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The problem reads as follows:

Find a map $f: [0,1] →ℝ$ such that $f$ is differentiable everywhere but $f'$ is unbounded.

Obviously $f$ is continuous, and $f'$ is unbounded if

$∀M \geq 0 \ ∃x$ such that $|f'(x)|>M$

How would you go about finding this function?

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    It should be $|f'(x)|>M$. – 2017-01-24
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    Of course.. Typo – 2017-01-24

2 Answers 2

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Try $$f(x) = x^2\sin\left(\frac{1}{x^2}\right).$$ and $f(0) = 0$.

For $x>0$ the derivative is $$f'(x)= 2x\sin(1/x^2) -\frac{2}{x}\cos(1/x^2)$$ which is unbounded as $x\rightarrow 0.$ The derivative at zero also exists at zero and is $$ \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x} = \lim_{x\rightarrow 0} x\sin\left(\frac{1}{x^2}\right)=0$$ so the function is differentible on $[0,1].$

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    Makes sense. I take it you meant to have 1/x^2 in the derivative as well? – 2017-01-24
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    @Helene Yes, I sure did. – 2017-01-24
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    Also, it seems to be bounded above? Would it still be unbounded? – 2017-01-24
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    @Helene The derivative is unbounded near the origin ( graph it or look at the second term) – 2017-01-24
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The famous example here is something like $f(x) = x^2\sin (1/x^2), x\ne 0,$ with $f(0)=0.$ Check using the definition of the derivative that $f'(0)=0.$ At any other point, use the standard formulas to calculate the derivative. I'm finding $f'(1/\sqrt {2\pi n}) \to - \infty.$