Problem 70 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function. CAS Use a CAS to do Problems 71 and \(72 .\)

Step-by-Step Solution

Answer

The derivative of an odd function is even, and the derivative of an even function is odd.

Step 1: Understanding Odd and Even Functions

An odd function satisfies the condition \(f(-x) = -f(x)\) for all \(x\), and an even function satisfies \(g(-x) = g(x)\). Our goal is to show that if \(f\) is odd, then \(f'\) is even, and if \(g\) is even, then \(g'\) is odd.

Step 2: Derivative of an Odd Function

Consider the odd function \(f\). We need to find \(f'(-x)\). By the definition of the derivative, \(f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\). Applying this to \(-x\):\[ f'(-x) = \lim_{{h \to 0}} \frac{f(-x+h) - f(-x)}{h} \]Using the odd property \(f(-x) = -f(x)\), this becomes:\[ f'(-x) = \lim_{{h \to 0}} \frac{-f(x-h) + f(x)}{h} \]If we consider \(k = -h\) (note \(k \to 0\) as \(h \to 0\)), we have:\[ f'(-x) = \lim_{{k \to 0}} \frac{-f(x) + f(x+k)}{-k} = \lim_{{k \to 0}} \frac{f(x+k) - f(x)}{k} = f'(x)\]Thus, \(f'(-x) = f'(x)\), proving \(f'\) is even.

Step 3: Derivative of an Even Function

Consider the even function \(g\). We need to find \(g'(-x)\). Using the definition of the derivative, \(g'(x) = \lim_{{h \to 0}} \frac{g(x+h) - g(x)}{h}\). Applying this to \(-x\):\[ g'(-x) = \lim_{{h \to 0}} \frac{g(-x+h) - g(-x)}{h} \]Using the even property \(g(-x) = g(x)\), this becomes:\[ g'(-x) = \lim_{{h \to 0}} \frac{g(-(x-h)) - g(x)}{h} = \lim_{{h \to 0}} \frac{g((h-x)) - g(x)}{h} \]If we now consider \(k = -h\) (note \(k \to 0\) as \(h \to 0\)), we have:\[ g'(-x) = \lim_{{k \to 0}} \frac{g(-(x-k)) - g(x)}{-k} = -\lim_{{k \to 0}} \frac{g(x-k) - g(x)}{k} = -g'(x)\]Thus, \(g'(-x) = -g'(x)\), proving \(g'\) is odd.

Key Concepts

Odd Function

When we describe a function as 'odd', it means the function has a specific type of symmetry. For any odd function, flipping the input (i.e., replacing it with its negative) will flip the output:- In mathematical terms, an odd function satisfies the condition \( f(-x) = -f(x) \) for all values of \( x \).This symmetry is around the origin on a graph. If you rotate the graph of an odd function 180 degrees about the origin, it will look the same.Examples of odd functions include:

• The function \( f(x) = x^3 \).
• The function \( f(x) = sin(x) \).

Odd functions have interesting calculus properties. Specifically, the derivative of an odd function is even, which conveys how evenness and symmetry play into calculus principles.

Even Function

Even functions come with a different kind of symmetry than odd functions. With an even function, flipping the input does not change the output:- Mathematically, an even function is defined by the property \( g(-x) = g(x) \) for all \( x \).This symmetry is across the y-axis. So, if you fold the graph of an even function along the y-axis, the two halves match perfectly.Here are some common examples of even functions:

• The function \( g(x) = x^2 \).
• The function \( g(x) = cos(x) \).

In calculus, the derivative of an even function is odd, highlighting how these two forms of symmetry influence derivative properties.

Derivative Properties

Derivative properties involve the rules that describe how the slopes of functions behave. When we differentiate functions, particularly odd and even functions, interesting relationships emerge:- **For Odd Functions**: The derivative of an odd function is even. This means if \( f \) is odd, then \( f' \) satisfies \( f'(-x) = f'(x) \).

This property arises because taking the derivative of an odd function means the changes in slopes are exactly mirrored at each side of the origin, which aligns with being even.

- **For Even Functions**: The derivative of an even function is odd. If \( g \) is even, then \( g' \) satisfies \( g'(-x) = -g'(x) \).

The mirrored slopes of even functions when differentiated become shifted and mirrored, giving rise to an odd function as its derivative.

Understanding these properties not only enhances comprehension of calculus but also deepens insights into the symmetrical nature of mathematical functions.