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Chapter 3: Problem 77

Extending the Power Rule to \(n=\frac{1}{2}, \frac{3}{2}\) and \(\frac{5}{2}\) With Theorem 3.3 and Exercise \(76,\) we have shown that the Power Rule, \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\) applies to any integer \(n .\) Later in the chapter, we extend this rule so that it applies to any rational number \(n\). a. Explain why the Power Rule is consistent with the formula \(\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}\) b. Prove that the Power Rule holds for \(n=\frac{3}{2}\). (Hint: Use the definition of the derivative: \(\frac{d}{d x}\left(x^{3 / 2}\right)=\lim _{h \rightarrow 0} \frac{(x+h)^{3 / 2}-x^{3 / 2}}{h}\) c. Prove that the Power Rule holds for \(n=\frac{5}{2}\). d. Propose a formula for \(\frac{d}{d x}\left(x^{n / 2}\right)\), for any positive integer \(n\).

Short Answer

Expert verified
**Short Answer:** The Power Rule for the fractional exponents \(n = \frac{1}{2}, \frac{3}{2}, \frac{5}{2}\) is consistent and can be extended to any positive integer \(n\). The general formula for the derivative of \(x^{\frac{n}{2}}\) is \(\frac{d}{d x}\left(x^{\frac{n}{2}}\right) = \frac{n}{2} x^{\frac{n}{2} - 1}\).

Step by step solution

01

Part a: Explain Consistency with \(\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}\)

To show that the Power Rule is consistent with the formula \(\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}\), we can rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\). Now, using the Power Rule, calculate the derivative: The Power Rule states: \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) For \(n = \frac{1}{2}\), \(\frac{d}{d x}\left(x^{\frac{1}{2}}\right)=\frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-\frac{1}{2}}\) Since \(x^{-\frac{1}{2}} = \frac{1}{\sqrt{x}}\), the result is: \(\frac{d}{d x}(\sqrt{x}) = \frac{1}{2 \sqrt{x}}\) Therefore, the Power Rule is consistent with the given formula.
02

Part b: Prove Power Rule for \(n = \frac{3}{2}\)

To prove the Power Rule for \(n = \frac{3}{2}\), we will use the definition of the derivative: \(\frac{d}{d x}\left(x^{\frac{3}{2}}\right) = \lim_{h \rightarrow 0} \frac{(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}}{h}\) Let's simplify the expression inside the limit: \((x+h)^{\frac{3}{2}} - x^{\frac{3}{2}} = x^{\frac{1}{2}}((x+h)-x) = x^{\frac{1}{2}}h\) Now, substitute this expression back into the limit: \(\lim_{h \rightarrow 0} \frac{x^{\frac{1}{2}}h}{h} = x^{\frac{1}{2}} \lim_{h \rightarrow 0}\frac{h}{h}\) \(\lim_{h \rightarrow 0}\frac{h}{h} = 1\) Therefore, \(\frac{d}{d x}\left(x^{\frac{3}{2}}\right) = x^{\frac{1}{2}}\) Using the Power Rule with \(n = \frac{1}{2}\), \(nx^{n-1} = \frac{3}{2}x^{\frac{1}{2}}\) Thus, the Power Rule is proven for \(n = \frac{3}{2}\).
03

Part c: Prove Power Rule for \(n = \frac{5}{2}\)

Proving the Power Rule for \(n = \frac{5}{2}\) follows a similar approach as in part b. Use the definition of the derivative: \(\frac{d}{d x}\left(x^{\frac{5}{2}}\right) = \lim_{h \rightarrow 0} \frac{(x+h)^{\frac{5}{2}} - x^{\frac{5}{2}}}{h}\) Let's simplify the expression inside the limit: \((x+h)^{\frac{5}{2}} - x^{\frac{5}{2}} = x^{\frac{3}{2}}((x+h)^2-x^2) = x^{\frac{3}{2}}(2xh+h^2)\) Now, substitute this expression back into the limit: \(\lim_{h \rightarrow 0} \frac{x^{\frac{3}{2}}(2xh+h^2)}{h} = x^{\frac{3}{2}}(2x \lim_{h \rightarrow 0}\frac{h}{h} + \lim_{h \rightarrow 0}\frac{h^2}{h})\) \(\lim_{h \rightarrow 0}\frac{h}{h} = 1\) and \(\lim_{h \rightarrow 0}\frac{h^2}{h} = 0\) Therefore, \(\frac{d}{d x}\left(x^{\frac{5}{2}}\right) = 2x^{\frac{3}{2}}\) Using the Power Rule with \(n = \frac{5}{2}\), \(nx^{n-1} = \frac{5}{2}x^{\frac{3}{2}}\) Thus, the Power Rule is proven for \(n = \frac{5}{2}\).
04

