I'm trying to find $\frac{d}{dt}[f(\textbf{c}(t))]\rvert_{t=0}$
Where $\textbf{c}(t)$ is such that $\frac{d}{dt}\textbf{c}(t) = \textbf{F}(c(t))$, with $\textbf{F} = \nabla f$, $\textbf{c}(0) = (-\frac{\pi}{2}, \frac{\pi}{2})$ and $f(x,y) = \sin(x + y)xy$
At the moment I'm thinking that I would need to use the chain rule to find it, that is:
$\frac{d}{dt}[f(\textbf{c}(t))] = \frac{df}{d\textbf{c}}\frac{d\textbf{c}}{dt}$
I think that $\frac{d\textbf{c}}{dt} = \textbf{F}(\textbf{c}(t)) = \nabla f$
However I am not sure exactly how to find $\frac{df}{d\textbf{c}}$, what would I need to do from here? Am I correct up tiil this point?
Thanks very much in advance.
The chain rule can be extremely misleadingly taught and/or written down. I sympathize. Here, an application of the chain rule should look like
$\frac{d}{dt} f(\mathbf{c}(t)) = \nabla f (\mathbf{c}(t)) \cdot \frac{d\mathbf{c}}{dt}(t)$