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. 

The simpler form is a linear map. Regardless of the setting, if you have $G : X \to Y$ which is differentiable at $x$, you will have
$$G(y)=G(x)+G'_x(y-x)+o(\| y - x \|)$$
where $G'_x$ is the derivative of $G$ at $x$, which is a linear map from $X$ to $Y$. When $X=Y=\mathbb{R}$, all linear maps are just multiplication by a real number, so derivatives correspond directly to real numbers. When $X=\mathbb{R}^n,Y=\mathbb{R}^m$, we identify $G'_x$ with a matrix, which we call the Jacobian matrix.
Because of the theorem that the Jacobian of a differentiable function from $\mathbb{R}^n$ to $\mathbb{R}^m$ is the matrix of partial derivatives, there is an analogue of the limit formula in $\mathbb{R}^n$, where each entry is the limit of a particular partial derivative. There is no direct analogue simply because there's no way to make sense of division by a vector (in general).
