Problem 122 - Precalculus

Question

Recall that the derivative of \(f\) can be found by letting \(h \rightarrow 0\) in the difference quotient \(\frac{f(x+h)-f(x)}{h} .\) In calculus we prove that \(\frac{e^{h}-1}{h}=1,\) when \(h\) approaches \(0 ;\) that is, for really small values of \(h, \frac{e^{h}-1}{h}\) gets very close to 1. Use this information to find the derivative of \(f(x)=e^{2 x}\).

Step-by-Step Solution

Answer

The derivative of \( f(x) = e^{2x} \) is \( 2e^{2x} \).

Step 1: Understanding the Problem

We need to find the derivative of the function \( f(x) = e^{2x} \) using the limit definition of a derivative. We'll use the information that \( \frac{e^{h} - 1}{h} = 1 \) as \( h \to 0 \) to assist in our calculation.

Step 2: Set Up the Difference Quotient

The difference quotient for the function \( f(x) = e^{2x} \) is \( \frac{f(x+h) - f(x)}{h} \). Substitute \( f(x+h) = e^{2(x+h)} \) and \( f(x) = e^{2x} \) into this expression: \[ \frac{e^{2(x+h)} - e^{2x}}{h} = \frac{e^{2x+2h} - e^{2x}}{h}. \]

Step 3: Factor Out Common Terms

Notice that we can factor \( e^{2x} \) out of the expression. This gives: \[ \frac{e^{2x}(e^{2h} - 1)}{h}. \] This separates the expression into terms we can evaluate individually.

Step 4: Apply the Limit

We apply the limit \( h \to 0 \) to the expression from Step 3: \[ \lim_{{h \to 0}} \frac{e^{2x}(e^{2h} - 1)}{h}. \] This simplifies to \[ e^{2x} \cdot \lim_{{h \to 0}} \frac{e^{2h} - 1}{h}. \]

Step 5: Substitute Known Limit Result

From the problem information, we know \( \lim_{{h \to 0}} \frac{e^{2h} - 1}{h} = 2 \) because \( \lim_{{u \to 0}} \frac{e^{u} - 1}{u} = 1 \) and we have \( u = 2h. \) Therefore, \[ \lim_{{h \to 0}} \frac{e^{2h} - 1}{h} = 2. \] Substituting this into our previous result, we get: \[ e^{2x} \cdot 2 = 2e^{2x}. \]

Step 6: Write the Final Derivative

The derivative of \( f(x) = e^{2x} \) is thus found to be \( f'(x) = 2e^{2x}. \)

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It allows us to find the rate of change, or slope, of a function at any given point. To understand this, we rely on the concept of the difference quotient:

• For a function \( f(x) \), the difference quotient is expressed as \( \frac{f(x+h)-f(x)}{h} \).
• This equation represents the average rate of change of the function over a small interval \( h \).

As \( h \) approaches zero, this average rate of change becomes the instantaneous rate of change, which is the derivative:

• Formally, this is written as \( \lim_{{h \to 0}} \frac{f(x+h)-f(x)}{h} \).
• The result gives us \( f'(x) \), the derivative of the function at \( x \).

Calculating derivatives using this limit approach forms the basis for understanding how functions behave locally, and it's a tool that helps solve a variety of problems in calculus.

Exponential Function

An exponential function is one where the variable is in the exponent. It is a powerful and unique type of function, and one of the most common forms is \( e^x \):

• Here, \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
• The function \( e^x \) has a distinct property: its rate of growth is proportional to its current value.

When working with exponential functions, especially when differentiating, several special characteristics come into play:

• The derivative of \( e^x \) with respect to \( x \) is \( e^x \) itself. This is unique to exponential functions with base \( e \).
• For a function like \( e^{2x} \), as in the exercise, the chain rule is applied during differentiation.

These properties, especially in calculus, make exponential functions a crucial concept to grasp for understanding continuous growth models in science, economics, and many other fields.

Differentiation Using Limits

Differentiation using limits is a technique that leverages the limit definition to find the derivative of functions that might not be easily covered by simple derivative rules. This process can involve more complex expressions where direct rules might not apply:

• For functions and expressions where simple derivative rules like power or product might not readily apply, we often revert to the foundational limit definition.
• In the exercise, the function \( e^{2x} \) is differentiated using this method.

The step-by-step method includes:

• Set up the difference quotient for the function, substituting accordingly based on the expression given.
• Simplify the expression, factor out common terms, and apply the limit.
• Integrate any known limits and related theorems, such as \( \lim_{{h \to 0}} \frac{e^{u} - 1}{u} = 1 \), into the process.

Taking the function \( e^{2x} \), applying these steps results in a derivative of \( 2e^{2x} \). This showcases the power and flexibility of using limits in differentiation, allowing us to derive results even with complex functions.