Problem 9 - Applied Calculus

Question

Let \(f(x)=5^{x}\). Use a small interval to estimate \(f^{\prime}(2)\). Now improve your accuracy by estimating \(f^{\prime}(2)\) again, using an even smaller interval.

Step-by-Step Solution

Answer

Estimated \( f'(2) \approx 10125.44 \) using the smaller interval. The smaller the interval, the more accurate the approximation.

Step 1: Understanding Derivative Estimate

To estimate the derivative of a function at a point, we use the limit definition of the derivative. This involves using small increments, \( h \), to approximate \( f'(x) \) as \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

Step 2: First Estimation Using a Small Interval

Start with a relatively small value for \( h \). Let \( h = 0.1 \). Use the formula \( \frac{f(2+0.1) - f(2)}{0.1} \). This gives: \[ \frac{5^{2.1} - 5^2}{0.1} \]. Calculate \( 5^{2.1} \) and \( 5^2 \): \( 5^{2.1} \approx 133.184 \) and \( 5^2 = 25 \). Therefore, \[ \frac{133.184 - 25}{0.1} = 1081.84 \].

Step 3: Improving Accuracy with a Smaller Interval

To improve accuracy, make \( h \) smaller. Let \( h = 0.01 \). Use \( \frac{f(2+0.01) - f(2)}{0.01} \). This gives: \[ \frac{5^{2.01} - 5^2}{0.01} \]. Calculate \( 5^{2.01} \approx 126.2544 \) and \( 5^2 = 25 \). Therefore, \[ \frac{126.2544 - 25}{0.01} = 10125.44 \].

Step 4: Analyzing the Results

The derivatives calculated with smaller \( h \) should be closer to the actual derivative. Initially, we calculated \( f'(2) \approx 1081.84 \) with \( h=0.1 \) and then \( f'(2) \approx 10125.44 \) with \( h=0.01 \). The latter interval yields a more accurate estimate.

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It helps us understand how functions behave at a specific point. Given a function \( f(x) \), the derivative at a point \( x = a \) is defined as the limit of the difference quotient as the increment \( h \) approaches zero. This is mathematically represented as:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]The difference quotient \( \frac{f(a+h) - f(a)}{h} \) measures the average rate of change over the interval from \( a \) to \( a+h \).
As \( h \) becomes smaller, this average rate of change more closely approximates the instantaneous rate of change, or the derivative, at \( a \).
For practical purposes, especially in numerical calculations, we choose small values of \( h \) that are computationally feasible to get a reasonable estimate of the derivative.

Exponential Functions

Exponential functions are powerful mathematical expressions where the variable appears in the exponent. The general form of an exponential function is \( f(x) = a^x \). In our example, we used \( f(x) = 5^x \).
These functions are noted for their rapid growth or decay, depending on the base \( a \). If \( a > 1 \), the function grows as \( x \) increases. If \( 0 < a < 1 \), the function decays.
Exponential functions have derivatives that are proportional to their value. For our function \( f(x) = 5^x \), this means its derivative is also an exponential function. It can be expressed as:\[ f'(x) = 5^x \ln(5) \]This relationship reflects the logarithmic constant \( \ln(5) \), indicating the rate at which the exponential function grows or decays.

Numerical Methods

Numerical methods are used to approximate solutions or values in mathematics when exact solutions are difficult or impossible to find. In the realm of derivatives, numerical estimation is essential when a function's derivative isn't easily derived analytically.
One common numerical method involves choosing a small interval \( h \) to calculate an approximate derivative. For instance, we estimated \( f'(2) \) for \( f(x) = 5^x \) using \( h = 0.1 \) and \( h = 0.01 \).
Here, smaller value of \( h \) helped improve the accuracy of the estimation, as smaller steps get us closer to the actual limit. However, it is crucial to balance between a smaller \( h \) and computational precision, as too small a value can lead to numerical errors due to machine precision limitations.

• Start with a reasonably small \( h \).
• Calculate the difference quotient \( \frac{f(x+h) - f(x)}{h} \).
• Adjust \( h \) for better accuracy, observing any computational limitations.