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Chapter 1: Problem 23

Find \(\frac{d y}{d x}\) $$ y=\frac{3 x}{4} $$

Short Answer

Expert verified
\( \frac{d y}{d x} = \frac{3}{4} \)

Step by step solution

01

Identify the Components

First, recognize that this function, \( y = \frac{3x}{4} \), is a linear function of the form \( y = mx + b \) where \( m = \frac{3}{4} \) and \( b = 0 \). The slope \( m \) is the coefficient of \( x \).
02

Differentiate the Function

Differentiate the function with respect to \( x \). The derivative of \( y = mx + b \) with respect to \( x \) is simply \( m \). So in this case, we differentiate \( y = \frac{3}{4}x \) to find the slope. The derivative is \( \frac{d y}{d x} = \frac{3}{4} \).

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Key Concepts

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

Understanding Linear Functions
A linear function is a mathematical expression that describes a straight line when graphed on a coordinate plane. It has the general form of \( y = mx + b \). In this formula:
  • \( m \) represents the slope of the line
  • \( b \) is the y-intercept, the point where the line crosses the y-axis
In our exercise, the function given is \( y = \frac{3x}{4} \), which is a perfect example of a linear function with a slope of \( \frac{3}{4} \) and an intercept \( b \) of 0. This is because there is no additional constant added to the function, thus indicating the line crosses through the origin point \((0,0)\). This simplicity makes linear functions easy to recognize and work with in various mathematical problems.
The Process of Differentiation
Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. When we differentiate a function, we're essentially calculating its derivative. The derivative gives us a function's slope at a specific point. For a linear function like \( y = mx + b \), the derivative is particularly straightforward.
The derivative of \( y = mx + b \) with respect to \( x \) is simply \( m \). This is because the slope of a straight line remains constant. To differentiate our exercise's function, \( y = \frac{3}{4}x \), we observe that the derivative is the slope, \( \frac{3}{4} \). Differentiation here involves recognizing the structure of a linear function and identifying the constant coefficient that acts as the slope.
What is the Slope?
The slope is a measurement of how steep a line is. It determines how much \( y \) changes with a change in \( x \). In mathematical terms, it's known as "rise over run," or the vertical change divided by the horizontal change between two points on a line.
  • A positive slope means the line is inclining upwards
  • A negative slope means it declines downwards
In our exercise with the line \( y = \frac{3x}{4} \), the slope is \( \frac{3}{4} \). This slope indicates that for every 4 units you move along the x-axis, you move 3 units on the y-axis. This direct measurement is crucial in understanding how linear functions behave and how they are affected by changes in their variables.

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Most popular questions from this chapter

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. $$ \lim _{x \rightarrow 10}\left(\frac{x^{2}+x-110}{x-10}\right) $$

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