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Finding the Derivative by the Limit Process In Exercises \(11-24,\) find the derivative of the function by the limit process. $$ g(x)=-3 $$

Short Answer

Expert verified
The derivative of the function \(g(x) = -3\) is \(g'(x) = 0\).

Step by step solution

01

Prepare the Equation

The derivative of a function \(f(x)\) can be defined by the limit process:\n\[f'(x)=\lim_{{h -> 0}} \frac{f(x + h) - f(x)}{h}\]\nSubstitute \(g(x) = -3\) into this formula to obtain the difference quotient.
02

Substitute the Function

Substitute \(g(x) = -3\) to the difference quotient:\n\[\lim_{{h -> 0}} \frac{g(x + h) - g(x)}{h} = \lim_{{h -> 0}} \frac{-3 - (-3)}{h}\]
03

Simplify the Difference Quotient

Simplifying above expression gives: \n\[\lim_{{h -> 0}} 0 = 0.\] The difference quotient simplifies to zero, because the numerator and denominator are the same, leading to the fraction being equal to zero.
04

The Final Derivative

The derivative \(g'(x)\) is 0 as the limit of the difference quotient for \(g(x)\) as \(h\) approaches 0 is 0.\n\[\ g'(x) = 0.\]

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Key Concepts

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

Difference Quotient
The difference quotient is a crucial concept in calculus that serves as the foundation for finding derivatives. It measures the rate at which a function's output changes with respect to a change in its input. To calculate the difference quotient, you need the function value at a certain point and its value at a point a small distance away, typically expressed as \( x+h \).

In the example of \( g(x)=-3 \), we use the difference quotient formula to set up the structure for finding the derivative: \[ \lim_{{h \rightarrow 0}} \frac{g(x + h) - g(x)}{h}.\] However, with a constant function like \( g(x)=-3 \), the output value does not change as 'x' changes, leading to the numerator of the difference quotient being zero whenever \( h \) is not zero. This simplicity is what ultimately allows us to determine the derivative of a constant function so directly.
Limit Definition of Derivative
Understandably, the limit definition of the derivative can seem abstract at first glance. It calculates the derivative of a function at a particular point based on the difference quotient and the limit as \( h \) approaches zero. The formal definition echoes the fundamental principle of calculus, where the derivative at \( x \) is the limit of the difference quotient: \[ f'(x) = \lim_{{h \rightarrow 0}} \frac{f(x + h) - f(x)}{h}. \]

In simple terms, this process measures how the function's rate of change at a specific point converges as the interval \( h \) becomes infinitesimally small. So when the function \( g(x) \) is constant, as in the illustrative exercise, we observe that the rate of change is nonexistent, hence, the derivative equals zero.
Simplifying Expressions
An integral part of working with derivatives is the ability to simplify expressions accurately and effectively. Simplifying can range from straightforward calculations to more complex algebraic manipulations. When applying the limit process to find a derivative, simplification often involves evaluating or canceling terms within the difference quotient.

In the context of the given exercise, simplifying involves noticing that the constant terms in the numerator cancel each other out: \( -3 - (-3) = 0 \). This results in the simplification of the difference quotient to zero, and thus the limit as \( h \) approaches zero also equals zero, since any term divided by \( h \) that results in zero signifies no change as \( h \) grows smaller.
Constant Function Derivative
A constant function is one where the output value does not depend on the input value; it remains the same regardless of the input. The derivative of a constant function, understandably, is always zero. This is because derivatives measure the change in a function's output relative to a change in the input, and with a constant function, there is no change to measure.

In the exercise where \( g(x) \) is defined as a constant function: \( g(x) = -3 \), the derivative \( g'(x) \) found through the limit process is zero. This reinforces the core concept that constant functions do not vary, and their slopes on a graph are always horizontal, corresponding to a derivative of zero.

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Most popular questions from this chapter

A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat (see figure). (a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock? (b) Suppose the boat is moving at a constant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when there is a total of 13 feet of rope out. What happens to the speed at which the winch pulls in rope as the boat gets closer to the dock?

Proof Let \(u\) be a differentiable function of \(x .\) Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\)

Determining Differentiability In Exercises \(75-80\) , describe the \(x\) -values at which \(f\) is differentiable. $$ f(x)=\left\\{\begin{array}{ll}{x^{2}-4,} & {x \leq 0} \\ {4-x^{2},} & {x>0}\end{array}\right. $$

Think About It Let \(f\) and \(g\) be functions whose first and second derivatives exist on an interval I. Which of the following formulas is (are) true? (a) \(f g^{\prime \prime}-f^{\prime \prime} g=\left(f g^{\prime}-f^{\prime} g\right)^{\prime} \quad\) (b) \(f g^{\prime \prime}+f^{\prime \prime} g=(f g)^{\prime \prime}\)

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\tan x ; \frac{d x}{d t}=3 \text { feet per second }} \\\ {\begin{array}{llll}{\text { (a) } x=-\frac{\pi}{3}} & {\text { (b) } x=-\frac{\pi}{4}} & {\text { (c) } x=0}\end{array}}\end{array} $$

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