Answer:
a. Let f(x) be an even function, then we have f(-x) = f(x) for all x in the domain of f. We want to show that the derivative of f(x), denoted as f'(x), is an odd function.
Using the chain rule, we have:
f'(-x) = (f(-x))' = (f(x))' = f'(x)
Now, let's evaluate f'(-x) + f'(x):
f'(-x) + f'(x) = f'(x) + f'(x) = 2f'(x)
Since f'(-x) + f'(x) simplifies to an expression that involves only f'(x), we can conclude that f'(x) is an even function.
b. Let g(x) be an odd function, then we have g(-x) = -g(x) for all x in the domain of g. We want to show that the derivative of g(x), denoted as g'(x), is an even function.
Using the chain rule, we have:
g'(-x) = (g(-x))' = (-g(x))' = -g'(x)
Now, let's evaluate g'(-x) + g'(x):
g'(-x) + g'(x) = -g'(x) + g'(x) = 0
Since g'(-x) + g'(x) simplifies to zero, we can conclude that g'(x) is an even function.
Explanation: