Can someone give me an example of a function $g:\Bbb R \to \Bbb R$ that is differentiable but the derivative $g'$ is not differentiable and prove there assertion.
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Duplicate – DHMO Apr 14 '17 at 19:28
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@DHMO Duplicate? Really? – zhw. Apr 14 '17 at 19:30
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@zhw. If it isn't continuous how can it be differentiable? – DHMO Apr 14 '17 at 19:30
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Come now @Gibberish, surely you have some ideas on this. – zhw. Apr 14 '17 at 19:33
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There exists a continuous $f:\mathbb R\to \mathbb R$ that is nowhere differentiable. Let $g(x)=\int_0^x f(t)dt$. – DanielWainfleet Apr 14 '17 at 19:43
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is the weierstrass function? – Apr 14 '17 at 19:50
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here is an example: $f(x) = \int_0^x |t| dt.$ by the fundamental theorem of calculus, $f$ is differentiable and its derivative is $|x|$ but $|x|$ is not differentiable at $x = 0.$
abel
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Set $g(x)=x^2 \sin(\frac{1}{x} ) $ when $x\neq 0$ and $g(0)=0$ .
then $g'$ exists at all point and is given by $f'(x)=2x \sin(\frac{1}{x})-\cos(\frac{1}{x}) ,g'(0)=0 $
But $g'$ is not continuous at $x=0$ as $\cos(\frac{1}{x})$ is not continuous at $0$ . As $g'$ is not continuous ,it is not differentiable .
Suman Kundu
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