# If a function is continuously differentiable and bounded, is its derivative also bounded?

Two questions:

1. Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuously differentiable and bounded. Is its derivative $f'$ bounded as well?

2. What about in the case of $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$, where $f$ is continuously differentiable, and its total derivative $f'$?

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What do you mean by total derivative? In any case, for (2) take a function $\mathbb R^n\to\mathbb R^n$ which is the identity in all coordinates but one, and there make it act like the example from (1). – Mariano Suárez-Alvarez Sep 20 '12 at 21:35

No, e.g. $\sin (x^2){}{}{}{}$.
Thanks (I'll accept as soon as the webpage allows me!). What if $f:\mathbb{R}\rightarrow\mathbb{R}$ is strictly increasing (sorry I know its a different question...)? – jkn Sep 20 '12 at 21:38
@JuanKuntz I think the answer is still no. Intuitively, the derivative must be very small for most large $x$, but could have sharp peaks. – Alex Becker Sep 20 '12 at 21:51
A strictly increasing example: Define $g:\mathbb R_{\geq 0}\to\mathbb R_{\geq 0}$ by $$g(x)=\begin{cases} n^4x-n^5+n &\text{if }n-\frac{1}{n^3}\leq x+1\leq n\\ n+n^5-n^4x &\text{if }n\leq x+1\leq n+\frac{1}{n^3}\\ 0 &\text{otherwise} \end{cases}$$ which intuitively is just a continuous function which is $0$ almost everywhere but has very narrow, tall spikes at each natural number. Note that $g(n-1)=n$, so $g$ is unbounded, yet $$\int_0^\infty g(x)dx = \sum\limits_{n=1}^\infty n\frac{1}{n^3}=\frac{\pi^2}{6}$$ which is finite. Define $f:\mathbb R\to\mathbb R$ by $$f(x)=\begin{cases} \arctan x+\int_0^x g(y)dy &\text{if } x\geq 0\\ \arctan x-\int_0^{-x} g(y)dy &\text{if } x\leq 0\\ \end{cases}$$ which is continuously differentiable, bounded and strictly increasing. Since the derivative of $\arctan x$ is bounded, the derivative of $f$ differs from $g$ by a bounded function, so must be unbounded.