Is the derivative function typically "worse" than the original function?

Yes, that is completely right. And inversely, integration makes functions nicer.

One way of measuring how "nice" a function is, it by how many derivatives it has. We say a function $f\in C^k$ if it is $k$ times continuously differentiable. The more times differentiable it is, the nicer a function is. It is "smoother". So if a function is $k$ times differentiable then its derivative is $k-1$ times differentiable. A function is "as nice" as its derivative if and only if its smooth (infinitely differentiable). These are functions like $\sin(x), e^x$, polynomials, etc.

Inversely, integration makes things nicer. For example integrating even a non continuous function results in a continuous function: Is an integral always continuous?

The concept you're talking about is called smoothness (Wikipedia, MathWorld).

Functions like $e^x$ and $\sin(x)$ and polynomials are called "smooth" because their derivatives, and $n$th derivatives are all continuous. Smooth functions have derivatives all the way down, so they're as nice as their derivatives.

But functions like $\operatorname{abs}(x)$ and $\operatorname{sgn}(x)$ aren't smooth since there are discontinuities in either them or their derivatives. They're nicer than their derivatives.

A function is in class $C^k$ if its derivatives up to, and including, $k$ are continuous. So the number of levels of nice-ness will depend on $k$. Think about how integrating $\operatorname{sgn}(x)$ gives you $\operatorname{abs}(x)$, which in turn gives $\operatorname{sgn}(x)x^2$, and so on. As Zachary Selk points out, you can make functions nicer by integrating them.

In fact, for most functions, its more likely that they can be integrated than differentiated. Not only is being "nice" a rare trait, being differentiable at all is too.

You can view it as a consequence of convolution being at least as nice as the involved functions. Because an integral of a function $f$ is exactly that, a convolution :

$$\int_{-\infty}^t f(\tau)d\tau = \int_{-\infty}^\infty f(\tau) H(t-\tau)d\tau = (f*H)(t)$$

Where $H(t)$ is the Heaviside step function:

$$H(t) = \cases{0 &, t < 0 \\ 0.5 &, t = 0 \\1 &, t > 0}$$