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Chapter 3: Problem 69

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The derivative of \(\log _{2} 9=1 /(9 \ln 2)\) b. \(\ln (x+1)+\ln (x-1)=\ln \left(x^{2}-1\right),\) for all \(x\) c. The exponential function \(2^{x+1}\) can be written in base \(e\) as \(e^{2 \ln (x+1)}\) d. \(\frac{d}{d x}\left(\sqrt{2}^{x}\right)=x \sqrt{2}^{x-1}\) e. \(\frac{d}{d x}\left(x^{\sqrt{2}}\right)=\sqrt{2} x^{\sqrt{2}-1}\)

Short Answer

Expert verified
Question: Determine whether the following statement is true or false: The derivative of \(\log_{2} 9 = \frac{1}{9\ln 2}\). Answer: False. Explanation: The given statement is false because \(\log_2 9\) is a constant, and the derivative of a constant is 0. The correct statement would be \(\frac{d}{dx}(\log_2{9}) = 0\).

Step by step solution

01

a. The derivative of \(\log _{2} 9=1 /(9 \ln 2)\)

Let's find the derivative of \(\log_2 9\) using the chain rule and properties of logarithms. Note that \(\log_2 9\) is a constant, not a function of \(x\). The derivative of a constant is 0, so the statement is false. The correct statement would be: $$\frac{d}{dx}\left(\log_2{9}\right) = 0$$
02

b. \(\ln (x+1)+\ln (x-1)=\ln \left(x^{2}-1\right),\) for all \(x\)

Let's use the properties of logarithms to simplify the left side of the equation: $$\ln(x+1)+\ln(x-1) = \ln((x+1)(x-1)) = \ln(x^2-1)$$ The statement is true because this is a valid property of logarithms, which states that \(\ln{a} + \ln{b} = \ln{(a \cdot b)}\). This only applies for domains of the functions where \(x+1 > 0\) and \(x-1 > 0\), which restricts the domain to \(x > 1\).
03

c. The exponential function \(2^{x+1}\) can be written in base \(e\) as \(e^{2 \ln (x+1)}\)

The given statement is incorrect. To write the exponential function \(2^{x+1}\) in base \(e\), we should use the property \(a^x = e^{x \ln a}\): $$2^{x+1} = e^{(x+1)\ln 2}$$ Thus, the correct expression in base \(e\) should be \(e^{(x+1) \ln 2}\).
04

d. \(\frac{d}{d x}\left(\sqrt{2}^{x}\right)=x \sqrt{2}^{x-1}\)

To find the derivative of \(\sqrt{2}^x\), we first rewrite it using base \(e\): $$\sqrt{2}^x = e^{x \ln{\sqrt{2}}} = e^{x\frac{1}{2}\ln{2}}$$ Now, we apply the chain rule to find the derivative: $$\frac{d}{dx}\left(\sqrt{2}^x\right) = e^{x\frac{1}{2}\ln{2}} \cdot \frac{1}{2}\ln{2} = \frac{1}{2}\ln{2}\cdot \sqrt{2}^x$$ The given statement is false. The correct derivative should be \(\frac{1}{2}\ln{2}\cdot \sqrt{2}^x\).
05

e. \(\frac{d}{d x}\left(x^{\sqrt{2}}\right)=\sqrt{2} x^{\sqrt{2}-1}\)

To find the derivative of \(x^{\sqrt{2}}\), we apply the power rule (for any function of the form \(x^a\), the derivative is \(ax^{a-1}\)): $$\frac{d}{dx}\left(x^{\sqrt{2}}\right) = \sqrt{2} x^{\sqrt{2}-1}$$ The statement is true, as the power rule applies to this case.

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Key Concepts

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

Logarithm Properties
Understanding the properties of logarithms can greatly simplify complex expressions. An important property is that the logarithm of a product is the sum of the logarithms. This property is stated as \( \ln{a} + \ln{b} = \ln{(a \cdot b)} \).

Similarly, the logarithm of a quotient is the difference of the logarithms: \( \ln{\frac{a}{b}} = \ln{a} - \ln{b} \). These rules make it easier to solve equations involving logarithms by simplifying them into more manageable forms.

Here are some practical applications of these properties:
  • If you have \( \ln(x+1) + \ln(x-1) \), you can use the product property to combine these into \( \ln((x+1)(x-1)) = \ln(x^2 - 1) \).
  • This property is valid across any logarithmic base, whether natural logarithms (base \(e\)) or common logarithms (base 10).
Understanding these principles is critical when solving logarithmic equations or integrating logarithmic functions in calculus.
Exponential Functions
Exponential functions are foundational in calculus and appear frequently in various branches of science and mathematics. An exponential function typically has the form \( f(x) = a^x \), where \( a \) is a positive constant.

The most important exponential function in calculus is the natural exponential function \( e^x \), where \( e \) is Euler's number, approximately 2.718. Exponential functions are characterized by their constant rate of growth or decay, making them pivotal in modeling real-life scenarios such as population growth and radioactive decay.

A practical way to convert any exponential expression into the base \(e\) is via the formula \( a^x = e^{x \ln a} \). This conversion is useful when taking derivatives, especially when the expression involves bases other than \(e\).
  • An example is converting \(2^{x+1}\) into \(e^{(x+1) \ln 2}\), which simplifies differentiation.
  • Understanding the transformation between different bases allows for flexibility in solving equations and computing derivatives.
With practice, handling exponential functions becomes second nature, aiding in solving advanced calculus problems.
Derivative of Power Functions
The derivative of power functions is a fundamental concept in calculus, leading to the power rule. This rule states that the derivative of \( x^n \) with respect to \( x \) is \( n x^{n-1} \), where \( n \) is any real number.

This rule simplifies finding derivatives of polynomials and is pivotal for calculus applications. When the exponent is any real number, this generalized rule still applies. For example, the derivative of \( x^{\sqrt{2}} \) is \( \sqrt{2} x^{\sqrt{2}-1} \), demonstrating the power rule's versatility.

The power rule makes it straightforward to:
  • Compute derivatives of standard polynomial functions, enhancing efficiency.
  • Determine slopes and rates of change at any point along polynomial curves.
Mastery of the power rule is essential for tackling more complex calculus tasks involving polynomial and non-integer exponents.

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Most popular questions from this chapter

A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance \(x\) (in inches) of the mass from its equilibrium position after \(t\) seconds is given by the function \(x(t)=10 \sin t-10 \cos t,\) where \(x\) is positive when the mass is above the equilibrium position. a. Graph and interpret this function. b. Find \(\frac{d x}{d t}\) and interpret the meaning of this derivative. c. At what times is the velocity of the mass zero? d. The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?

Tangent lines and exponentials. Assume \(b\) is given with \(b>0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, 28, Mar 1997).

Derivative of \(u(x)^{v(x)}\) Use logarithmic differentiation to prove that $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln u(x)+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$.

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

Suppose you own a fuel-efficient hybrid automobile with a monitor on the dashboard that displays the mileage and gas consumption. The number of miles you can drive with \(g\) gallons of gas remaining in the tank on a particular stretch of highway is given by \(m(g)=50 g-25.8 g^{2}+12.5 g^{3}-1.6 g^{4},\) for \(0 \leq g \leq 4\). a. Graph and interpret the mileage function. b. Graph and interpret the gas mileage \(m(g) / \mathrm{g}\). c. Graph and interpret \(d m / d g\).
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