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I need a little help from the math community on a homework problem. The problem states the following: Assume that $g$ is a bounded function such that $|g(x)|< B$ for all $x$. Let $f(x) = (x^2)g(x)$. Why doesn't the product rule for derivatives apply to $f(x)$?

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I assume, you are talking about the derivative $f'(0)$? –  martini Nov 10 '12 at 14:18
    
That is actually a follow up question. Show that f(x) is differentiable at 0 and find f'(0). –  user42864 Nov 10 '12 at 14:21
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Well, it certainly DOES apply (in the same way the rule "If an animal is a bird, then it has two legs" applies to crocodiles: crocodiles are not birds, so the rule just doesn't give any conclusion about them). –  fedja Nov 10 '12 at 14:30
    
@fedja What if a cat ate one of the bird's legs? –  Matt N. Nov 10 '12 at 14:32
    
@fedja I am not making the claim. The authors are. –  user42864 Nov 10 '12 at 14:33

1 Answer 1

up vote 3 down vote accepted

Use the definition of $f'(0)$ as a limit. That way you don't need to assume $g(x)$ is differentiable, or even continuous. $$f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0},$$ which for your function is $$f'(0)=\lim_{x \to 0} \frac{x^2g(x)}{x}=\lim_{x \to 0}xg(x).$$ Since $g$ is bounded, the latter limit is clearly $0$.

For your question "why does the product rule not apply?": The short answer is that g is not said to be differentiable, and the product rule assumes two functions $f(x),g(x)$ are differentiable at $x=a$ , and then the usual formula holds. If this isn't an answer I think the question is "philosophical", not mathematical.

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I had that part. Thank you. –  user42864 Nov 10 '12 at 14:34
    
So I guess your question is "why does the product rule not apply?" If so the short answer is that $g$ is not said to be differentiable, and the product rule assumes two functions $f(x),g(x)$ are differentiable at $x=a$, and then the usual formula holds. –  coffeemath Nov 10 '12 at 14:50
    
I was leaning towards that claim. Like others, I couldn't see a reason why the product rule wouldn't hold. Thank you for your help and reinforcing the claim. –  user42864 Nov 10 '12 at 14:54

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