Derivative of dot product Let $\frac{dy}{dt}=Ay+g(y)$ and consider $\lVert y(t)\rVert^2=\langle y,y\rangle$. I would like to prove that
$$
\frac{d}{dt}\langle y,y\rangle= 2\langle\frac{dy}{dt},y\rangle.
$$
To do so, I made the following start:
$$
\langle y+\Delta y,y+\Delta y\rangle=\langle y,y\rangle+2\langle\Delta y,y\rangle+o(\lVert \Delta y\rVert)\text{ as }\Delta y\to 0.
$$
So the derivative seems to arise from the summand
$$
2\langle\Delta y,y\rangle.
$$
But how? It seems to be
$$
2\langle\Delta y,y\rangle=2\langle\dot{y},y\rangle\Delta y?
$$
 A: It is straight forward:
\begin{align}
\frac{d}{dt} (y \cdot y) 
&= \frac{d}{dt}\sum_i y_i^2 \\
&= \sum_i 2 y_i \dot{y_i} \\
&= 2 \sum_i y_i \dot{y_i} \\
&= 2(y \cdot \dot{y})
\end{align}
A: The derivative $\frac{d}{dt}\left<y(t),y(t)\right>$ is defined as $$\frac{d}{dt}\left<y(t),y(t)\right> \equiv \lim_{\Delta t\to 0}\frac{1}{\Delta t}\left[\left<y(t+\Delta t),\,y(t+\Delta t)\right> - \left<y(t),y(t)\right>\right]$$ 
To prove the result in the question we can try to rewrite the right hand side above (forget about the limit until the end) using the laws of an inner-product 


*

*Bi-linearity: $\left<x+y,z\right>=\left<x,z\right> + \left<y,z\right>$

*Multiplication by scalar: $c\left<x,z\right>=\left<cx,z\right>$ for $c\in\mathbb{C}$.

*Symmetry: $\left<x,y\right>=\left<y,x\right>$


to try to put it on a form resembling $2\left<\frac{dy(t)}{dt},y(t)\right>$. Doing this we find
$$\frac{1}{\Delta t}\left[\left<y(t+\Delta t),\,y(t+\Delta t)\right> - \left<y(t),y(t)\right>\right] = \left<\frac{y(t+\Delta t) - y(t)}{\Delta t},y(t)\right>  + \left<y(t),\frac{y(t+\Delta t)-y(t)}{\Delta t}\right> + \Delta t\left<\frac{y(t+\Delta t) - y(t)}{\Delta t},\frac{y(t+\Delta t) - y(t)}{\Delta t}\right>$$
And from this form it's just a matter of taking the limit $\Delta t\to 0$ to arrive at the result (you will need to use that "the limit of a product equals the product of the limits" for the last term). Also note that the inner-product is a continuous function in both arguments so the limit can be moved inside the brackets.
