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Use the limit definition to find the derivative of the function. $$ g(s)=\frac{1}{3} s+2 $$

Short Answer

Expert verified
The derivative of the function \(g(s) = \frac{1}{3}s+2\) is \(\frac{1}{3}\).

Step by step solution

01

Define the function

The function is given as \(g(s) = \frac{1}{3}s+2\).
02

Apply the limit definition

Applying the limit definition of the derivative to this function gives \(\lim_{h \to 0} \frac{g(s+h)-g(s)}{h} = \lim_{h \to 0} \frac{(\frac{1}{3}(s+h)+2) - (\frac{1}{3}s+2)}{h}\)
03

Simplify the expression

Simplify the fraction to get \(\lim_{h \to 0} \frac{1}{3}\). The constant factor \(\frac{1}{3}\) can be moved outside the limit, so \(\lim_{h \to 0} \frac{1}{3} = \frac{1}{3}\)
04

Take the limit

As \(h\) approaches zero, the expression \(\frac{1}{3}\) stays constant. Therefore, the derivative of the function \(g(s) = \frac{1}{3}s+2\) is \(\frac{1}{3}\).

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Key Concepts

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

Limit Definition of Derivative
The limit definition of a derivative is a fundamental concept in calculus which helps us determine the rate at which a function is changing at any given point. The derivative represents the slope of a function's tangent line at a specific point. To find it, we use the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Here, \(h\) is a small increment that approaches zero.
  • It's about finding how much \(f(x)\) changes as \(x\) itself changes by a small amount \(h\).
  • This formula allows for calculating the derivative for any function, ensuring we capture how it scales.
When applying this concept to a linear function like \(g(s) = \frac{1}{3}s + 2\), we're calculating how much the function \(g(s)\) changes as \(s\) changes slightly. Always start by substituting \(g(s + h)\) and \(g(s)\) into the limit definition, then simplify. Remember that the result of the limit gives you the function's derivative, which, for linear functions, is just the slope of the line.
Linear Functions
Linear functions are among the simplest types of functions to work with. They are represented by the equation \(y = mx + b\), where:
  • \(m\) is the slope of the line, indicating the rate of change, or how steep the line is.
  • \(b\) is the y-intercept, indicating where the line crosses the y-axis.
For the function \(g(s) = \frac{1}{3}s + 2\):
  • The slope \(m\) is \(\frac{1}{3}\), showing that for every unit increase in \(s\), \(g(s)\) increases by \(\frac{1}{3}\).
  • The y-intercept \(b\) is 2, so the line crosses the y-axis at 2.
Linear functions are straightforward because their slope \(m\) is constant. This means whenever you differentiate a linear function using the limit definition, the answer is the slope \(m\) itself. In this way, linear functions act as a great starting point for understanding derivatives, as they constantly change at a uniform rate.
Basic Differentiation Rules
Differentiation is a key operation in calculus, offering powerful tools to analyze how functions behave. There are several basic rules that simplify the process of finding derivatives:
  • **Constant Rule:** The derivative of a constant is zero. For example, the derivative of 2 in \(g(s) = \frac{1}{3}s + 2\) is zero, because constants don’t change.
  • **Power Rule:** For any term \(ax^n\), the derivative is \(n \cdot ax^{n-1}\). This rule becomes handy when differentiating polynomial expressions.
  • **Sum/Difference Rule:** The derivative of a sum or difference is just the sum or difference of the derivatives. This makes calculating derivatives term by term viable.
These basic rules provide a straightforward method to differentiate functions without always resorting to the limit definition. However, understanding how to find derivatives using the limit definition builds a solid foundation, making it easier to apply these rules effectively.

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Most popular questions from this chapter

Marginal Profit When the price of a glass of lemonade at a lemonade stand was \(\$ 1.75,400\) glasses were sold. When the price was lowered to \(\$ 1.50,500\) glasses were sold. Assume that the demand function is linear and that the variable and fixed costs are \(\$ 0.10\) and \(\$ 25\), respectively. (a) Find the profit \(P\) as a function of \(x,\) the number of glasses of lemonade sold. (b) Use a graphing utility to graph \(P,\) and comment about the slopes of \(P\) when \(x=300\) and when \(x=700\). (c) Find the marginal profits when 300 glasses of lemonade are sold and when 700 glasses of lemonade are sold.

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Find the slope of the tangent line to \(y=x^{n}\) at the point \((1,1) .\) $$ \text { (a) } y=x^{3 / 2} \quad \text { (b) } y=x^{3} $$

Find \(f^{\prime}(x)\) $$ f(x)=\left(3 x^{2}-5 x\right)\left(x^{2}+2\right) $$

Use a graphing utility to graph the function and find its average rate of change on the interval. Compare this rate with the instantaneous rates of change at the endpoints of the interval. $$ f(x)=x^{3 / 2} ;[1,4] $$

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