand such functions graphically, we imagine a surface in 3D space. The height of this surface at any point \( (x, y) \) depends on the value of the function.
- Partial derivatives help us find slopes of this surface along the x or y directions.
- This is crucial for identifying peaks, valleys, and saddle points.
By using partial derivatives, multivariable calculus provides tools for optimizing functions and modeling real-world phenomena that involve multiple changing factors.
The Process of Function Differentiation
Function differentiation is a fundamental concept in calculus, and it involves finding the derivative, which measures how a function changes as its input changes. In the context of partial derivatives, this concept is extended to functions of several variables.
When differentiating functions like \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \), the differentiation process involves treating one variable as the main variable and the others as constants. Here’s a quick refresher on the rules used:
By applying these rules, we can systematically determine the rate of change for each variable independently, aiding in a comprehensive understanding of the behavior of complex functions.
When differentiating functions like \( f(x, y) = \frac{2x}{y} - \frac{3}{xy^2} \), the differentiation process involves treating one variable as the main variable and the others as constants. Here’s a quick refresher on the rules used:
- Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- Constant Rule: If \( f(x) = c \), where \( c \) is a constant, then \( f'(x) = 0 \).
- Product and Quotient Rules: These are used when dealing with products or quotients of functions.
By applying these rules, we can systematically determine the rate of change for each variable independently, aiding in a comprehensive understanding of the behavior of complex functions.
Other exercises in this chapter
Problem 1
In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=x^{3} y^{2} $$
View solution Problem 1
1\. Let \(f(x, y)=x^{2}+y^{2}\) with \(x(t)=3 t\) and \(y(t)=t^{2}\). Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=1\).
View solution Problem 2
Use the properties of limits to calculate the following limits: \(\lim _{(x, y) \rightarrow(-1.1)}\left(2 x y+y^{2}\right)\)
View solution Problem 2
Show that $$ c(x, t)=\frac{1}{\sqrt{2 \pi t}} \exp \left[-\frac{x^{2}}{2 t}\right] $$ solves $$ \frac{\partial c(x, t)}{\partial t}=\frac{1}{2} \frac{\partial^{
View solution