Derivative being linear

Question

Very straight forward question, so I'm studying differentiation between (infinite) normed vector spaces and when considering the very basic example of $f(x)=x^2+2x$ from reals to real we have the usual derivative being $f'(x)=2x+2$. But this isn't a linear map from $\mathbb{R}$ to $ \mathbb{R}$ but rather an affine transformation. The frechet derivative has to be a bounded linear map, but this isn't a linear map, what's going on?

Answer 1

The differential operator is $D: \mathbb{R}^{\mathbb{R}} \to \mathbb{R}^{\mathbb{R}}$, and it's linear, because $\forall f,g \in \mathbb{R}^{\mathbb{R}}$ and $a \in \mathbb{R}$, we have that $D(af+g)=aDf+Dg$. It does not matter if $Df$ is linear function or not.

We call $f: \mathbb{R}^n \to \mathbb{R}^m$ Fréchet differentiable at $a\in (\text{dom}(f))'$ if there exists a linear transformation $L:\mathbb{R}^n \to \mathbb{R}^m$ so that $f$ has the following form: $$f(x)=f(a)+L(x-a)+o(x)$$ And we call $L$ the derivative of $f$ at $a$. In the case of $\mathbb{R}\to\mathbb{R}$ functions, $L$ is just a scalar, and multiplication by a scalar is linear.