Consider a two-valued function $f(x,y) : R^2 \rightarrow R$.
Define $f_x(x,y) = \frac{\partial f(x,y)}{\partial x}=\lim_{\epsilon\rightarrow 0}\frac{f(x+\epsilon,y)-f(x,y)}{\epsilon}$ and $f_y(x,y)$ as well.
Additionally, there is a function $c(t):R\rightarrow R$.
I want to track the value of function $f$ on the $c(t)$, that is, $f(t,c(t))$.
By differentiating with respect to $t$, I obtain
\begin{eqnarray} \frac{d f(t,c(t))}{d t}=\frac{\partial f(x,y)}{\partial x}\frac{d x}{dt}|_{x=t}+\frac{\partial f(x,y)}{\partial y}\frac{d y}{d t}|_{y=c(t)}\\ =f_x(t,c(t))+f_y(t,c(t))\frac{d c(t)}{d t} \end{eqnarray}
If I take a partial differentiation with respect to $t$ and $c(t)$,repectively, Can I write $$ \frac{\partial f(t,c(t))}{\partial t}=f_x(t,c(t)) $$ and $$ \frac{\partial f(t,c(t))}{\partial c(t)}=f_y(t,c(t)) $$
My question arises here. $\frac{\partial f(t,c(t))}{\partial c(t)}=f_y(t,c(t))$ is well defined? Apparently, there is no condition for $c(t)$. I ask for the conditions on $c(t)$.
If this is well defined, second question is about exchange order of partial differentiation and integration.
When is it possible
$$ \frac{\partial}{\partial c(t)}\int_0^s f(t,c(t))dt=\int_0^s \frac{\partial f(t,c(t))}{\partial c(t)}dt = \int_0^s f_y(t,c(t)) dt $$
???
Thanks in advance.
Gerw, I accepted your comment. Then, I could say
By differentiating with respect to y
$$ \frac{\partial}{\partial y}\int_0^s f(t,c(t))dt=\int_0^s \frac{\partial f(t,c(t))}{\partial y}dt = \int_0^s f_y(t,c(t)) dt $$ . This result is just a matter of notation... am I correct?
In my opinion, this is bad notation. In fact, you can take partial derivatives of your function only w.r.t. the arguments, which you called $x$ and $y$. It is, however, perfectly valid (and just by definition), to write $$\frac{\partial f}{\partial x} (t, c(t)) = f_x(t,c(t))$$ and $$\frac{\partial f}{\partial y} (t, c(t)) = f_y(t,c(t)).$$
For your last question, a definition of $$\frac{\partial}{\partial c(t)} \int_0^s f(t, c(t)) \, \mathrm{d} t$$ is missing.