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It is easy to verify by definition that the function $f(x)=x^2 \sin(1/x), x\neq0,f(0)=0 $is differentiable at the origin ,that is, $f'(0)=0$.But by the formula we can not calculate this.$f'(x)=2x\sin(1/x)-\cos(1/x)$. How can we explain this?Do you know any other function excluding $x^2\cos(1/x)$ that behaves like this.

Thanks for your help

Yegan

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A differentiable function need not be of class $C^1$. If your second approach worked, then it is easy to check that $f$ would be $C^1$.

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Any function $f(x)$ such that $|f(x)|\leq C x^2$ for some $C>0$ and $x$ in a neighborhood of $0$ is differentiable at zero with $f'(0)=0$. Simply use the difference quotient so see this:

Note that $f(0)=0$. Thus $$|f'(0)|\leftarrow\frac{|f(x)-f(0)|}{|x-0|}\leq C |x| \to 0 $$ as $x\to 0$

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Consider

$$\left| \frac{f(x) - f(0)}{x- 0} - 0 \right|= |x||\sin(1/x)| \leq |x|\cdot 1,$$

which certainly goes to $0$ as $x \to 0$. Thus $f'(0) = 0$.

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