Any function defined from $\Bbb R$ (the set of real numbers) to $\Bbb R$ is monotonic iff its derivative never changes sign, yes. But $\tan(x)$ is not a function from $\Bbb R$ to $\Bbb R$, since it's not defined on all real numbers.
In fact, the thing about derivatives is only true when the domain is a connected set (i.e. an interval)! (Remember that $\Bbb R=(-\infty,\infty)$ and is thus an interval.)
The domain of $\tan(x)$ is:
$$\left\{x\in\Bbb R:x\ne k\pi+\frac\pi2\right\}$$
This is not an interval! Therefore, the theorem linking derivatives to monotonicity does not hold.
In fact, $\tan(x)$ is not monotonic. To see this, note that:
$$\frac\pi3<\frac{2\pi}3$$
but:
$$\tan\Big(\frac\pi3\Big)>\tan\Big(\frac{2\pi}3\Big)$$
(The former is $\sqrt3$; the latter is $-\sqrt3$.)
However, $\tan(x)$ is monotonic over the interval $(-\pi/2,\pi/2)$. And it is monotonic over any interval on which it's defined. But it's not monotonic over its entire domain.
Basically, there are the following inclusions
$(\star)$ $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n) \subset \mathcal{E}(\mathbb{R}^n)$ and $\mathcal{E}'(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$.
Also $L^p(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n)$. Almost all of these inclusions are also continuing, i.e. $\mathcal{D}_K(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n)$, or also $L^p(\mathbb{R}^n) \hookrightarrow \mathcal{S}'(\mathbb{R}^n)$, this means that the inclusion operator $\iota : L^p(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ is continuous with respect to topology defined in those spaces, which basically may be a topology induced by a separable seminorms family or by norm. Where
(a) $\mathcal{D}(\mathbb{R}^n)$ is the space test functions
(b) $\mathcal{S}(\mathbb{R}^n)$ is the Schartz space
(c) $\mathcal{E}(\mathbb{R}^n)$ is the space of the regular functions
(d) $\mathcal{E}'(\mathbb{R}^n)$ is the space of the distribution with compact support
(e) $\mathcal{S}'(\mathbb{R}^n)$ is the space of the tempered distributions
(f) $\mathcal{D}'(\mathbb{R}^n)$ is the space of the distributions
For example, we have the inclusions $\mathcal{E}'(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$ because the inclusions $\mathcal{D}(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow \mathcal{E}(\mathbb{R}^n)$ are continuous and dense with respect to topology of these spaces and then, for example, the application $v \in \mathcal{E}'(\mathbb{R}^n) \longrightarrow v=u_{\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one. Therefore each distribution determines a continuous linear functional $v : \mathcal{E}(\mathbb{R}^n) \longrightarrow \mathbb{C}$ (with respect to convergence in $\mathcal{E}(\mathbb{R}^n))$ which it is a compact support distribution, likewise $v : \mathcal{S}(\mathbb{R}^n) \longrightarrow \mathbb{C}$ is a temperate distribution.
Look, here are a lot of theorems to prove, I'm doing you a summary, and basically $\mathcal{S}'(\mathbb{R}^n)$, that is the space of tempered distributions, it's a good space to define the Fourier transform for duality, because essentially the Fourier transform of Schwartz functions satisfies the very useful properties.
Most of these I have studied in several books, there is not just one. "Real Analysis: Modern Techniques and Their Applications" by Folland and "Linear Functinoal Analysis" by "J. Cerda" cover a large part of these topics.
Best Answer
FYI this was problem B5 in the 2010 Putnam exam, so you can find it here: http://amc.maa.org/a-activities/a7-problems/putnamindex.shtml
They had a pretty succinct solution. Suppose $f$ is strictly increasing. Then for for any $y_0$ you can define an inverse funciton $g(y)$ for $y>y_0$ such that $x=g(f(x))$. Differentiating, we get $1=g'(f(x))f'(x)=g'(f(x))f(f(x))$, so that $g'(y)=\dfrac{1}{f(y)}$. We know that $g$ obtains arbitrarily large values since it is the inverse function of $f$ and $f$ is defined for all $x$, which means $g(z) - g(y_0) = \displaystyle \int_{y_0}^zg'(y)dy = \int_{y_0}^z\frac{dy}{f(y)}$ must diverge as $z\rightarrow\infty$.
Now all we have to do is show that $f$ is bounded below by a function that causes the integral to converge. For $x>g(y_0)\equiv g_0$, we have $f'(x)>g_0$, so we can assume that for some $\beta$ and $x$ large enough, $f(x)>\beta x$. Iterating this argument, we get that $f(x)>\alpha x^2$ for some $\alpha$ and $x$ large enough. So we can assume that $f(x)$ is asymptotically greater than $\alpha x^2$. But then the integral above converges, contradicting that $g(z$) is unbounded as $z\rightarrow\infty$. Thus, we conclude that $f$ cannot be strictly increasing.