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Show that $f$ is differentiable on $\mathbb{R}$ and give the value of $f'(x)$ for all $x$ in $\mathbb{R};$ justify any assertions you make. $$f(x)=\begin{cases}x^2\sin\left(\frac{1}{x^2}\right)\cos(x^4),\quad > &x\not=0 \\ 0, &x=0\end{cases}. $$

How do I go about this?

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closed as off-topic by GEdgar, user99914, Math1000, Ethan Bolker, zz20s May 17 '18 at 12:46

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  • 2
    $\begingroup$ Use the definition of the derivative (at the origin, elsewhere it's kinda clear). Observe that $-x^2\le f(x)\le x^2$, so the squeeze theorem bites. $\endgroup$ – Jyrki Lahtonen May 16 '18 at 12:18
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HINT

For $x\neq 0$ we can calculate directly $f'(x)$.

For $x=0$ let consider the definition

$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\,x \sin\left(\frac{1}{x^2}\right)\cos(x^4)$$

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