prove using the definition of derivative that: a)the derivative of every even fu
ID: 3002191 • Letter: P
Question
prove using the definition of derivative that:a)the derivative of every even function is an odd function
b)the derivative of every odd function is an even function
Explanation / Answer
sol: suppose that, You can either use the definition of the derivative or the chain rule. so. Recall, f'(x) = lim (h -> 0) [f(x+h) - f(x)]/h Now, suppose f(x) is odd. Therefore, f(-x) = -f(x). We know for an even function, f(-x) = f(x). So let's examine the behavior of f'(-x) given that f(x) is odd; f'(-x) = lim (h -> 0) [f(-x+h) - f(-x)]/h ; this is just the definition of the derivative so , Now, since f(x) is odd, then f(-x + h) = -f(x + h) and again, f(-x) = -f(x) for odd functions. So f'(-x) = lim (h -> 0) [-f(x+h) - -f(x)]/h f'(-x) = lim (h -> 0) [-f(x+h) + f(x)]/h ; two negatives make a positive, Now, suppose we multiply everything top and bottom -1/-1. This is multiplying by 1, so we aren't actually changing anything. So, f'(-x) = lim (h -> 0) [-f(x+h) + f(x)]/h * -1/-1 ; this changes all the signs f'(-x) = lim (h -> 0) [f(x+h) - f(x)]/-h so, Now, suppose we called -h the letter u. Note that as h -> 0, so does u. We can rewrite the limit as f'(-x) = lim (u -> 0) [f(x+h) - f(x)]/u Well, this is just the definition of the derivative, we're just using the letter u instead of h. So the limit is equal to f'(-x) = f'(x) Wait! Notice that the derivative evaluated at -x is just the derivative itself; in other words, this function is even. We just proved that the derivative of an odd function has to be even. A quicker proof is just to start with the definition of an odd function: f(-x) = -f(x) I can bring the negative over on the left, -f(-x) = f(x) so, Now, differentiate both sides with respect to x, -f'(-x) * -1 = f'(x) ; I multiply by -1 since I apply the Chain Rule on the left: the derivative of -x is just -1 Simplify and f(-x) = f(x) Which is just an even function.
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