Derivative of scalar product of two curves The usual scalar product on $\mathbb{R}^n$ is $\langle x,y \rangle = \sum_{i = 1}^n x_iy_i$. 
Given two curves $\alpha,\beta:I \to \mathbb{R}^n$ I want to show that $\langle \alpha(t),\beta(t)\rangle' = \langle \alpha'(t),\beta(t) \rangle + \langle \alpha(t),\beta'(t) \rangle$
$\langle \alpha(t),\beta(t)\rangle' = (\sum a(t)_i \beta(t)_i)' = \sum (\alpha'(t)_i\beta(t)_i+\alpha(t)_i\beta'(t)_i) = \langle \alpha'(t),\beta(t) \rangle + \langle \alpha(t),\beta'(t) \rangle$
Therefore, this proof is quite easy but since this property is used through my Curves and Surfaces course I wonder if it remains true for general scalar products. 
Please reason why it would hold or provide a counterexample.
 A: By definition
$$\left< \alpha(t),\beta(t)\right>' = \lim_{h\rightarrow 0} \frac{\left<\alpha(t+h),\beta(t+h)\right>-\left<\alpha(t),\beta(t)\right>}{h}=$$
$$=\lim_{h\rightarrow 0} \frac{\left<\alpha(t+h),\beta(t+h)\right>- \left<\alpha(t),\beta(t+h)\right>+\left<\alpha(t),\beta(t+h)\right>-\left<\alpha(t),\beta(t)\right>}{h}$$
Using linearity we have
$$=\lim_{h\rightarrow 0} \left<\frac{ \alpha(t+h)-\alpha(t)}{h},\beta(t+h)\right>+\left<\alpha(t),\frac{ \beta(t+h)-\beta(t)}{h}\right>$$
Now if it is possible to insert the limit inside the inner product then by continuity of $\beta$ we have, 
$$=\left<\alpha'(t),\beta(t)\right>+\left<\alpha(t),\beta '(t)\right>$$
why it is possible to insert the limit inside the inner product
Let $a_h\rightarrow a$ and $b_h\rightarrow b$ I claim that $\left<a_h,b_h\right>\rightarrow \left<a,b\right>$  
Proof: add and subtract $\left<a,b_h\right>$ and use the triangle inequality to get  $|\left<a_h,b_h\right>-\left<a,b\right>| = |\left<a_h,b_h\right>-\left<a,b_h\right> + \left<a,b_h\right> - \left<a,b\right>|\leq |\left<a_h,b_h\right>-\left<a,b_h\right>| + |\left<a,b_h\right>-\left<a,b\right>| $
Using the Cauchy Schwartz inequality we have
$$|\left<a_h,b_h\right>-\left<a,b_h\right>| = |\left<a_h-a,b_h\right>| \leq \|a_h-a\|\|b_h\|$$ as $\|b_h\|$ is bounded this converges to zero and the same holds for the other term.
A: It remains true with any scalar product on $\mathbb R^n$. To see it, you can use the fact that, denoting by $A = [a_{i,j}]$ the (constant) matrix of the scalar product, we have $\displaystyle \langle x,y\rangle = \sum_{i,j} x_ia_{i,j}y_j$ for all vectors $x,y\in\Bbb R^n$. Now, you can apply your idea with this formula and conclude...
