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$$f(x) = \begin{cases}x^4\left(2 + \sin\frac1x\right) & x\neq 0\\ 0 & x=0.\end{cases}$$

Prove that $f$ is differentiable on $\Bbb R$.

I know how to prove that $f$ is differentiable at $x = 0$ by showing that $$\lim_{h \to 0} \frac {f(0 + h) - f(0)}h = 0$$ and solving it, but how do i prove that it is differentiable if $x$ doesnt equal 0? Is it showing $$\lim_{h \to 0} \frac {f(x + h) - f(x)}h?$$ Stuck can someone show me how?

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    As you can probably see, you definition of $\,f(x)\,$ came out messed up. Try to fix this.2012-11-06
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    You can even find a formula for the derivative of $f$ away from zero using the product rule and the chain rule. Products and compositions of differentiable functions are differentiable.2012-11-06
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    Just to let you know, Mantan, you never defined $f(0)$ in your original post (hence the question marks I put in there). I suspect it's supposed to be $0$ (since otherwise, it *can't* be differentiable at $x=0$.2012-11-06
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    "I know how to prove that $f$ is differentiable at $x=0$ by showing that $\lim_{h\to0}\frac{f(0+h)−f(0)}{h}=0$ and solving it [...]" — FYI this is a common misuse of language. One doesn't solve a limit, one computes it. Essentially your "and solving it" is not only unnecessary, it's also gibberish.2012-11-06

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