Consistency of exp(x) being its own derivative as a special 1-dimensional case in vector differentiation Recently I learnt the more general definition for differentiation of a vector as: 

*

*If c is an interior point of S $\subset \mathbb{R}^n $ and $\mathbf{f}: S \rightarrow \mathbb{R}^m$, then f is differentiable at the point c if there exists a linear transformation $ \mathbf{M}: \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that in some open ball centred on c we have
$$ \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{c}) + \mathbf{\mathsf{M}}*(\mathbf{x-c}) + \boldsymbol{\eta}$$
where $\boldsymbol{\eta} = o(\mathbf{x-c})$ as $\mathbf{x} \rightarrow \mathbf{c}$, and  $\mathbf{\mathsf{M}}$ is the matrix dependent on c, $\mathbf{\mathsf{M}}:= \mathbf{\mathsf{Df}}(\mathbf{c})$. (I think also that $\mathbf{M}(\mathbf{x}) = \mathbf{\mathsf{M}}\mathbf{x}) $ 
And was introduced to linear transformations $\mathbf{L} :\mathbb{R}^n \rightarrow \mathbb{R}^m$ being their own derivative s.t. $\mathbf{\mathsf{DL}}=\mathbf{\mathsf{L}}$. I didn't find this result particularly hard to prove, but it doesn't seem to be consistent, with m=n=1, with the real analysis result that exp{x} is the unique function being its own derivative, since: 

*

*exp{x} is not a linear transformation of x, as neither exp{x+y}=exp{x}+exp{y} nor exp{ax}=a*exp{x} for real a is true. Hence there should be no relevant 1x1 $\mathbf{\mathsf{M}}$ matrix representative of the linear transformation, and so no derivative. Is that correct?

*I think, by doing the algebra, that here $\mathbf{\mathsf{M}}=\exp{c}$ which would imply that $M(x)=\exp{x}=\exp{c}*x$ which is pretty clearly not true for most x.

*The function $f(x)=ax$ is a linear transformation in $\mathbb{R}$ which would imply the uniqueness is not true. Here $\mathbf{\mathsf{M}}=a$.

My linear algebra is really rusty so I'm sure some of the things above are wrong, the application of linear transformations especially. I'd appreciate if someone could explain how exactly the exponential is consistent with the linear transformation derivative property, and also possibly give some remarks on the accuracy of my thoughts.
Thanks for your help
 A: The result is still true for $n=m=1$. There is a difference between derivatives and derivative functions. Linear maps are the only functions which are their own derivative. The exponential function is the only function which is its own derivative function.
The derivative is just a linear map. The derivative function is a function mapping each point to a linear map. In the case where $n=m=1$, we identify linear maps with numbers. The linear map $x\mapsto ax$ is identified with the number $a$. And lo! the derivative of this map is $a$ at all points, which is exactly the number with which we identified the map.
For $n\neq1$, it doesn't make sense to talk about a function being its own derivative function. The derivative is a linear map $\mathbb R^n\to\mathbb R^m$, and the derivative function is then a function $S\to\operatorname{Lin}(\mathbb R^n,\mathbb R^m)$ into the space of linear maps. Under certain conditions ($n=1$, for instance), this space can be identified with $\mathbb R^m$, and then we can say that a function is its own derivative function. But that's a different thing from saying that it is its own derivative.
