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I have the problem $\frac{d}{dt}[p(x(t),t)] = \frac{\partial p}{\partial x}\frac{dx}{dt}+\frac{\partial p}{\partial t}$

Could someone please help me understand why this equality is the case? I imagine it has to do with the chain rule or something but I am having a hard time understanding this idea.

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    $\begingroup$ Are you sure that first partial derivative is with respect to $t$ and not $x$? $\endgroup$
    – Brian Tung
    Jul 4, 2018 at 2:07
  • $\begingroup$ You are right. Sorry I will correct it. I am still not sure why though... $\endgroup$
    – MathIsHard
    Jul 4, 2018 at 2:09

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For a function P(x,y) where x and y are functions of t we have

$$ \frac {d}{dt} P(x,y) = \frac {\partial p }{ \partial x} \frac {dx }{dt}+ \frac {\partial p }{ \partial y} \frac {dy}{dt}$$

your problem is a special case where y=t and as a result you have

$$ \frac {d}{dt} P(x,t) = \frac {\partial p }{ \partial x} \frac {dx }{dt}+ \frac {\partial p }{ \partial t} \frac {dt}{dt}=\frac {\partial p }{ \partial x} \frac {dx }{dt}+ \frac {\partial p }{ \partial t} $$

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  • $\begingroup$ Thank you so very much. I appreciate your time and help. :) $\endgroup$
    – MathIsHard
    Jul 4, 2018 at 2:21
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    $\begingroup$ @MathIsHard Thanks for your attention and understanding. $\endgroup$ Jul 4, 2018 at 2:23
  • $\begingroup$ @ Mohammad Riazi-Kermani is this a case of a product rule of some kind? $\endgroup$
    – MathIsHard
    Jul 4, 2018 at 2:26
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    $\begingroup$ Chain rule not product rule. $\endgroup$ Jul 4, 2018 at 2:27

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