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Consider following function:

$f(x)=x^2 \sin{\frac{1}{x} }$ if $x\neq 0$ and $f(0)=0$.

Why does the derivative of $f(x)$ exist? Find the deriviative and determine whether or not it is continous.

Now it is easy to find the derivative:

$f'(x)=2x \sin{\frac{1}{x}}-\cos{\frac{1}{x}}$. $(*)$

One can further find that the derivative exists at $x=0$ but is not continous at that point.

All that is clear to me but the explenation given for its existance is the chain rule 1. I was taught the rule as a means for computing derivatives of computing dervivatives of products. How does the chain rule convey existance of a derivative?

Edit: Typo: Changed $\cos{x}$ in $(*)$ to $\cos{\frac{1}{x}}$

www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch4.pdf

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  • $\begingroup$ if you have two functions that are differentiable on some open region $D$, the chain rule tells you that their product $f\cdot g$ is differentiable on that region AND how to explicitly calculate the derivative of $f\cdot g$ assuming you know the derivatives of $f$ and $g$, respectively. $\endgroup$ – user159517 Jan 20 '16 at 19:28
  • $\begingroup$ I have looked at your class notes, theorem 4.20 (Chain Rule). It says "If $f$ is differentiable at $c$ and $g$ is differentiable at $f(c)$, then $g◦f: A→\mathbb{R}$ is differentiable at $c$." Thus, as set forth in your school notes, the chain rule conveys the existence of a derivative. $\endgroup$ – ForgotALot Jan 20 '16 at 19:31
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    $\begingroup$ And it's not about derivatives of products but about derivatives of composites. Those derivatives (of composites) happen to be products. $\endgroup$ – Justpassingby Jan 20 '16 at 19:53
  • $\begingroup$ and $f'(x) = 2 x \sin(1/x) - \cos(1/x)$ $\endgroup$ – reuns Jan 20 '16 at 20:00
  • $\begingroup$ @Justpassingby I admit I got a bit confused: I know that the chain rule and product rule are two different things $\endgroup$ – Thomas Windisch Jan 20 '16 at 20:06
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It's hard to see how a general differentiation rule can justify the existence of $f'(0)$. Instead, you have to resort to the definition: $$f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h-0}=\lim_{h\to0}\frac{f(h)}{h}=\lim_{h\to0}h\sin(1/h)=0,$$ where the last equality may be justified by noting that $-h\le h\sin(1/h)\le h$ and applying the squeeze law.

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  • $\begingroup$ – or did I misread the quesetion? It looked to me that you were asking about justification for the existence of $f'(0)$. If that is not what you meant, perhaps you could clarify the question. $\endgroup$ – Harald Hanche-Olsen Jan 20 '16 at 20:34

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