/*! 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 3 Find \(\frac{d y}{d x}\). $$ y... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 3

Find \(\frac{d y}{d x}\). $$ y=-0.5 x $$

Short Answer

Expert verified
The derivative \( \frac{dy}{dx} = -0.5 \).

Step by step solution

01

Understand the Problem

We are given the function \( y = -0.5x \) and need to find the derivative \( \frac{dy}{dx} \). The derivative represents the rate of change of \( y \) with respect to \( x \).
02

Identify the Function Type

The function \( y = -0.5x \) is a linear function in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In this case, \( m = -0.5 \) and \( b = 0 \).
03

Apply the Derivative Rule for Linear Functions

For a linear function of the form \( y = mx + b \), the derivative is simply the coefficient \( m \) because the rate of change of a linear function is constant. According to this rule, \( \frac{d}{dx}(mx + b) = m \).
04

Compute the Derivative

Using the derivative rule from the previous step, the derivative of \( y = -0.5x \) is simply the slope \( m = -0.5 \). Thus, \( \frac{dy}{dx} = -0.5 \).

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Key Concepts

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

Linear Functions
Linear functions form the backbone of basic algebra and calculus. A linear function is represented by the equation \[ y = mx + b \] Here,
  • m is the slope of the line.
  • b is the y-intercept, where the line crosses the y-axis.
The slope, represented by m, indicates how steep the line is. It is essentially the "rise over run" which signifies how much y changes with a unit change in x. Because the function is linear, this rate of change is constant.
In the given exercise, we have a linear function with the equation \[ y = -0.5x \] This means that the slope m is \(-0.5\). Thus, for every 1 unit increase in x, y decreases by \(0.5\) units, implying a downward slope. The intercept b is \(0\), so the line passes through the origin.
Derivative Rules
The concept of derivatives is a foundational block in calculus, which quantifies how a function changes as its input changes. Specifically, when we talk about derivative rules for linear functions, we rely on a very straightforward principle: The derivative of a linear function \( y = mx + b \) is constant and equal to the slope \( m \).
  • Constant Rule: The derivative of a constant is \(0\). Since \(b\) is a constant, its derivative is \(0\).
  • Slope Rule: The derivative of \( mx + b \) is \( m \), the coefficient of \( x \). The derivative captures the consistent rate of change of \( y \) relative to \( x \).
Applying these rules, the derivative of our function \( y = -0.5x \) is straightforwardly the slope, \( m = -0.5 \). This process underscores the simplicity and elegance of linear functions in calculus, as their derivatives are simply their slopes.
Rate of Change
The rate of change is a vital concept in understanding how one variable affects another. In mathematics, particularly calculus, the rate of change of a function is determined by its derivative. For linear functions, like in our given exercise, this rate is constant.
In the context of the function \( y = -0.5x \), the constant rate of change is \(-0.5\), which means:
  • For each unit increase in \( x \), \( y \) decreases by \(0.5\) units.
  • This predictable change applies uniformly across all values of \( x \).
Understanding the constant nature of rate of change in linear functions equips students to predict and analyze patterns in data. The derivative thus acts as a lens through which we can view the dynamic nature of mathematical relationships, even in their simplest forms, providing powerful insights into how systems behave.

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Most popular questions from this chapter

At the beginning of a trip, the odometer on a car reads \(30,680,\) and the car has a full tank of gas. At the end of the trip, the odometer reads \(31,077 .\) It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=5 x^{2}-3 x+8 $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-0.01 x^{2}-0.5 x+70 $$

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=-0.01 x^{2}+2 x $$
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