Problem 80 - Calculus - AP Edition

Question

Recall that a function \(f\) is even if \(f(-x)=f(x)\) and odd if \(f(-x)=-f(x),\) for all \(x\) in the domain of \(f .\) Assuming that \(f\) is differentiable, prove: (a) \(f^{\prime}\) is odd if \(f\) is even (b) \(f^{\prime}\) is even if \(f\) is odd.

Step-by-Step Solution

Answer

If \(f\) is even, then \(f'\) is odd; if \(f\) is odd, then \(f'\) is even.

Step 1: Interpret the Given Information

We need to show that the derivative of a function, denoted as \(f'\), is odd if the original function \(f\) is even, and that \(f'\) is even if \(f\) is odd. We will use the definitions of even and odd functions: \(f(-x) = f(x)\) implies \(f\) is even, and \(f(-x) = -f(x)\) implies \(f\) is odd.

Step 2: Prove (a) – If \(f\) is even, then \(f'\) is odd

Given \(f\) is even, we have \(f(-x) = f(x)\). Differentiating both sides with respect to \(x\) using the chain rule, we get:\[-f'(-x) = f'(x)\]This implies \(f'(-x) = -f'(x)\), showing that \(f'\) is odd.

Step 3: Prove (b) – If \(f\) is odd, then \(f'\) is even

Given \(f\) is odd, we have \(f(-x) = -f(x)\). Differentiating both sides with respect to \(x\) using the chain rule, we get:\[-f'(-x) = -f'(x)\]Simplifying this, we obtain \(f'(-x) = f'(x)\), showing that \(f'\) is even.

Key Concepts

Even and Odd Functions

Understanding even and odd functions is key to grasping the behavior of a function under various transformations. A function is considered ***even*** if for every value of \(x\) in its domain, \(f(-x) = f(x)\). This symmetry occurs around the y-axis. Think of it like a mirror image. Typical examples of even functions include polynomials like \(x^2\) and cosines like \(\cos(x)\).

On the flip side, a function is ***odd*** if for any \(x\), \(f(-x) = -f(x)\). This type of function has rotational symmetry around the origin. If you rotate the graph 180 degrees about the origin, it looks the same. Some examples are \(x^3\) and \(\sin(x)\).

These symmetries give even and odd functions unique properties, especially when evaluating their derivatives, leading us right into the realm of calculus.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to differentiate compositions of functions. If you have a function composed of two functions, like \(g(f(x))\), the chain rule states that you need to take the derivative of the outer function and multiply it by the derivative of the inner function. Mathematically, this is written as \((g(f(x)))' = g'(f(x)) \cdot f'(x)\).

This rule is especially useful when tackling the differentiation of even and odd functions, as seen in the step-by-step solution.

When differentiating an even function \(f(x)\), expressed as \(f(-x) = f(x)\), the chain rule helps show that the derivative \(f'(-x) = -f'(x)\), proving that \(f'\) is odd. Conversely, for odd functions \(f(-x) = -f(x)\), applying the chain rule gives \(f'(-x) = f'(x)\), establishing that \(f'\) is even.

• Remember: Always involve the chain rule when differentiating compositions of functions!
• Keep track of negative signs, as these indicate the symmetry of the function.

Properties of Derivatives

Derivatives have several important properties, especially when dealing with function symmetries like even and odd. One key property is how the derivative of a function reflects its original symmetry.

1. **Derivative of Even Functions:** When you differentiate an even function, it produces an odd derivative. This happens because the derivative reflects the function's rate of change, which now behaves oppositely at symmetrically opposite points. As demonstrated through differentiation using the chain rule, if \(f(x)\) is even, \(f'(-x) = -f'(x)\) makes \(f'(x)\) odd.

2. **Derivative of Odd Functions:** Conversely, differentiating an odd function results in an even derivative. This is due to the way the original function's symmetry influences the derivative. For an odd \(f(x)\), the derivative satisfies \(f'(-x) = f'(x)\), which confirms evenness.

• These symmetric properties of derivatives are powerful as they provide insights into the function's behavior without directly plotting it.
• Understanding these properties helps predict and analyze the curves effectively.

Derivatives tell us not just about slopes, but also about the symmetries and transformations happening in the graph of the function!