Problem 63 - Calculus: Early Transcendentals

Question

Recall that a function \( f \) is called \textit{even} if \( f(-x) = f(x) \) for all \( x \) in its domain and \textit{odd} if \( f(-x) = -f(x) \) for all such \( x \). Prove each of the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function.

Step-by-Step Solution

Answer

(a) The derivative of an even function is odd. (b) The derivative of an odd function is even.

Step 1: Understanding Even and Odd Functions

To determine if a function is even or odd, we consider their definitions: a function \( f \) is even if \( f(-x) = f(x) \) for all \( x \) in its domain, and odd if \( f(-x) = -f(x) \). We will use these definitions to explore the properties of the derivatives.

Step 2: Prove Derivative of an Even Function is Odd

Let \( f(x) \) be an even function, meaning \( f(-x) = f(x) \) for all \( x \). We will show the derivative \( f'(x) \) is odd, i.e., \( f'(-x) = -f'(x) \). First, compute \( f'(-x) \) using the definition of a derivative:\[ f'(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]Since \( f(x) \) is even, \( f(-x+h) = f(x-h) \) and \( f(-x) = f(x) \), substitute these into the derivative:\[ f'(-x) = \lim_{h \to 0} \frac{f(x-h) - f(x)}{h} \]Change variable \( h \to -h \), then\[ f'(-x) = \lim_{h \to 0} \frac{f(x) - f(x-h)}{-h} = -\lim_{h \to 0} \frac{f(x-h) - f(x)}{h} = -f'(x) \]Thus, \( f'(x) \) is odd.

Step 3: Prove Derivative of an Odd Function is Even

Let \( f(x) \) be an odd function, meaning \( f(-x) = -f(x) \) for all \( x \). We want to show that \( f'(x) \) is even, i.e., \( f'(-x) = f'(x) \). Start with the derivative definition:\[ f'(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]Substitute the identity of an odd function:\[ f(-x+h) = -f(x-h) \] and \(-f(-x) = f(x) \), leading to:\[ f'(-x) = \lim_{h \to 0} \frac{-f(x-h) - (-f(x))}{h} = \lim_{h \to 0} \frac{-f(x-h) + f(x)}{h} \]\[ f'(-x) = \lim_{h \to 0} \frac{f(x) - f(x-h)}{h} = \lim_{h \to 0} \frac{f(x-h) - f(x)}{-h} = f'(x) \]Hence, \( f'(x) \) is even.

Key Concepts

Derivative of Even Function

When we explore derivatives, we often start by understanding the nature of the function itself, specifically whether it is even or odd. For an even function, the defining characteristic is that it remains unchanged when the input variable is replaced by its negative counterpart, which is expressed as \( f(-x) = f(x) \). Now, we aim to see how this property affects its derivative.
Let's delve into the process. Consider an even function \( f(x) \). We need to find its derivative using the limit definition:\[ f '(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]Using the property of even functions \( f(-x+h) = f(x-h) \) and \( f(-x) = f(x) \), substitution leads to:\[ f '(-x) = \lim_{h \to 0} \frac{f(x-h) - f(x)}{h} \]When you change the variable \( h \to -h \) in the limit, you will get:\[f '(-x) = -\lim_{h \to 0} \frac{f(x) - f(x-h)}{h} = -f'(x) \]Thus, the derivative of an even function is indeed an odd function, since \( f'(-x) = -f'(x) \). This is a significant revelation as it points out that applying differentiation to an even function transforms it in a specific structured way.

Derivative of Odd Function

Odd functions shine through their symmetric twist around the origin. Mathematically, an odd function satisfies \( f(-x) = -f(x) \) for every input \( x \). Now that we've explored even functions, let's turn our attention to odd functions and their derivatives.
The task here is to demonstrate that the derivative of an odd function is even. We begin with the definition of a derivative:\[ f '(-x) = \lim_{h \to 0} \frac{f(-x+h) - f(-x)}{h} \]For an odd function, where \( f(-x+h) = -f(x-h) \) and \( f(-x) = -f(x) \), substitution yields:\[ f '(-x) = \lim_{h \to 0} \frac{-f(x-h) + f(x)}{h} \]After a bit of algebraic manipulation, you're left with:\[f '(-x) = \lim_{h \to 0} \frac{f(x) - f(x-h)}{h} = f'(x) \]Therefore, the derivative of an odd function is indeed an even function, satisfying the relation \( f'(-x) = f'(x) \). This result offers a captivating insight into how differentiation impacts the symmetry properties of functions.

Properties of Derivatives

The intriguing dance of derivatives extends beyond just even and odd functions. The myriad of properties of derivatives can help you peel back layers of mathematical complexity. Here are some vital considerations and highlights:

• Linearity: The derivative of a sum is the sum of the derivatives. Simply put, \( (f + g)' = f' + g' \).
• Product Rule: For the product of two functions \( f(x) \) and \( g(x) \), the derivative is \( (fg)' = f'g + fg' \).
• Quotient Rule: For dividing two functions, the derivative becomes \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \).
• Chain Rule: This allows differentiation of composite functions, providing \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \).

Each property can be elegantly applied to solve derivatives based on the function's form. Additionally, understanding the derivative provides critical insights into the function's behavior, such as its slope, rate of change, and points of inflection.
Knowing these properties doesn't just help in mathematical calculations. They equip you to analyze and solve more complex problems efficiently.