Question

Even and Odd Functions (a) Show that if \(f\) is a differentiable even function, then \(f^{\prime}\) is an odd function. (b) Show that if \(f\) is a differentiable odd function, then \(f^{\prime}\) is an even function.

Short answer

The derivative of an even function is an odd function, and the derivative of an odd function is an even function. This is demonstrated by applying the limit definition of the derivative to an even function and an odd function, respectively, and using the properties of even and odd functions.

Step-by-step solution

Show that the derivative of an even function is odd

Let \(f\) be an even function. This means \(f(x) = f(-x)\) for all x. If \(f\) is differentiable, then by the definition of the derivative, \(f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). If \(f\) is even, then \(f(x+h) = f(-(x+h))\). Therefore, \(f^{\prime}(x) = \lim_{h \to 0} \frac{f(-(x+h)) - f(-x)}{-h}\), which simplifies to \(f^{\prime}(x) = -f^{\prime}(-x)\), thus showing that \(f^{\prime}\) is an odd function.

Show that the derivative of an odd function is even

Let \(f\) be an odd function. This means \(f(x) = -f(-x)\) for all x. If \(f\) is differentiable, then by the definition of the derivative, \(f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). If \(f\) is odd, then \(f(x+h) = -f(-(x+h))\). Therefore, \(f^{\prime}(x) = \lim_{h \to 0} \frac{-f(-(x+h)) + f(-x)}{-h}\), which simplifies to \(f^{\prime}(x) = f^{\prime}(-x)\), thus showing that \(f^{\prime}\) is an even function.