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?

  • $\begingroup$ Does your question relate to the thread: Euler-Lagrange In any case you should have a look on that as it also likely solves your issues with partial derivatives. Let me know wether it helped. :) $\endgroup$ – C-Star-W-Star Nov 2 '15 at 18:00

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.

  • $\begingroup$ I changed my question according to your comment, Gerw. $\endgroup$ – user155214 Nov 2 '15 at 10:53
  • $\begingroup$ Now, a definition of $\frac{\partial}{\partial y} \int_0^s f(t, c(t)) \, \mathrm{d} t$ is missing. The integral does not depend on $y$, hence you cannot take partial derivatives w.r.t. $y$. $\endgroup$ – gerw Nov 2 '15 at 11:34
  • $\begingroup$ By definition, $\frac{\partial f(t,c(t))}{\partial c(t)}=\lim_{h\rightarrow 0} \frac{f(t,c(t)+h)-f(t,c(t))}{h}=f_y(t,c(t))$. So is it well defined? $\endgroup$ – user155214 Nov 2 '15 at 12:31
  • $\begingroup$ And what is the definition of the partial derivative of the integral? $\endgroup$ – gerw Nov 2 '15 at 13:02

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