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Chapter 4: Problem 98

Explain why it is obvious, without any calculation, that \(\frac{d}{d x} \ln e^{x}=1\).

Short Answer

Expert verified
The derivative is 1 because \( \ln(e^x) = x \) and \( \frac{d}{dx} x = 1 \).

Step by step solution

01

Recognize Basic Properties

Recall that the natural logarithm function, \( \ln(x) \), is the inverse of the exponential function, \( e^{x} \). This implies that for any real number \( x \), \( \ln(e^{x}) = x \).
02

Determine Function Generalized Form

The expression \( \ln(e^{x}) \) essentially simplifies to \( x \) because the logarithm and the exponential are inverse functions. This means that \( \ln(e^{x}) \) represents the identity function, \( f(x) = x \).
03

Apply Differentiation to Simplified Form

When you differentiate the function \( f(x) = x \) with respect to \( x \), you get \( \frac{d}{dx} x = 1 \). This is a fundamental derivative result because differentiation of \( x \) with respect to itself always yields \( 1 \).

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Key Concepts

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

Natural Logarithm Properties
The natural logarithm, denoted as \(\ln(x)\), is a logarithmic function that uses the base \(e\), where \(e\) is approximately equal to 2.718. One of the key properties of natural logarithms is how they relate to the exponential function. Specifically, \(\ln(x)\) is the inverse of the exponential function \(e^{x}\). This means that applying the natural logarithm to \(e^{x}\) will simply return \(x\).
  • \(\ln(e^{x}) = x\)
  • This property emphasizes how logarithms can "cancel" the effect of exponentials.
  • Understanding this can simplify working with complex algebraic expressions.
Overall, learning the basic properties of \(\ln(x)\) can significantly help in calculus, especially when you're dealing with functions involving \(e\). It not only clarifies the operations but also simplifies processes like differentiation.
Inverse Functions
Inverse functions are pairs of functions that "undo" each other's actions. If you have a function \(f\) and its inverse \(f^{-1}\), applying \(f\) and then \(f^{-1}\) will return the original input. For example, if \(f(x) = e^{x}\), then its inverse is \(f^{-1}(x) = \ln(x)\). When you apply \(\ln\) to \(e^{x}\), you get the initial value back, hence, \(\ln(e^{x}) = x\).
  • This concept greatly simplifies the process when working with differentiable functions.
  • It helps to focus on what the combinations of functions truly achieve.
Understanding inverse functions is essential in calculus because many derivative problems require the ability to switch between a function and its inverse geometrically and analytically.
Fundamental Derivative Rules
The fundamental derivative rules are building blocks for calculus, providing shortcuts for finding derivatives. One of the most basic rules is the derivative of \(x\) with respect to \(x\), which is always 1. This is because \(f(x) = x\) is a linear function with a constant rate of change.
  • The derivative of \(x\) with respect to \(x\) is \(\frac{d}{dx} x = 1\).
  • This rule is simple yet powerful, helping to solve more complex functions efficiently.
Applying this rule to is key in this exercise, as recognizing \(\ln(e^{x}) = x\) lets us directly find that \(\frac{d}{dx} \ln(e^{x}) = 1\). This principle is fundamental, as it appears repeatedly when mastering calculus and beyond.

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Most popular questions from this chapter

The weekly sales (in thousands) of a new product are predicted to be \(S(x)=1000-900 e^{-0.1 x}\) after \(x\) weeks. Find the rate of change of sales after: a. 1 week. b. 10 weeks.

For each demand function \(D(p)\) : a. Find the elasticity of demand \(E(p)\). b. Determine whether the demand is elastic, inelastic, or unit-elastic at the given price \(p\). $$ D(p)=\sqrt{175-3 p}, \quad p=50 $$

For each demand function \(D(p)\) : a. Find the elasticity of demand \(E(p)\). b. Determine whether the demand is elastic, inelastic, or unit-elastic at the given price \(p\). $$ D(p)=\sqrt{100-2 p}, \quad p=20 $$

A supply function \(S(p)\) gives the total amount of a product that producers are willing to supply at a given price \(p\). The elasticity of supply is defined as $$ E_{s}(p)=\frac{p \cdot S^{\prime}(p)}{S(p)} $$ Elasticity of supply measures the relative increase in supply resulting from a small relative increase in price. It is less useful than elasticity of demand, however, since it is not related to total revenue. Use the preceding formula to find the elasticity of supply for a supply function of the form \(S(p)=a e^{c p}\), where \(a\) and \(c\) are positive constants.

For each function, find the indicated expressions. \(f(x)=x^{2} \ln x-x^{2}, \quad\) find \(\quad\) a. \(f^{\prime}(x) \quad\) b. \(f^{\prime}(e)\)
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