TITLE: how to compute a differential?
QUESTION [6 upvotes]: Let $$l:\text{Imm}(S^{1},\mathbb{R^2}) \to \mathbb{R}$$ be $$l(c)=\int_{S^1}{|c_{\theta}|d\theta}$$ and its diffrential is $$dl(c)(h)=\int_{S^1}{\frac{<h_{\theta},c_{\theta}>}{|c_{\theta}|}}d\theta$$
Can you please, explain how this differential is computed?

REPLY [5 votes]: I would proceed as follows:
\begin{align}
dl(c)(h)&=\left.\frac{d}{dt}\right|_{t=0}\int_{S^{1}}|c_{\theta}+th_{\theta}|d\theta\\
&=\int_{S^{1}}\left.\frac{d}{dt}\right|_{t=0}\sqrt{\langle c_{\theta}+th_{\theta},c_{\theta}+th_{\theta}\rangle}d\theta\\
&=\int_{S^{1}}\left.\frac{d}{dt}\right|_{t=0}\sqrt{f(t)}d\theta,
\end{align}
where we set $f(t):=\langle c_{\theta}+th_{\theta},c_{\theta}+th_{\theta}\rangle$. Now
$$
\left.\frac{d}{dt}\right|_{t=0}\sqrt{f(t)}=\frac{f'(0)}{2\sqrt{f(0)}},
$$
where
\begin{align*}
f'(0)&=\left.\frac{d}{dt}\right|_{t=0}\langle c_{\theta}+th_{\theta},c_{\theta}+th_{\theta}\rangle\\
&=\left\langle \left.\frac{d}{dt}\right|_{t=0}(c_{\theta}+th_{\theta}),c_{\theta}\right\rangle+\left\langle c_{\theta},\left.\frac{d}{dt}\right|_{t=0}(c_{\theta}+th_{\theta})\right\rangle\\
&=\langle h_{\theta},c_{\theta}\rangle+\langle c_{\theta},h_{\theta}\rangle\\
&=2\langle h_{\theta},c_{\theta}\rangle
\end{align*}
and
$$
2\sqrt{f(0)}=2\sqrt{\langle c_{\theta},c_{\theta}\rangle}=2|c_{\theta}|.
$$