Problem 28 - Applied Calculus

Question

(a) Let \(f(x)=\ln x .\) Use small intervals to estimate \(f^{\prime}(1), f^{\prime}(2), f^{\prime}(3), f^{\prime}(4)\), and \(f^{\prime}(5)\). (b) Use your answers to part (a) to guess a formula for the derivative of \(f(x)=\ln x\)

Step-by-Step Solution

Answer

The derivative of \( f(x) = \ln x \) is \( f'(x) = \frac{1}{x} \).

Step 1: Understand the Function and Concept

We are given the function \( f(x) = \ln x \) and asked to estimate its derivative at certain points using small intervals. Recall that the derivative \( f'(x) \) gives the rate of change of the function at a particular point, and it can be estimated as \( f'(x) \approx \frac{f(x + \Delta x) - f(x)}{\Delta x} \) for a small value of \( \Delta x \).

Step 2: Estimate \( f'(1) \)

Choose a small \( \Delta x \), for instance, \( \Delta x = 0.01 \). Calculate \( f'(1) \approx \frac{\ln(1.01) - \ln(1)}{0.01} = \frac{\ln(1.01)}{0.01} \). Use a calculator to find \( \ln(1.01) \approx 0.00995 \), so \( f'(1) \approx \frac{0.00995}{0.01} = 0.995. \)

Step 3: Estimate \( f'(2) \)

Using \( \Delta x = 0.01 \), calculate \( f'(2) \approx \frac{\ln(2.01) - \ln(2)}{0.01} \). Find \( \ln(2) \approx 0.6931 \) and \( \ln(2.01) \approx 0.6981 \). Thus, \( f'(2) \approx \frac{0.6981 - 0.6931}{0.01} = 0.5. \)

Step 4: Estimate \( f'(3) \)

With \( \Delta x = 0.01 \), calculate \( f'(3) \approx \frac{\ln(3.01) - \ln(3)}{0.01} \). Use \( \ln(3) \approx 1.0986 \) and \( \ln(3.01) \approx 1.1012 \). So, \( f'(3) \approx \frac{1.1012 - 1.0986}{0.01} = 0.26. \)

Step 5: Estimate \( f'(4) \)

With \( \Delta x = 0.01 \), calculate \( f'(4) \approx \frac{\ln(4.01) - \ln(4)}{0.01} \). Use \( \ln(4) \approx 1.3863 \) and \( \ln(4.01) \approx 1.3889 \). Therefore, \( f'(4) \approx \frac{1.3889 - 1.3863}{0.01} = 0.26. \)

Step 6: Estimate \( f'(5) \)

Using \( \Delta x = 0.01 \), calculate \( f'(5) \approx \frac{\ln(5.01) - \ln(5)}{0.01} \). Compute \( \ln(5) \approx 1.6094 \) and \( \ln(5.01) \approx 1.6113 \). Thus, \( f'(5) \approx \frac{1.6113 - 1.6094}{0.01} = 0.19. \)

Step 7: Formulate a General Expression

Based on the estimates: \( f'(1) \approx 1, f'(2) \approx 0.5, f'(3) \approx 0.33, f'(4) \approx 0.25, f'(5) \approx 0.2 \), observe that these values are approximately \( \frac{1}{x} \). This suggests that for \( f(x) = \ln x \), the derivative \( f'(x) = \frac{1}{x} \).

Step 8: Verify the Pattern

You can verify the pattern through further evaluations or by using the derivative rule for logarithmic functions, which indeed states that \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).

Key Concepts

Natural Logarithm

The natural logarithm, commonly denoted as \(\ln x\), is a powerful mathematical function. It's the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828.

• Natural logarithms are particularly useful in calculus, statistics, and numerous scientific applications because of their unique properties.
• The natural logarithm of a number \(x\), \(\ln x\), is the power to which \(e\) must be raised to get \(x\).
• For example, \(\ln(e) = 1\) because \(e^1 = e\), and \(\ln(1) = 0\) because \(e^0 = 1\).

Understanding the natural logarithm is crucial for studying its derivative, which involves calculating the rate of change of \(\ln x\).

Rate of Change

The derivative of a function is a measure of how a function value changes as its input changes. In simpler terms, it tells us the rate of change of a function.

• For the function \(f(x) = \ln x\), the derivative \(f'(x)\) represents the rate at which \(\ln x\) changes with respect to \(x\).
• The concept of rate of change is vital for understanding how functions behave. It's often visualized as the slope of a tangent line to the curve of the function at any point.
• In the context of estimating derivatives at specific points, we use the formula \(f'(x) \approx \frac{f(x + \Delta x) - f(x)}{\Delta x}\) where \(\Delta x\) is a small number.

This allows us to understand how quickly or slowly \(\ln x\) increases as \(x\) itself increases.

Approximation Methods

Approximation methods are essential tools in both mathematics and applied sciences. They allow us to estimate values of complicated functions or expressions, such as the derivative of \(\ln x\), without exact calculations.

• To estimate the derivative of \(\ln x\) at a point, choose a small \(\Delta x\), like 0.01. This will help in calculating the rate of change over tiny intervals.
• The difference quotient \(\frac{f(x + \Delta x) - f(x)}{\Delta x}\) is a primary method used for deriving approximations.
• For instance, to find \(f'(1)\), putting \(f(x) = \ln x\), use \(\ln(1.01) - \ln(1)\) and divide by 0.01.

Such methods provide a practical approach to grasping complex mathematical behavior without needing a calculator for every small detail.

Logarithmic Functions

Logarithmic functions like \(\ln x\) are inverse operations of exponential functions. They play an incredible role across different fields.

• A logarithmic function \(y = \ln x\) is used to determine the exponent that \(e\) must be raised to yield a given number \(x\).
• The derivative of a logarithmic function, particularly \(\ln x\), is \(\frac{1}{x}\). This shows how the function's growth rate slows as \(x\) increases.
• Understanding the properties of logarithms helps one apply these concepts in calculus, such as differentiation and integration.

Logarithmic functions therefore provide a critical bridge between exponential growth and multiplicative processes, offering insights into natural phenomena and growth patterns.