Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 10
Find the gradient \(\nabla f\). $$ f(x, y, z)=x z \ln (x+y+z) $$
Short Answer
Expert verified
The gradient \( \nabla f \) is \( \left( z \ln(x+y+z) + \frac{xz}{x+y+z}, \frac{-xz}{x+y+z}, x \ln(x+y+z) + \frac{xz}{x+y+z} \right) \).
Step by step solution
01
Identify the Function
The function given is \( f(x, y, z) = xz \ln(x+y+z) \). Our goal is to find the gradient \( abla f \), which consists of the partial derivatives \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), and \( \frac{\partial f}{\partial z} \).
02
Partial Derivative with respect to x
To find \( \frac{\partial f}{\partial x} \), apply the product rule: differentiate \( xz \) to get \( z \), and differentiate \( \ln(x+y+z) \) to get \( \frac{1}{x+y+z} \cdot 1 \). This gives us: \[ \frac{\partial f}{\partial x} = z \ln(x+y+z) + \frac{xz}{x+y+z}. \]
03
Partial Derivative with respect to y
The derivative \( \frac{\partial f}{\partial y} \) is found by differentiating \( f \) with respect to \( y \). The partial derivative of \( \ln(x+y+z) \) with respect to \( y \) is \( \frac{1}{x+y+z} \), and since \( f \) includes \( xz \) as a coefficient, we use: \[ \frac{\partial f}{\partial y} = \frac{-xz}{x+y+z}. \]
04
Partial Derivative with respect to z
For \( \frac{\partial f}{\partial z} \), again use the product rule: the partial derivative of \( x \) with respect to \( z \) is \( x \). This results in: \[ \frac{\partial f}{\partial z} = x \ln(x+y+z) + \frac{xz}{x+y+z}. \]
05
Combine to Form the Gradient
Combine these partial derivatives to get the gradient vector: \[ abla f = \left( z \ln(x+y+z) + \frac{xz}{x+y+z}, \frac{-xz}{x+y+z}, x \ln(x+y+z) + \frac{xz}{x+y+z} \right). \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In multivariable calculus, partial derivatives are crucial for understanding how a function changes as we modify one variable while keeping the others constant. In our exercise, the function is \( f(x, y, z) = xz \ln(x+y+z) \). Partial derivatives measure the rate of change of \( f \) with respect to \( x \), \( y \), and \( z \), individually.
- Partial Derivative with Respect to \( x \): This involves focusing on how \( f \) changes as \( x \) changes, keeping \( y \) and \( z \) constant. It involves applying rules like the product rule and chain rule to differentiate complex expressions.
- Partial Derivative with Respect to \( y \): Here, we differentiate with respect to \( y \) considering it as the variable and \( x \), \( z \) as constants. The result tells us how sensitive \( f \) is to small changes in \( y \).
- Partial Derivative with Respect to \( z \): Similarly, this derivative highlights how \( f \) responds to changes in \( z \), while \( x \) and \( y \) remain unchanged.
Multivariable Calculus
The exercise is an excellent showcase of multivariable calculus, a field of mathematics that extends the concepts of single-variable calculus to functions with several variables. When dealing with functions like \( f(x, y, z) = xz \ln(x+y+z) \), multivariable calculus allows us to explore how changes in multiple inputs affect the output. Here are some essential aspects:
- Functions of Several Variables: These are functions that depend on more than one variable鈥攊n our case, three: \( x \), \( y \), and \( z \).
- Gradient: The gradient is the vector composed of all the partial derivatives. It points in the direction of the greatest rate of increase of the function. For \( f \), it consolidates the partial derivatives determined for each variable.
- Application: Multivariable calculus is crucial in fields such as physics and engineering where systems are often dependent on several variables simultaneously.
Product Rule
The product rule is a fundamental principle in calculus used to differentiate expressions formed by the multiplication of two or more functions. In the given function \( f(x, y, z) = xz \ln(x+y+z) \), applying the product rule is essential. Let's see how it works:
- When to Use: Use the product rule when differentiating a product of two functions. In our case, \( xz \) and \( \ln(x+y+z) \) are multiplied together.
- How to Apply: The rule states that \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). Apply this to find the partial derivatives with respect to each variable.
- Differentiate \( xz \), keeping \( \ln(x+y+z) \) constant, and vice versa.
- For \( xz \ln(x+y+z) \), treating \( xz \) as \( u \) and \( \ln(x+y+z) \) as \( v \), the partial derivative results from differentiating both parts separately and combining them appropriately.
- Example: For \( \frac{\partial f}{\partial x} \), use the product rule to differentiate \( xz \) and \( \ln(x+y+z) \) individually, then sum them as per the rule's guidance.
