Problem 59 - Precalculus

Question

Refer to the following: In calculus, we find the derivative, \(f^{\prime}(x),\) of a function \(f(x)\) by allowing \(h\) to approach 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of functions involving exponential functions. Use the fact that \(\frac{e^{h}-1}{h}=1\) when \(h\) is close to zero to find the derivative of \(f(x)=e^{x}+x\).

Step-by-Step Solution

Answer

The derivative of \( f(x) = e^x + x \) is \( f'(x) = e^x + 1 \).

Step 1: Set up the Difference Quotient

Begin by writing the difference quotient for the function \( f(x) = e^x + x \). The difference quotient is the expression \( \frac{f(x+h) - f(x)}{h} \). First, find \( f(x+h) \):\[ f(x+h) = e^{x+h} + (x + h) = e^x \cdot e^h + x + h. \] Plug this into the difference quotient to get\[ \frac{(e^x \cdot e^h + x + h) - (e^x + x)}{h}. \]

Step 2: Simplify the Difference Quotient

Simplify the expression from Step 1. Subtract \( e^x + x \) from \( e^x \cdot e^h + x + h \):\[ e^x \cdot e^h + x + h - e^x - x = e^x \cdot e^h - e^x + h. \] So the difference quotient becomes\[ \frac{e^x \cdot e^h - e^x + h}{h}. \]

Step 3: Factor and Separate the Quotient

Factor \( e^x \) out of the first two terms in the numerator:\[ \frac{e^x (e^h - 1) + h}{h}. \] Now separate this quotient into two parts:\[ \frac{e^x (e^h - 1)}{h} + \frac{h}{h}. \]

Step 4: Apply the Limit as h Approaches 0

Apply the limit \( \lim_{h \to 0} \) to the expression from Step 3. We have\[ \lim_{h \to 0} \left( \frac{e^x (e^h - 1)}{h} + 1 \right). \] Since \( \frac{e^h - 1}{h} = 1 \) as \( h \to 0 \), this becomes\[ \lim_{h \to 0} e^x \cdot 1 + \lim_{h \to 0} 1 = e^x + 1. \]

Step 5: Conclusion

Summarize the findings. By applying the limit, we find that the derivative of \( f(x) = e^x + x \) is \( e^x + 1 \). Therefore, \( f^{\prime}(x) = e^x + 1 \).

Key Concepts

Difference Quotient

The difference quotient is a key concept in calculus, primarily used to calculate the derivative of a function. It provides a way to find the slope of the tangent line to the curve of a function at any given point. This is highly useful in understanding the behavior of functions.

For a function \( f(x) \), the difference quotient is expressed as:

• \( \frac{f(x+h) - f(x)}{h} \)

Here, \( h \) represents a small increment to \( x \), and \( f(x+h) \) gives us the function's value at this slightly larger input. The expression measures the average rate of change of the function between \( x \) and \( x+h \).

As \( h \) approaches 0, this rate of change approaches the exact rate, or the derivative, \( f'(x) \), of the function at that point. Understanding the difference quotient is crucial because it lays the foundation for understanding limits and derivatives.

Exponential Functions

Exponential functions play a significant role in calculus, especially in the context of derivatives. An exponential function is generally of the form \( f(x) = a^x \), where \( a \) is a constant, and \( x \) is the exponent. A particularly important exponential function is \( e^x \), where \( e \) is Euler's number, approximately 2.718.

These functions are unique because their rate of growth is proportional to their current value. When calculating derivatives, exponential functions have properties that simplify the process:

• The derivative of \( e^x \) is \( e^x \) itself, which makes it easy to work with.
• For a function like \( e^{x+h} \), it can be rewritten as \( e^x \cdot e^h \), showing how the function evolves over an interval.

Exponential functions appear frequently in real-world applications, such as compound interest, population growth, and natural phenomena, making them highly relevant in various fields.

Limits

The concept of limits is foundational in calculus. In the context of finding derivatives using the difference quotient, limits help us determine what happens to a function as it approaches a certain point.

When we write \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \), we are interested in the value of the expression as \( h \) becomes infinitesimally small. This process allows us to find the derivative of a function, which tells us the instantaneous rate of change or the slope of the tangent line to the function at a specific point.

In our original problem, we used the fact

• \( \frac{e^h - 1}{h} = 1 \) as \( h \to 0 \)

This result shows how limits are used to evaluate expressions that seem indeterminate by simple substitution. Understanding limits enables deeper insights into how functions behave near particular points, bridging the gap between algebraic expressions and their graphical interpretations.