Question

(a) Show that the derivative of an odd function is even. That is, if \(f(-x)=-f(x),\) then \(f^{\prime}(-x)=f^{\prime}(x)\) (b) Show that the derivative of an even function is odd. That is, if \(f(-x)=f(x),\) then \(f^{\prime}(-x)=-f^{\prime}(x)\)

Short answer

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

Step-by-step solution

Understanding the properties of odd and even functions

An even function is defined as a function that satisfies \(f(-x) = f(x)\) for every \(x\) in the function's domain. Conversely, an odd function is a function that satisfies \(f(-x) = -f(x)\) for every \(x\) in the function's domain. It's important to remember these definitions for understanding step 2.

Demonstrating derivative of odd function is even

If the function \(f(x)\) is odd, then \(f(-x) = -f(x)\). We're going to prove that the derivative of \(f(x)\) is even. We start by defining the derivative at a point \(-x\), so \(f'(-x) = \lim_{h \to 0} \frac{f(-x+h)-f(-x)}{h}\). Since \(f\) is odd we can substitute \(f(-x+h) = -f(x-h)\) and \(f(-x) = -f(x)\). We find that \(f'(-x) = \frac{-f(x-h)+f(x)}{h} = \frac{f(x) - f(x-h)}{h}\), which is the negative of the definition of derivative at \(x\), thus \(f'(-x) = f'(x)\), and so the derivative of an odd function is even.

Demonstrating derivative of even function is odd

If the function \(f(x)\) is even, then \(f(-x) = f(x)\). We're going to prove that the derivative of \(f(x)\) is odd. We start by defining the derivative at a point \(-x\), so \(f'(-x) = \lim_{h \to 0} \frac{f(-x+h)-f(-x)}{h}\). Since \(f\) is even, we can substitute \(f(-x+h) = f(x-h)\) and \(f(-x) = f(x)\). We find that \(f'(-x) = \frac{f(x-h) - f(x)}{h} = \frac{-f(x) + f(x-h)}{h}= -\frac{f(x)-f(x-h)}{h}\), which is just the definition of derivative at \(x\), thus \(f'(-x) = -f'(x)\), and so the derivative of even function is odd.