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 $f(x)$?
Homework Help: Bounded function and differentiation
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calculus
real-analysis
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0Well, it is a fairly sloppy claim regardless of who ma$k$es it. That's all I wanted to say... – 2012-11-10
1 Answers
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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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0I 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. – 2012-11-10