Show that the function $g(x) = x^2 \sin\left(\frac{1}{x}\right) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$.
Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$
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calculus
real-analysis
derivatives
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0possible duplicate of [Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$](http://math.stackexchange.com/questions/393602/differentiability-of-fx-x2-sin-frac1x-and-f) – 2014-03-01
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0@TuckerRapu: The dependency is the other way around; this question is older than the one you've linked. – 2014-04-09