Total derivative in multi-variable calculus I read the definition of differentiability as given in Tom M. Apostol's Mathematical Analysis as :
The function $f$ is said to be differentiable at $c$ if there exists a linear
function $T_c : \mathbb R^n \to \mathbb R^m$ such that
$f(c + v) = f(c) + T_c (v) + ||v|| E_c (v)$, where $E_c (v) \to 0$ as $v \to 0$.
There $T_c()$ is called a linear function and called the total derivative of $f$ at $c$, $c$ is a point in $\mathbb R^n$.
I can relate to the "error term" $E_c$ as I am familiar with Taylor's formula and this is a first order approximation of $f$. But I can't make a connection with functions in one variable. Is there a total derivative equivalent in one dimension? And also, everywhere on the internet this definition seems to be obscure and I only find total derivative defined as $df/dt$, like here. What connection am I missing? 
 A: Take $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ (domain can also be an open subset of $\mathbb{R}^n$).  We say $f$ is differentiable at $c$ if there exists a linear transformation $T_\mathbf{c}: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that
$$
\lim_{\mathbf{h} \to 0} \ \frac{|f(\mathbf{c}+\mathbf{h})-f(\mathbf{c})-T_\mathbf{c}(\mathbf{h})|}{|\mathbf{h}|} = 0.
$$ 
This is equivalent to what you wrote. 
At any rate, we write $df_\mathbf{c} = T_\mathbf{c}$ when it exists.  By definition, $df_\mathbf{c}$ is a linear transformation (input variable $\mathbf{h}$) and so is representable by an $m \times n$ matrix $A$.  In the general case this is the matrix of partials of the component functions.  
In the 1-variable case, this is a $1 \times 1$ matrix, which is just a scalar and the ordinary derivative you're used to. 
Example:  let $f(x)=x^2$ and let's examine $f'(3)$ (which will be 6, trust me).  The linear transformation $T: \mathbb{R} \to \mathbb{R}$ via  $T(h)=6h$ satisfies my condition:
$$
\lim_{h \to 0} \ \frac{f(3+h)-f(3)-6h}{h} = 0.
$$
(I don't need norms anymore since my $h$ is now a scalar, not a vector:  single variable calculus!)
Hence $T(h)=6h$ is the magic transformation, and as $T(1)=6$ we write the derivative as $f'(3)=6$, which is what you've always known from Calc I.
