Frechet Derivative of Evaluation function This is an exercise from Jack K.Hales book on ODEs 
Let $X=C^1([0,1],\mathbb{R}^n) \times [0,1]$ and consider
$$
w:X\longrightarrow \mathbb{R}^n
$$
where $w(f,t)=f(t)\in \mathbb{R}^n$. The point is to show that $w$ is Frechet differentiable and calculate it's derivative.
Some thoughts are that if we fix $f\in C^1([0,1],\mathbb{R}^n)$ then
$$
|f(t+h)-f(t)+D_{t}(f)h|\leq r(t,h)
$$
with $\frac{ r(t,h)}{h}\longrightarrow 0$ for some function $r$ and $D_{t}(f)$ denoting the total derivative of the vector valued $f$ at $t$.
Now would it be correct to define the operator
$$
D:X\longrightarrow \mathbb{R}^n
$$
with $D(f,t)=D_{t}(f)$ to be the Frechet derivative of $w$?
 A: The way you started is unfortunately wrong. The Frechet derivative of $w$ is
$
Dw\left(\left[\begin{smallmatrix}f \\ t\end{smallmatrix}\right]\right) = \begin{bmatrix}\delta_t & f'(t)\end{bmatrix}
$
(a linear operator).
Just for clarification how to interpret this equality:
$$
Dw\left(\begin{bmatrix}f\\t\end{bmatrix}\right)
\begin{bmatrix}g \\ s\end{bmatrix}
= \begin{bmatrix}\delta_t & f'(t)\end{bmatrix}
\begin{bmatrix} g \\ s\end{bmatrix}
= g(t) + f'(t)s.
$$

Calculating the Frechet derivative
First of all we need a norm on $X$. I suggest
$$\|(f,t)\|_X := \|f\|_\infty + \|f'\|_\infty + |t|$$
where $f'$ denotes the derivate of $f$ and $\|g\|_\infty = \sup_{s\in[0,1]} \|g(s)\|_p$ for any $p$-norm on $\mathbb{R}^n$.

Recall an equivalence of the definition of Frechet derivative:

Let $\phi: X \to Z$ and $D\phi : X \to L(X,Z)$ then $D\phi$ is the
Frechet derivate of $\phi$ if $$\|\phi(x+y) - \phi(x)- D\phi(x)y\|_Z \in o(\|y\|_X)\tag{1}$$

In our case we have that each $x\in X$ has the representation $(f,t)$.
An educated guess for the Frechet Derivate of $w$ is
$$
Dw(x)y
= Dw\left(\begin{bmatrix}f\\t\end{bmatrix}\right)
\begin{bmatrix}g \\ s\end{bmatrix}
= \begin{bmatrix} \delta_t & f'(t) \end{bmatrix}
\begin{bmatrix}g \\ s\end{bmatrix} 
= g(t) + f'(t)s
$$
This comes from regarding $w(x+y)-w(x)$.
In order to show that our guess is in fact the Frechet derivative we will show the condition $(1)$. Let $x = (f,t) \in X$ and $y=(g,s) \in X$
\begin{align}
\|w(x+y)-w(x) - Dw(x)y \|_p &= \|(f+g)(t+s)-f(t) - g(t) - f'(t)s\|_p \\
&= \|f(t+s) + g(t+s) - f(t) - g(t) - f'(t)s\|_p \\
&\leq \|f(t+s) - f(t) - f'(t)s\|_p + \|g(t+s) - g(t)\|_p
\end{align}
Since $\|y\|_X = \|g\|_\infty + \|g'\|_\infty + |s|$ we have from the well-known properties of differentiable functions that
\begin{align}
\|f(t+s) - f(t) - f'(t)s\|_p \in o(|s|) \subseteq o(\|(g,s)\|_X)
\end{align}
and
\begin{align}
\|g(t+s) - g(t)\|_p = \|g'(t)s + o(s)\|_p \in o(\|(g,s)\|_X)
\end{align}
which gives us that also the sum is in $o(\|(g,s)\|_X) = o(\|y\|_X)$. Hence we showed $(1)$ and consequently the desired statement.
