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I'm struggling to prove from first principles that the function $f(x)$ is differentiable.

Definition of differentiable:

Let $S$ be an open interval and let $f : S \rightarrow \Bbb R$ be a function. Let $x_0 \in S$. we say that $f$ is differentiable at $x_0$ if the limit $$\lim_{x\to x_0} \frac {f(x)-f(x_0)} {x-x_0}$$ exists. If this is the case, then the limit i denoted by $f'(x_0)$ and called the derivative of $f$ at $x_0$.

Prove from first principles that the function $g : \Bbb R \rightarrow \Bbb R$ defined by $$g(x) = \begin{cases} 0, & 0 \\ x^3\sin\left(x^{-4}\right), & x \ne 0 \end{cases} $$ is differentiable.

gt6989b
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Joel Biffin
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  • If this is from a textbook, I'm curious as to which one. If this was assigned in a class, it seems rather pointless and sadistic to me since I don't see the skills being developed of any use. Separately obtaining the derivative of $x^3$ and maybe $1/x^4$ by first principles are reasonable, as well as $\sin x$ (for which you'll have to make use of a certain well known limit), but what you have seems going overboard. (continued) – Dave L. Renfro Jan 16 '18 at 17:30
  • And for what it's worth, I say this despite my answer to Differentiation using first principles with rational powers. (At least what I did there illustrates several sometimes useful methods. I don't see this happening in your case.) I wonder if what you really want is to show the function is differentiable at $x=0$ from first principles? – Dave L. Renfro Jan 16 '18 at 17:31

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