Let $v,w$ vector fields along the curve $\alpha:I\rightarrow S$, where $S$ is a regular surface. I need to prove that $$\frac{d}{dt}\left<v(t),w(t)\right>=\left<\frac{Dv}{dt}(t),w(t)\right>+\left<v(t),\frac{Dw}{dt}(t)\right>.$$
What I tried:
Let $\mathbb{x}(u,v)$ a parametrization of $S$. So, we can write $$w(t)=a\mathbb{x}_{u}+b\mathbb{x}_{v}, v(t)=c\mathbb{x}_{u}+d\mathbb{x}_{v}, $$ where $a,b,c,d$ are differentiable functions. So, we have
$$\left<v(t),w(t)\right>=ac+bd\implies \frac{d}{dt}\left<v(t),w(t)\right>=a'c+ac'+b'd+bd'. $$
Well, I don't know how to get the expressions for the covariant derivatives, since they are given by the Christofell symbols.
What can I do?
The intuition is to remeber that $v(t),w(t)$ are vector fields tangents to the surface $S$. See that
$$\frac{d}{dt}\left<v(t),w(t)\right>=\left<v'(t),w(t)\right>+\left<v(t),w'(t)\right>. $$ Not necessarily $v'$ and $w'$ are tangents to the surface $S$, so, write
$$v'(t)=\frac{Dv}{dt}(t)+v_{n}(t), $$ where $v_{n}(t)$ denotes the normal component of $v'(t)$ and $\frac{Dv}{dt}(t)$ is the tangencial component of $v'(t)$. So, $\left<v_{n}(t),w(t)\right>=0,$ since $v_{n}$ are orthogonal to $w$. So,
$$\frac{d}{dt}\left<v(t),w(t)\right>=\left<v'(t),w(t)\right>+\left<v(t),w'(t)\right>=\left<\frac{Dv}{dt}+v_{n}(t),w(t)\right>+\left<v(t),\frac{Dw}{dt}+w_{n}(t)\right> \\ =\left<\frac{Dv}{dt},w(t)\right>+\left<v_{n},w(t)\right>+\left<v(t),\frac{Dw}{dt}\right>+\left<v(t),w_{n}(t)\right>=\left<\frac{Dv}{dt},w(t)\right>+\left<v(t),\frac{Dw}{dt}\right>. $$