Are derivatives linear maps? I am reading Rudin and I am very confused what a derivative is now. I used to think a derivative was just the process of taking the limit like this
$$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}$$
But between Apostol and Rudin, I am confused in what sense total derivatives are derivatives. 
Partial derivatives much more resemble the usual derivatives taught in high school 
$$f(x,y) = xy$$
$$\frac{\partial f}{\partial x} = y$$
But the Jacobian doesn't resemble this at all. And according to my books it is a linear map.  
If derivatives are linear maps, can someone help me see more clearly how my intuitions about simpler derivatives relate to the more complicated forms? I just don't understand where the limits have gone, why its more complex, and why the simpler forms aren't described as linear maps. 

I'll just answer based on what linearity of a derivative means; the basic derivative is linear, when viewed from the right context.
As I mentioned in the comment, if we have a vector space $V$ (let's say it's over the field of real numbers, $\Bbb R$), then we say a map $T: V \to V$ is linear if, given vectors $f, g \in V$ and a scalar $c \in \Bbb R$, the following hold:
\begin{align*}
T(f + g) &= T(f) + T(g) \\
T(cf) &= cT(f).
\end{align*}
So, the trick is to think of a function $f: \Bbb R \to \Bbb R$ as belonging to the vector space of functions from $\Bbb R$ to $\Bbb R$ (if you need to, convince yourself that this really is a vector space; we can add them, multiply by scalars, we have a $0$ function, etc). 
If we call this vector space of functions $V$, then the derivative \begin{align*}\frac{d}{dx}: V &\to V\\
f &\mapsto \frac{df}{dx} = f'
\end{align*}
is a map from $V$ to $V$ (we apply the derivative to a function, and get a function in return), and it satisfies the linearity requirements, as immortalized by the phrase "the derivative of a sum is the sum of derivatives" and the fact that we can "pull constants out" while taking a derivative;
$$\frac{d(f + g)}{dx} = \frac{df}{dx}+\frac{dg}{dx},$$ and
$$\frac{d(cf)}{dx} = c\frac{df}{dx}.$$
