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Chapter 4: Problem 19

1-44. Find the derivative of each function. $$ f(x)=\ln e^{2 x} $$

Short Answer

Expert verified
The derivative, \( f'(x) \), is \( 2 \).

Step by step solution

01

Simplify the Function

The function provided is \( f(x) = \ln e^{2x} \). Recognize that using the logarithmic identity \( \ln a^b = b \ln a \), we can simplify the function. Since \( \ln e = 1 \), the function simplifies to \( f(x) = 2x \cdot \ln e = 2x \cdot 1 = 2x \). So, we have \( f(x) = 2x \).
02

Differentiate the Simplified Function

Now that we have simplified the function to \( f(x) = 2x \), find the derivative with respect to \( x \). Using the basic derivative rule where the derivative of \( ax \) with respect to \( x \) is \( a \), the derivative of \( 2x \) is \( 2 \). So, the derivative \( f'(x) = 2 \).

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Key Concepts

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

Logarithmic Identity
Understanding logarithmic identities is essential when working with exponential and logarithmic functions. A key identity that comes into play is \( \ln a^b = b \ln a \). This identity helps simplify expressions where a power is inside a logarithm.
For the function in the exercise, \( f(x) = \ln e^{2x} \), you can use the identity by recognizing that \( e^{2x} \) can be rewritten as \( 2x \ln e \). Knowing that \( \ln e = 1 \), the expression further simplifies to \( 2x \).
Helpful tips while using logarithmic identities:
  • The power of the expression moves as a coefficient in front of the logarithm.
  • It is crucial to be familiar with basic logarithmic properties such as \( \ln e = 1 \) for simplification.
  • Accuracy in rearranging terms is important to avoid mistakes in calculation.
Simplification
Simplification makes solving math problems more manageable and intuitive. In this case, simplifying \( f(x) = \ln e^{2x} \) to \( f(x) = 2x \) is crucial before finding the derivative.

Simplification strategies include:
  • Recognizing identities like \( \ln a^b = b \ln a \).
  • Using properties of logarithms such as \( \ln e = 1 \) to transform expressions.
  • Breaking down complex expressions into simpler, more understandable parts.
By simplifying early on, you reduce the complexity of differentiation and other calculus operations. Simplification can save time and effort by transforming the problem into a more familiar and straightforward form.
Basic Derivative Rule
Differentiation is a core concept in calculus that deals with finding the rate at which a function changes. The problem asks for the derivative of the simplified function \( f(x) = 2x \). Using the basic derivative rule, you can find the derivative quite easily.

In basic differentiation, one common rule is that the derivative of \( ax \), where \( a \) is a constant, is simply \( a \). This means:
  • For a linear function such as \( f(x) = 2x \), the derivative \( f'(x) \) is \( 2 \).
  • This basic rule allows for quick calculation of the rate of change.
  • Such rules form the foundation of more complex calculus operations and understanding.
Remembering and understanding these basic rules is crucial for tackling more complicated functions and applying calculus concepts effectively.

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Most popular questions from this chapter

73-76. Use your graphing calculator to graph each function on the indicated interval, and give the coordinates of all relative extreme points and inflection points (rounded to two decimal places). [Hint: Use NDERIV once or twice together with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=\frac{x}{e^{x}} \text { for }-1 \leq x \leq 5 $$

The population dynamics of many fish (such as salmon) can be described by the Ricker curve \(y=a x e^{-b x}\) for \(x \geq 0\) where \(a>1\) and \(b>0\) are constants, \(x\) is the size of the parental stock, and \(y\) is the number of recruits (offspring). Determine the size of the equilibrium population for which \(y=x\).

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\(61-62 .\) By calculating the first few derivatives, find a formula for the \(n\) th derivative of each function \((k\) is a constant \()\). $$ f(x)=e^{-k x} $$

BIOMEDICAL: Population Growth The Gompertz growth curve models the size \(N(t)\) of a population at time \(t \geq 0\) as \(N(t)=K e^{-a e^{-b t}}\) where \(K\) and \(b\) are positive constants. Show that \(\frac{d N}{d t}=b N \ln \left(\frac{K}{N}\right)\) and interpret this derivative to make statements about the population growth when \(NK\)
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