Understanding linearity of derivatives I am a little bit confused about the statement that the derivative of a function $f$ (no matter if it is a partial derivative, the total derivative or simply a derivative in $\mathbb{R}$) is a linear transformation. 
1.) By definition the derivative is always defined at a given point say $a$, so let $Df(a)$ be the derivative at this point $a$. But if you consider the derivative as a function of any point $x$ of the domain of $Df$ then $Df(x)$ is not necessarily linear.
2.) If you consider a derivative at a given point $a$, $Df(a)$ and you then plug in a further variable or vector $v$ then $Df(a)(v)$ is a linear transformation with respect to $v$.
Are those thoughts on derivatives correct?
 A: When we say "The derivative is linear", we most commonly mean something like "The derivative operator, which takes a function to its derivative, is a linear transformation on a vector space of functions". In other words, $$D(f+g)(a)(v) = Df(a)(v) + Dg(a)(v)$$ and $$D(k\cdot f)(a)(v) = k\cdot Df(a)(v)$$ for any differentiable functions $f, g: \Bbb R^m\to \Bbb R^n$, any point $a\in \Bbb R^m$ and any scalar $k\in \Bbb R$.
Yes, it's true that with respect to the point we are differentiating at, the derivative is not linear at all; in general we do not have $Df(x+y) = Df(x) + Df(y)$. I would call this something like "The derivative of $f$ is not a linear transformation".
It is also true that with a function $f$ and a point $a$, $Df(a)$ can be seen as a linear map: the linear map which, together with adding the constant $f(a)$, closest resembles the behaviour of $f$ in the immediate vicinity to $a$. I would call this something like "The derivative of $f$ at $a$ is a linear transformation".
A: You may be misunderstanding what is meant by "the derivative".  In the very simple, one dimensional, case, with $f(x)= x^3$, "The derivative", $f'(x)= 3x^2$, is certainly NOT a linear function.  But the derivative operator, that takes f(x) to f'(x), IS a linear functional.  That is d(af(x)+ bg(x))/dx= a df(x)/dx+ b dg(x)/dx.
