/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the indicated partial d... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the indicated partial derivatives at the given points. $$z=3 x y-x^{2},\left.\frac{\partial z}{\partial x}\right|_{(1,-2,-7)}$$

Short Answer

Expert verified
The partial derivative \( \left.\frac{\partial z}{\partial x}\right|_{(1,-2,-7)} \) is \(-8\).

Step by step solution

01

Expression for Partial Derivative

To find the partial derivative of the function \( z = 3xy - x^2 \) with respect to \( x \), we will differentiate \( z \) with respect to \( x \), while treating \( y \) as a constant. The expression for the partial derivative \( \frac{\partial z}{\partial x} \) is given by differentiating \( z \) in terms of \( x \).
02

Differentiate with Respect to x

The function \( z = 3xy - x^2 \) is differentiated with respect to \( x \). The derivative of \( 3xy \) with respect to \( x \) is \( 3y \) because \( y \) is treated as a constant. The derivative of \( -x^2 \) with respect to \( x \) is \( -2x \). Therefore, the partial derivative is: \[ \frac{\partial z}{\partial x} = 3y - 2x \]
03

Substitute the Point into the Partial Derivative

We need to evaluate \( \frac{\partial z}{\partial x} \) at the point \((1, -2, -7)\). Substitute \( x = 1 \) and \( y = -2 \) into the expression \( 3y - 2x \). This yields: \[ 3(-2) - 2(1) = -6 - 2 \]
04

Evaluate the Expression

Simplify the expression from the previous step: \[ -6 - 2 = -8 \] The value of the partial derivative of \( z \) with respect to \( x \) at the point \((1, -2, -7)\) is \( -8 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics that studies how things change. It provides the tools to examine and understand change through the concepts of differentiation and integration.
Differentiation focuses on finding the rate at which something changes, commonly applied as derivatives.
  • Derivatives measure how a function changes as its input changes. They can provide insight into rates of change, slopes, and tangents.
  • Integration, on the other hand, aggregates values, often used to find areas under curves or the total accumulated change.
Calculus lays the foundation for understanding complex mathematical relationships and is vital in fields such as physics, engineering, economics, and beyond, where it helps model and solve real-world problems.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions with more than one variable. It allows the examination and manipulation of multi-dimensional spaces where functions depend on several variables at once.
Multivariable calculus introduces the concept of partial derivatives:
  • Partial derivatives help us understand how a function changes with respect to one variable while keeping other variables constant. It is akin to taking a single slice out of a multi-dimensional object to study it individually.
  • This concept is pivotal when dealing with surfaces and rates of change in higher dimensions, enabling detailed analyses such as evaluating slopes and curvatures in three-dimensional space.
Applications include optimizing functions that depend on multiple inputs, such as finding the most efficient design parameters in engineering or determining maximum profits in economics.
Mathematical Functions
Mathematical functions describe relationships between sets of numbers or objects. They assign each input to exactly one output, making them fundamental in mathematics.
  • Functions can depend on one variable or several variables and are often expressed as equations, such as \(z = 3xy - x^2\) in our exercise.
  • They can depict simple relationships like linear functions or complex ones involving multiple variables or higher powers.
In our problem, understanding functions includes knowing variables: variables are placeholders for numbers we manipulate within these functions. Mastery of how functions work forms the groundwork for solving algebraic equations and is crucial in analyses involving dynamics, graphs, and data interpretation.

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