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?
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 $$ 2\sqrt{f(0)}=2\sqrt{\langle c_{\theta},c_{\theta}\rangle}=2|c_{\theta}|. $$