Computing the Frechet differentials I am new to differential calculus on normed spaces and I struggle with some easy things.

Let $- \infty <a < b< +\infty$ and $[a, b]$ denote a finite interval.
Let $C[a,b]$ denote the collection of all real-valued continuous
  functions defined on $[a,b]$. Then, endowed with the usual choice of
  norm $\|x\| = \max_{a\leq t \leq b} |x(t)|$, $C[a,b]$ is a Banach
  space.
Let $\phi \colon \mathbb{R} \to \mathbb{R}$ be a twice differentiable
  function, and suppose that its inverse $\phi^{-1} \colon \mathbb{R}
 \to \mathbb{R}$ exists and is still twice differentiable.
Let the kernel function $K \colon [a,b] \times [a,b] \to \mathbb{R}$ be continuous.
Does the Frechet derivative of the
  following operator $A$ exist? 
  $$ [A(x)](s) = 
  \int^b_a K(s,t) \, [ x(t)] \,\mathrm{d}t $$ for all
  $x \in C[a,b]$ and for all $s \in [a,b]$.

Any ideas or suggestions are much appreciated! Thanks in advance:)
 A: The other answer is correct, but shows only that the map is Gateaux differentiable as you have noted.
To see that it is Frechet differentiable (this is commonly just called "differentiable"), we write the map as a composition:
$$A=L^{-1}\circ \hat K\circ L,$$
here $L(f)=\phi\circ f$ (so $L^{-1}(f)=\phi^{-1}\circ f$) and $\hat K(f)\,(t)=\int_a^b K(t,y)f(y)\,dy$. The map $\hat K$ is linear and continuous, thus differentiable with differential $(d_f \hat K)\,(g)=\hat K(g)$.
We will verify differentiability of the map $L$, the map $L^{-1}$ is basically the same map, except with $\phi$ replaced by $\phi^{-1}$. Your conditions on $\phi$ and $\phi^{-1}$ are the same (both must be twice differentiable functions $\Bbb R\to \Bbb R$), so a proof of the differentiability of $L$ is also a proof of the differentiability of $L^{-1}$.
Once we have found all the differentials the chain rule will tell us:
$$d_fA=d_{\hat K(L(f))}(L^{-1})\cdot d_{L(f)}\hat K\cdot d_f L$$
So why is $L$ differentiable? Let $M_{\phi'\circ f}$ be the map $M_{\phi'\circ f}(g)\,(t)=\phi'(f(t))\cdot g(t)$. This is, for any $f\in C[a,b]$, a continuous linear map $C[a,b]\to C[a,b]$.
Now lets use the mean value theorem, which tells us that
$$\phi(f(t)+h(t))-\phi(f(t)) = \phi'(\xi)\cdot h(t)$$
for some $\xi\in [f(t),f(t)+h(t)]$. Now $\phi'$ is itself continuously differentiable, so
$$|\phi'(\xi)-\phi'(f(t))|≤\sup_{x\in[a,b]}|\phi''(f(x))|\cdot |f(t)-\xi|≤\|\phi''\circ f\|\,\|h\|.$$
It follows:
\begin{align}
\frac{\|L(f+h)-L(f)-M_{\phi'\circ f}(h)\|}{\|h\|} &=\frac{\sup_{t\in[a,b]}|\phi(f(t)+h(t))-\phi(f(t))-\phi'(f(t))\cdot h(t)|}{\|h\|}\\
&=\sup_{t\in[a,b]}\frac{|\phi'(\xi)-\phi'(f(t))|\,|h(t)|}{\|h\|}≤\|\phi''\circ f\|\,\|h\|.
\end{align}
So as $\|h\|\to0$ this expression goes to $0$. This is precisely the condition that $L$ is differentiable at $f$ with differential $d_f L=M_{\phi'\circ f}$. The differential of $L^{-1}$ at $f$ is $d_f(L^{-1})= M_{(\phi^{-1})'\circ f}=M_{\frac1{\phi'}\circ f}$. 
A: Let $\hat K : C[a,b] \to C[a,b]$ be the linear map defined by 
$$\hat K(f) = \int_a^b K(\cdot, t) \, f(t) \, dt.$$
Then we have 
$A(x) = \phi^{-1} \circ \hat K(\phi \circ x)$
and
$$
A(x+\lambda h) 
= \phi^{-1} \circ \hat K(\phi \circ (x+\lambda h))
$$
Expanding in $\lambda$ we now have
$
\phi \circ (x + \lambda \, h) \sim (\phi \circ x + \lambda \, h \, \phi' \circ x)
$
so
$$
\hat K(\phi \circ (x+\lambda h))
\sim \hat K(\phi \circ x + \lambda \, h \, \phi' \circ x)
= \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x) 
$$
by linearity of $\hat K.$
Then
$$
A(x+\lambda h)
\sim \phi^{-1} \circ \left( \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x) \right) \\
\sim \phi^{-1} \circ \hat K(\phi \circ x) + \lambda \, \hat K(h \, \phi' \circ x)
 \cdot (\phi^{-1})' \circ \hat K(\phi \circ x) \\
$$
Thus,
$$
\left< DA(x), h \right> 
= \hat K(h \, \phi' \circ x)
 \cdot (\phi^{-1})' \circ \hat K(\phi \circ x).
$$
