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Give the derivative formula for each function. \(\quad f(x)=10\left(1+\frac{0.05}{4}\right)^{4 x}\) (Hint: Use \(a b^{c x}=a\left(b^{c}\right)^{x}\) to rewrite as \(\left.a d^{x} .\right)\)

Short Answer

Expert verified
The derivative formula is \( f'(x) = 10 \cdot 1.0504^x \cdot 0.04937 \).

Step by step solution

01

Rewriting the Exponential Function

Given the function \( f(x) = 10\left(1 + \frac{0.05}{4}\right)^{4x} \), we need to rewrite it in a simpler exponential form. Start by evaluating the base: \( b = 1 + \frac{0.05}{4} = 1.0125 \). Then, rewrite the function as \( f(x) = 10 \cdot (1.0125)^{4x} \). Recognize that \( 10\cdot (1.0125)^{4x} \) fits the pattern \( a \cdot b^{cx} = a \cdot (b^c)^x \). Thus, use \( b = 1.0125 \) and \( c = 4 \) to simplify it to \( f(x) = 10 \cdot d^x \) where \( d = (1.0125)^4 \).
02

Simplifying the Constant Base

Calculate \( d = (1.0125)^4 \). To do this, first square \( 1.0125 \) twice: \( 1.0125^2 \approx 1.02515625 \) and then square again: \( 1.02515625^2 \approx 1.0504 \). Thus, \( d \approx 1.0504 \). Now the function simplifies to \( f(x) \approx 10 \cdot 1.0504^x \).
03

Derivative of Exponential Function

Using the simplified function \( f(x) = 10 \cdot 1.0504^x \), find the derivative. Recall the derivative formula for an exponential function \( a \cdot b^x \) is \( f'(x) = a \cdot b^x \cdot \ln(b) \). Apply this formula to \( f(x) = 10 \cdot 1.0504^x \): \( f'(x) = 10 \cdot 1.0504^x \cdot \ln(1.0504) \).
04

Writing the Final Result

The derivative of the function \( f(x) = 10 \cdot 1.0504^x \) is \( f'(x) = 10 \cdot 1.0504^x \cdot \ln(1.0504) \). Use a calculator to find \( \ln(1.0504) \), which is approximately \( 0.04937 \). Thus, the final expression for the derivative is \( f'(x) = 10 \cdot 1.0504^x \cdot 0.04937 \).

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Key Concepts

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

Exponential Function
Exponential functions are fundamental in mathematics and have the general form \( f(x) = a \cdot b^x \), where \( a \) is a constant, and \( b \) is the base of the exponential function. These functions are characterized by the rate of growth or decay they exhibit, depending on the value of \( b \). When the base \( b > 1 \), the function represents exponential growth, and when \( 0 < b < 1 \), it represents exponential decay.
To manipulate an exponential function, it is often helpful to rewrite it in a simpler form. For example, a common manipulation is to express an exponential as \( a \cdot (b^c)^x \), which simplifies computations and understanding. In this exercise, we saw how the function \( f(x) = 10\left(1 + \frac{0.05}{4}\right)^{4x} \) was transformed into \( f(x) = 10 \cdot d^x \) by evaluating the base \( b = 1.0125 \) and then raising it to the 4th power to find \( d = b^4 = 1.0504 \).
Exponential functions appear in many real-world situations, such as population growth, radioactive decay, and financial investments. Understanding how to manipulate and differentiate these functions is crucial for analyzing their behavior and predicting outcomes.
Derivative Formula
The derivative of a function measures how the function's output changes as the input changes. For exponential functions, the derivative gives us a powerful way to understand and predict the behavior of the function across its domain. The general formula for finding the derivative of an exponential function \( f(x) = a \cdot b^x \) is:

\( f'(x) = a \cdot b^x \cdot \ln(b) \)

This formula tells us that the derivative is directly proportional to the original function and the natural logarithm of the base \( b \).
In the exercise, we had \( f(x) = 10 \cdot 1.0504^x \). Using the derivative formula, we found:
  • \( f'(x) = 10 \cdot 1.0504^x \cdot \ln(1.0504) \)
  • After calculating \( \ln(1.0504) \), we approximated it to \( 0.04937 \).
  • Thus, the derivative result is \( f'(x) = 10 \cdot 1.0504^x \cdot 0.04937 \).
This step-by-step application of the derivative formula helps in understanding how the exponential function's growth rate is influenced by both its base and the coefficient in front of it. The derivative is particularly useful in analyzing growth models and understanding changes over time.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function with the base \( e \), a mathematical constant approximately equal to 2.71828. It is the inverse of the exponential function \( e^x \).
Natural logarithms are essential in calculus, especially when dealing with derivatives of exponential functions, as they simplify the differentiation process. In our context, the natural logarithm appears in the derivative formula of an exponential function. For a function \( f(x) = a \cdot b^x \), the derivative involves \( \ln(b) \) to indicate the rate of change given by the base \( b \).
In the original exercise, calculating \( \ln(1.0504) \) was crucial for obtaining the derivative of the function. The approximate value, \( 0.04937 \), reflects how much the function \( f(x) = 10 \cdot 1.0504^x \) changes as \( x \) increases by 1 unit.
Natural logarithms bridge the gap between exponential growth and linear log scales, allowing us to explore exponential behavior in a more manageable, linear way. They are found in numerous applications, from solving exponential equations to modeling real-life phenomena like sound intensity and earthquake magnitudes.

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Most popular questions from this chapter

College Tuition The tuition \(x\) years from now at a private four-year college is projected to be $$ t(x)=24,072 e^{0.056 x} \text { dollars. } $$ a. Write the rate-of-change formula for tuition. b. What is the rate of change in tuition four years from now?

Write derivative formulas for the functions. $$ f(x)=(x+5) e^{x} $$

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Evaluate the limit. If the limit is of an indeterminate form, indicate the form and use L'Hôpital's Rule to evaluate the limit. $$ \lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1} $$

High School Dropouts (Historic) The table shows the number of students enrolled in the ninth through twelfth grades and the number of dropouts from those same grades in South Carolina for each school year from \(1980-1981\) through \(1989-1990\) $$ \begin{aligned} &\text { South Carolina High School Enrollment and Dropouts }\\\ &\begin{array}{|c|c|c|} \hline \text { School year } & \text { Enrollment } & \text { Dropouts } \\ \hline 1980-81 & 194,072 & 11,651 \\ \hline 1981-82 & 190,372 & 10,599 \\ \hline 1982-83 & 185,248 & 9314 \\ \hline 1983-84 & 182,661 & 9659 \\ \hline 1984-85 & 181,949 & 8605 \\ \hline 1985-86 & 182,787 & 8048 \\ \hline 1986-87 & 185,131 & 7466 \\ \hline 1987-88 & 183,930 & 7740 \\ \hline 1988-89 & 178,094 & 7466 \\ \hline 1989-90 & 172,372 & 5768 \\ \hline \end{array} \end{aligned} $$ a. Find cubic models for enrollment and the number of dropouts. Align both models to the number of years since \(1980-81\) b. Use the two models found in part \(a\) to construct an equation for the percentage of high school students who dropped out each year. c. Find the rate-of-change formula of the percentage of high school students who dropped out each year. d. Look at the rate of change for each school year from \(1980-81\) through \(1989-90 .\) In which school year was the rate of change smallest? When was it greatest?

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