Problem 28 - The Calculus with Analytic Geometry

Question

Use the chain rule to prove that (a) the derivative of an even function is an odd function, and (b) the derivative of an odd function is an even function, provided that these derivatives exist.

Step-by-Step Solution

Answer

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

Step 1: Define Even and Odd Functions

An even function satisfies the condition: \( f(-x) = f(x) \) for all \( x \). An odd function satisfies the condition: \( g(-x) = -g(x) \) for all \( x \).

Step 2: Differentiate Even Function, f(x)

Use the chain rule to find the derivative of \( f(x) \) if \( f(x) \) is an even function: \( \frac{d}{dx}[f(-x)] = f'(-x) \cdot (-1) \). Since \( f(-x) = f(x) \), replace \( f(-x) \) with \( f(x) \), resulting in \( -f'(-x) \).

Step 3: Establish Odd Nature of the Derivative

We have \( \frac{d}{dx}[f(x)] = -f'(-x) \). This implies that \( f'(x) = -f'(-x) \), which means \( f'(x) \) is an odd function.

Step 4: Differentiate Odd Function, g(x)

Use the chain rule to find the derivative of \( g(x) \) if \( g(x) \) is an odd function: \( \frac{d}{dx}[g(-x)] = g'(-x) \cdot (-1) \). Since \( g(-x) = -g(x) \), replace \( g(-x) \) with \( -g(x) \), resulting in \( -g'(-x) \cdot (-1) \).

Step 5: Establish Even Nature of the Derivative

We have \( \frac{d}{dx}[g(x)] = g'(-x) \). This implies that \( g'(x) = g'(-x) \), which means \( g'(x) \) is an even function.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of compositions of functions. Let's consider two functions, say, \(u(x)\) and \(v(x)\), where \(u\) is a function of \(v\), and \(v\) is a function of \(x\). According to the chain rule, the derivative of \(u\) with respect to \(x\) is given by \(u'(v(x)) \times v'(x)\). In simpler terms, if you have \(f(g(x))\), its derivative will be \(f'(g(x)) \times g'(x)\). The chain rule helps us break complex differentiation problems into simpler parts, making it easier to understand and solve.

Even Function

An even function is a type of function that exhibits symmetry about the \(y\)-axis. For a function \(f\) to be even, it must satisfy the condition \(f(-x) = f(x)\) for all \(x\) in its domain. Typical examples of even functions include \(f(x) = x^2\) and \(f(x) = \text{cos}(x)\). When differentiating an even function using the chain rule, we find that the derivative is an odd function. For instance, if \(f(x)\) is even, then \(f(-x) = f(x)\), and differentiating using the chain rule yields \(-f'(-x)\). Hence, \(f'(x) = -f'(-x)\), indicating that \(f'(x)\) is an odd function.

Odd Function

An odd function is a function that satisfies \(g(-x) = -g(x)\) for all \(x\) in its domain. This means odd functions have rotational symmetry about the origin. Common examples include \(g(x) = x^3\) and \(g(x) = \text{sin}(x)\). Using the chain rule to differentiate an odd function, we obtain that the derivative is an even function. Specifically, if \(g(x)\) is odd, then \(g(-x) = -g(x)\). Applying the chain rule gives \(-g'(-x) \times (-1) = g'(-x)\), which implies \(g'(x) = g'(-x)\). Therefore, \(g'(x)\) is an even function.

Differentiation

Differentiation is the mathematical process used to find the rate at which one quantity changes with respect to another. It's a core idea in calculus. The derivative of a function \(f(x)\) is represented as \(f'(x)\) or \(\frac{df}{dx}\). Differentiation answers how a function changes as its input changes. For even and odd functions, differentiation has unique properties. Differentiating an even function results in an odd function, while differentiating an odd function results in an even function. These properties can be proven using the chain rule, highlighting the interconnectedness of these fundamental concepts.