Let $X$ be the vector space $X = C([0, 1], R)$ equipped with sup-norm. Let $P$ be a polynomial, let $x_0 ∈ [0, 1]$, and let $F : X → R$ be defined by $F(f) = P(f(x_0))$
I am supposed to show that F is differentiable at each element $f ∈ X$ and find the derivatives.
I've thought of starting with the directional derivative:
$\lim_{t\to 0} \frac{F(f+tr)-F(f)}{t} = \lim_{t\to 0} \frac{P(f(x_0)+tr(x_0))-P(f(x_0))}{t}$.
Can I then say that this equals
$\lim_{t\to 0} \frac{P(f(x_0))+P(tr(x_0))-P(f(x_0))}{t} = \lim_{t\to 0} \frac{P(tr(x_0))}{t} = \lim_{t\to 0} \frac{tP(r(x_0))}{t} = P(r(x_0))$?
First observe that the derivative of the point evaluation functional $E_{x_0}(f)=f(x_0)$ is its own derivative. In fact, $$\lim_{\|g\|\to0}\frac{\|E_{x_0}(f+g)-E_{x_0}(f)-E_{x_0}(g)\|}{\|g\|}=\lim_{\|g\|\to0}\frac{\|(f+g)(x_0)-f(x_0)-g(x_0)\|}{\|g\|}=0$$
Your operator $F$ is the composition of $E_{x_0}$ and $P:R\to R$. The latter having derivative (multiplication by) $P'$, defined as usual.
Then, you can apply the chain rule:
$$\begin{align}(DF)(f)&=D(P\circ E_{x_0})(f)\\&=(DP)(E_{x_0}(f))\circ (DE_{x_0})(f)\\&=P'(f(x_0))\circ E_{x_0}(f)\\&=P'(f(x_0))\cdot f(x_0)\end{align}$$