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. 
 A: If $f:M\to N$ is some (possibly nonlinear) function (here I have in mind a diffeomorphism), then the Jacobian $J_f$ can be viewed as a linear map taking tangent vectors at some point $p \in M$ and returning a tangent vector at $f(p) \in N$.
Let's consider the case where $M$ and $N$ are both $\mathbb R^3$.  $f$ is some vector field (or perhaps it could be viewed as a transformation of the vector space), and the Jacobian $J_f$ tells us about directional derivatives of $f$.  Given a direction $v$ at a point $p$, $J_f(v) = (v \cdot \nabla) f|_p$.
If $M$ and $N$ are both $\mathbb R^1$, then what do we have?  There's only one linearly independent tangent vector at each point, so $J_f$ is uniquely determined by some unit vector $\hat x$ and we get $J_f(\hat x) = \frac{df}{dx} \hat x$.  The Jacobian here just tells us how a unit length is stretched or shrunk when we view $f$ as a transformation of the real line.
This is the way in which a Jacobian is a linear map:  it tells us how directions in the domain correspond to directions in the range.  And even 1d derivatives can be seen in this way.  The components of the Jacobian are still the partial derivatives you're familiar with.  We're just using those partial derivatives to talk about transformations of directions under some function, some map.
A: 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
$$\frac{d(cf)}{dx} = c\frac{df}{dx}.$$
A: 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).
A: A derivative of a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ at $x \in \mathbb{R}^n$ is a linear map $L : \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that
$$\lim_{v \rightarrow 0} \frac{f(x+v)-f(x)-L(v)}{\|v\|} = 0$$
or alternatively $f(x+v) = f(x) + L(v) + o(\|v\|)$, i.e. $f(x)$ is the constant part of $f$ at $x$, and $L$ is the linear part of $f$ at $x$, and everything else is sub-linear.
In the one-dimensional case, the linear map is just multiplication by a scalar, and we call that scalar $f'(x)$, with $L(v) = f'(x)\, v$ for $v \in \mathbb{R}$.
For the case $f: \mathbb{R}^n \rightarrow \mathbb{R}$, the partial derivative at $x$ in the direction $u$ (a unit vector in $\mathbb{R}^n$ would be $L(u) \in \mathbb{R}$.
This generalizes directly to functions $f: X \rightarrow Y$ between Banach spaces $X$ and $Y$ (vector spaces with a norm which are additionally complete [so that limits are well-behaved]).
