Problem 21 - University Calculus: Early Transcendentals

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(3^{x}\) b. \(2^{-x}\) c. \(\left(\frac{5}{3}\right)^{x}\)

Step-by-Step Solution

Answer

The antiderivatives are: a. \(\frac{3^x}{\ln(3)} + C\), b. \(\frac{2^{-x}}{-\ln(2)} + C\), c. \(\frac{\left(\frac{5}{3}\right)^x}{\ln\left(\frac{5}{3}\right)} + C\).

Step 1: Recognize the Form of the Function

Each function given is an exponential function of the form \(a^x\). The general method for finding the antiderivative of \(a^x\) is to use the formula: \(\frac{a^x}{\ln(a)} + C\), where \(C\) is the constant of integration.

Step 2: Apply the Antiderivative Formula to \(3^x\)

Using the formula \(\int a^x \, dx = \frac{a^x}{\ln(a)} + C\), we substitute \(a = 3\). Thus, the antiderivative of \(3^x\) is \(\frac{3^x}{\ln(3)} + C\).

Step 3: Differentiate to Check the Solution for \(3^x\)

Differentiate \(\frac{3^x}{\ln(3)} + C\) with respect to \(x\). \(\frac{d}{dx}[\frac{3^x}{\ln(3)}] = \frac{3^x \ln(3)}{\ln(3)} = 3^x\), confirming correctness as differentiation yields the original function.

Step 4: Apply the Antiderivative Formula to \(2^{-x}\)

Rewrite \(2^{-x} = \left(\frac{1}{2}\right)^x\). Applying the antiderivative formula gives: \(\frac{\left(\frac{1}{2}\right)^x}{\ln\left(\frac{1}{2}\right)} + C = \frac{2^{-x}}{-\ln(2)} + C\).

Step 5: Differentiate to Check the Solution for \(2^{-x}\)

Differentiate \(\frac{2^{-x}}{-\ln(2)} + C\). Using the chain rule, \(\frac{d}{dx}[\frac{2^{-x}}{-\ln(2)}] = \frac{-2^{-x} \ln(2)}{-\ln(2)} = 2^{-x}\), confirming correctness.

Step 6: Apply the Antiderivative Formula to \(\left(\frac{5}{3}\right)^x\)

Utilize the antiderivative formula for the function \(\left(\frac{5}{3}\right)^x\). The result is \(\frac{\left(\frac{5}{3}\right)^x}{\ln\left(\frac{5}{3}\right)} + C\).

Step 7: Differentiate to Check the Solution for \(\left(\frac{5}{3}\right)^x\)

Differentiate the expression \(\frac{\left(\frac{5}{3}\right)^x}{\ln\left(\frac{5}{3}\right)} + C\). The derivative is \(\frac{\left(\frac{5}{3}\right)^x \ln\left(\frac{5}{3}\right)}{\ln\left(\frac{5}{3}\right)} = \left(\frac{5}{3}\right)^x\), confirming correctness.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent, typically denoted as \( a^x \). This structure makes exponential functions unique compared to polynomial or linear functions. These functions are prevalent in various real-world scenarios, such as modeling population growth, radioactive decay, and compound interest.

Some key characteristics of exponential functions include:

• The base \( a \) must be a positive real number, except for 1, because a base of 1 would lead a constant function, which isn't genuinely exponential.
• If \( a > 1 \), the function grows rapidly as \( x \) increases, depicting exponential growth.
• If \( 0 < a < 1 \), the function decreases, representing exponential decay.

The simplicity of their form leads to substantial applications in science, architecture, and finance. Continuous exploration of exponential functions helps in understanding natural processes and solvinig complex problems.

Differentiation

Differentiation is the process of finding the derivative of a function. It allows us to understand how a function changes at every point. In technical terms, the derivative of a function represents the rate of change or the slope of the tangent to its curve at any given point.

For exponential functions like \( a^x \), the process of differentiation uses a special formula. The derivative of \( a^x \) is expressed as \( a^x \ln(a) \). This formula arises because of the nature of exponential growth, where the rate of increase is proportional to its current value.

Key aspects of differentiation include:

• Power Rule: Used for derivatives of terms like \( x^n \).
• Product Rule: Applied when two functions are multiplied.
• Quotient Rule: Used to find the derivative when one function is divided by another.

Differentiation finds its use across many fields, whether it's in calculating the velocity in physics or finding the maxima and minima in optimization problems.

Chain Rule

The chain rule is a vital technique in calculus used to differentiate composite functions. When dealing with expressions involving nested functions, the chain rule simplifies the differentiation process by relating the derivative of the outer function to the inner function.

In mathematical terms, for a composite function \( f(g(x)) \), the chain rule states:\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
The chain rule becomes essential when differentiating more complex exponential functions like \( 2^{-x} \). By rewriting \( 2^{-x} \) as \( \left( \frac{1}{2} \right)^x \), it allows us to handle the differentiation using the familiar exponential differentiation formula, while accounting for the negative sign by applying the chain rule.

Here's how it works:

• The outer function is \( a^x \).
• The inner function is \( -x \), whose derivative is \( -1 \).

This stepwise approach manages variations in the components of functions, making the differentiation process coherent and feasible. Mastery of the chain rule provides solid groundwork for tackling sophisticated calculus problems.