Problem 64 - Calculus Single Variable

Question

Let \(g(x)=\mathrm{e}^{k x}\) where \(k\) is a constant. Using the definition of the derivative, show that $$ g^{\prime}(x)=g^{\prime}(0) g(x). $$

Step-by-Step Solution

Answer

The derivative is \(g'(x) = g'(0)g(x)\) with \(g'(0) = k\).

Step 1: Understand the Definition of the Derivative

The definition of the derivative of a function \(f(x)\) at a point \(x\) is given by \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). We'll use this definition to find \(g'(x)\).

Step 2: Apply the Definition to g(x)

We start by evaluating the derivative of \(g(x) = e^{kx}\) using the definition: \[g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{e^{k(x+h)} - e^{kx}}{h}.\]

Step 3: Simplify the Expression

Notice that \(e^{k(x+h)} = e^{kx}e^{kh}\). Substitute this into the limit: \[g'(x) = \lim_{h \to 0} \frac{e^{kx}e^{kh} - e^{kx}}{h} = \lim_{h \to 0} \frac{e^{kx}(e^{kh} - 1)}{h}.\]

Step 4: Factor Out Constants

Factor \(e^{kx}\) out of the limit since it does not depend on \(h\): \[g'(x) = e^{kx} \lim_{h \to 0} \frac{e^{kh} - 1}{h}.\]

Step 5: Evaluate the Remaining Limit

The limit \(\lim_{h \to 0} \frac{e^{kh} - 1}{h}\) is a standard derivative result which equals \(k\) because it is the derivative of \(e^{kx}\) evaluated at \(x=0\). Thus, \[\lim_{h \to 0} \frac{e^{kh} - 1}{h} = k.\]

Step 6: Combine Results

Combine the results to find the derivative: \[g'(x) = e^{kx} \cdot k = k e^{kx}.\] This can be rewritten as \(g'(x) = g'(0)g(x)\), where \(g'(0) = k\) since \(g(0) = e^0 = 1\).

Key Concepts

Definition of Derivative

The derivative of a function at a point essentially tells us how the function's output value changes as we slightly change the input. It is a fundamental concept in calculus. This change is described by the definition of the derivative:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

Here, the "\(f(x+h) - f(x)\)" in the numerator represents the change in the function's value, and "\(h\)" in the denominator is the small change in input. As \(h\) approaches zero, the fraction estimates the slope of the tangent line to the function at that point.
This concept is crucial for understanding how functions behave locally.

Limit Process

The limit process is an essential tool when calculating derivatives and understanding changes in functions. A limit looks at what happens to a function as the input approaches some value.In the context of differentiation, specifically when calculating derivatives, we care about the limit as \(h\) approaches zero. This helps us find the slope of the tangent to the curve. Consider the expression:

• \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

This formula helps determine how much \(f(x)\) changes per unit change in \(x\), capturing the idea of instantaneous rate of change.

Exponential Function

An exponential function is one where the variable is in the exponent. It's a central function in mathematics, given by:

• \( g(x) = e^{kx} \)

Here \(e\) is approximately 2.71828, known as Euler's number, and \(k\) is a constant. Exponential functions grow rapidly, and when we differentiate them, a unique property emerges:
The derivative of an exponential function maintains its exponential form. In specific application, this means:

• \( g'(x) = k e^{kx} \)

This result demonstrates how exponential growth behaves under differentiation.

Differentiation

Differentiation is the process of finding a derivative. By calculating the derivative, we determine the slope of the tangent line to the function at any point:

• For \( g(x) = e^{kx} \), applying the definition of derivative:\( g'(x) = \lim_{h \to 0} \frac{e^{k(x+h)} - e^{kx}}{h} \)

We're using the property that \(e^{k(x+h)} = e^{kx}e^{kh}\). Once we split the limit:

• \( g'(x) = e^{kx} \lim_{h \to 0} \frac{e^{kh} - 1}{h} \)

Then, using a known result—\(\lim_{h \to 0} \frac{e^{kh} - 1}{h} = k\)—it simplifies beautifully to:

• \( g'(x) = k e^{kx} \)

This process highlights the steps from function setup to finding its derivative.