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Find \(f^{\prime}(x)\) by using the definition of the derivative. [Hint: See Example 4.] $$ \text { } f(x)=\frac{x}{2} $$

Short Answer

Expert verified
The derivative \( f^{\prime}(x) = \frac{1}{2} \).

Step by step solution

01

Write the Definition

The definition of the derivative of a function \( f(x) \) at a point \( x \) is given by \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]
02

Substitute the Function

Substitute \( f(x) = \frac{x}{2} \) into the derivative definition formula:\[ f(x+h) = \frac{x+h}{2}. \] This gives:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{x+h}{2} - \frac{x}{2}}{h}. \]
03

Simplify the Expression

Simplify the expression inside the limit:\[ \frac{x+h}{2} - \frac{x}{2} = \frac{x+h - x}{2} = \frac{h}{2}. \] Substituting back:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{h}{2}}{h} = \lim_{h \to 0} \frac{1}{2}. \]
04

Evaluate the Limit

Evaluate the limit:\[ f^{\prime}(x) = \frac{1}{2}. \]

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Key Concepts

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

Definition of Derivative
The derivative of a function is a key concept in calculus. It represents the rate of change of a function as its input varies. To find the derivative, we use its formal definition. The derivative of a function \( f(x) \) at a point \( x \) is defined as:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]This formula essentially calculates the change in the function, \( f(x) \), over a very small interval around \( x \). The expression inside the limit, \( \frac{f(x+h) - f(x)}{h} \), is called the difference quotient. As \( h \)—the interval length—approaches zero, the difference quotient approaches the derivative at \( x \).

This limit process provides a precise mathematical way of describing how a function behaves around any point on its curve. The derivative tells us whether the function is increasing or decreasing at a specific point and gives insights into the function's overall behavior.
Limit Process
The limit process is an essential part of understanding derivatives. When we talk about limits in calculus, we're exploring what happens to a function as its input gets closer and closer to a particular value. In the context of derivatives, we are interested in what happens as the change in our input variable, \( h \), tends to zero.

The limit process in our derivative formula, \( \lim_{h \to 0} \), means we're considering values of \( h \) that are getting infinitesimally small, effectively zero. This allows us to see the behavior of the function exactly at point \( x \), without actually dividing by zero. In practice, it provides a way to calculate the exact slope of the tangent line to the curve of \( f(x) \) at any given point.
  • For each small value of \( h \), evaluate the difference quotient \( \frac{f(x+h) - f(x)}{h} \).
  • As \( h \to 0 \), observe how this quotient behaves.
This is how we harness the power of limits to define derivatives, granting us powerful tools for understanding change and motion.
Simplifying Expressions in Calculus
Simplifying expressions is a pivotal step when calculating derivatives. It transforms complex algebraic expressions into simpler ones, making them easier to evaluate, especially within a limit.

For the function \( f(x) = \frac{x}{2} \), when we substitute the function into the definition of the derivative:\[ f(x+h) = \frac{x+h}{2} \],
we set up a difference quotient:\[ \frac{\frac{x+h}{2} - \frac{x}{2}}{h} \].
To simplify it:
  • First, combine like terms in the numerator: \( \frac{x+h}{2} - \frac{x}{2} = \frac{h}{2} \). Here, the \( x \) terms cancel out.
  • Using this simplified expression, our limit becomes \( \lim_{h \to 0} \frac{\frac{h}{2}}{h} \).
  • Simplify further to \( \lim_{h \to 0} \frac{1}{2} = \frac{1}{2} \), since \( \frac{h}{h} = 1 \).
By breaking down the steps and simplifying each part of the expression, we can quickly and accurately evaluate limits and find derivatives. This process highlights the beauty of calculus in transforming complicated expressions into elegant solutions using basic algebraic principles.

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Most popular questions from this chapter

Suppose that \(E(x)\) is a function such that \(E^{\prime}(x)=E(x)\). Use the Chain Rule to show that the derivative of the composite function \(E(g(x))\) is \(\frac{d}{d x} E(g(x))=E(g(x)) \cdot g^{\prime}(x) .\)

A study estimated how a person's social status (rated on a scale where 100 indicates the status of a college graduate) depended on years of education. Based on this study, with \(e\) years of education, a person's status is \(S(e)=0.22(e+4)^{2.1}\). Find \(S^{\prime}(12)\) and interpret your answer.

a. Graph the function \(f(x)=x^{2}-3 x+5\) on the window \([-10,10]\) by \([-10,10]\). Then use the DRAW menu to graph the TANGENT line at \(x=2\). Your screen should also show the equation of the tangent line. (If you did Exercise 45, this equation for the tangent line should agree with the one you found there.) b. Add to your graph the tangent line at \(x=1\), and the tangent lines at any other \(x\) -values that you choose.

The temperature in an industrial pasteurization tank is \(f(x)=\) \(x^{2}-8 x+110\) degrees centigrade after \(x\) minutes (for \(0 \leq x \leq 12\) ). a. Find \(f^{\prime}(x)\) by using the definition of the derivative. b. Use your answer to part (a) to find the instantaneous rate of change of the temperature after 2 minutes. Be sure to interpret the sign of your answer. c. Use your answer to part (a) to find the instantaneous rate of change after 5 minutes.

Blood Flow It follows from Poiseuille's Law that blood flowing through certain arteries will encounter a resistance of \(R(x)=0.25(1+x)^{4}, \quad\) where \(x\) is the distance (in meters) from the heart. Find the instantaneous rate of change of the resistance at: a. 0 meters. b. 1 meter.

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