Part d: Propose Formula for \(x^{\frac{n}{2}}\) Derivative

Based on our findings in parts a, b, and c, we can propose a formula for the derivative of \(x^{\frac{n}{2}}\) for any positive integer \(n\): \(\frac{d}{d x}\left(x^{\frac{n}{2}}\right) = \frac{n}{2} x^{\frac{n}{2} - 1}\)

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Key Concepts

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

Derivative of Power Functions
Understanding the derivative of power functions is fundamental in calculus, as it extends the ability to work with a variety of functions. The Power Rule, a cornerstone of differentiation rules, states that for any power function of the form \(x^n\), the derivative is obtained by multiplying the exponent by the function itself and then subtracting one from the exponent. In mathematical terms, the Power Rule is expressed as \(\frac{d}{dx}(x^n) = nx^{n-1}\).

When the power is a fraction, the Power Rule still applies. For example, the derivative of \(\sqrt{x}\) is \(\frac{1}{2}x^{-\frac{1}{2}}\) or \(\frac{1}{2\sqrt{x}}\) when written using positive exponents. This rule is incredibly powerful because it simplifies the differentiation process for polynomial functions and makes it easier to calculate slopes of tangents, rates of change, and other quantities associated with calculus.
Proofs in Calculus
Proofs play a critical role in calculus, as they are used to validate theorems and formulas that are widely used in the subject. Proving the Power Rule for non-integer exponents requires a solid understanding of limits and algebraic manipulation.

For instance, to prove the Power Rule for \(n = \frac{3}{2}\), one uses the limit definition of the derivative, setting up a limit expression and simplifying it to show that the result matches the Power Rule's prediction. Proofs like these involve carefully taking limits as \(h\) approaches zero and often require creative algebraic techniques to simplify complex expressions. They provide the rigorous foundation upon which calculus is built and give students the confidence that the formulas they use are reliable.
Limit Definition of the Derivative
The limit definition of the derivative is the formal way of determining the derivative of a function at a point. This definition states that the derivative of a function \(f(x)\) at a point \(x\) is \(\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\), provided this limit exists.

The process requires the computation of a difference quotient (the ratio of the change in the function's value to the change in the variable) and the taking of a limit as the change in the variable approaches zero. The Power Rule for derivatives is often derived and proven utilizing this fundamental approach, ensuring that it is not simply a rule of thumb but a rigorously established calculus tool.
Differentiation Rules
Differentiation rules in calculus are like shortcuts for finding derivatives without going through the limit process every time. Alongside the Power Rule, other rules such as the Product Rule, Quotient Rule, and Chain Rule allow for efficient differentiation of more complex functions.

For example, the Product Rule is used when taking the derivative of the product of two functions, and the Quotient Rule is applied when differentiating the quotient of two functions. Understanding these rules is essential for tackling a wide array of problems in calculus and for building upon more advanced concepts such as implicit differentiation and higher-order derivatives.

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Most popular questions from this chapter

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x},\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this assumption implies that the rule is true for \(k=n+1\). (Hint: Write \(e^{(n+1) x}\) as the product of two functions and use the Product Rule.)

Assuming that \(f\) is differentiable for all \(x,\) simplify \(\lim _{x \rightarrow 5} \frac{f\left(x^{2}\right)-f(25)}{x-5}.\)

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no points of intersection. It follows that for some real number \(1

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?
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