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After seeing the following equation in a lecture about tensor analysis, I became confused.

$$ \frac{d\phi}{ds}=\frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds} $$

What exactly is the difference between $d$ and $\partial$?

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    $\begingroup$ Total derivative vs partial derivative. $\endgroup$
    – user2468
    Aug 16, 2012 at 18:28
  • $\begingroup$ @Jen, maybe elaborate a bit more in an answer? :) $\endgroup$ Aug 16, 2012 at 18:33
  • $\begingroup$ I know, one is the partial and the other one is a total derivative. But isn't $\frac{\partial f}{\partial x}$ the same as $\frac{df}{dx}$? $\endgroup$
    – iblue
    Aug 16, 2012 at 18:35
  • $\begingroup$ @iblue, the first one treats all the other independent variables as if they were constants. The second one doesn't. $\endgroup$ Aug 16, 2012 at 18:40
  • $\begingroup$ @iblue Read what Jennifer linked to, it will help. $\endgroup$
    – Pedro
    Aug 16, 2012 at 18:40

2 Answers 2

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The difference is whether the rest of the variables of $f$ are considered constants or variables in $x.$ Former is partial, latter is total.

  • $\frac{\partial f}{\partial x}$ is the partial derivative: $f$ is differentiated w.r.t. to $x$ while all other variables are considered constants in $x.$

  • $\frac{d f}{d x}$ the is total derivative: $f$ is differentiated w.r.t. to $x$ while nothing is assumed about the other variables; they are considered variables in $x.$ (some variables might be, in fact, constants in $x.$)

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As mentioned $d$ means total and $\partial$ partial derivative and are not the same. Total derivative also counts $x$ dependencies in other variables. For instance: $$ f(x,v) = x^2 + v(x) \\ \frac{\partial f}{\partial x} = 2x \\ \frac{\partial f}{\partial v} = 1 \\ \frac{d f}{d x} = 2x + \frac{\partial v(x)}{\partial x} $$

Your formula most probably uses Einstein notation and is only a shorter way to write $$ \frac{d\phi}{ds}= \sum_m \frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds} $$

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    $\begingroup$ Why it's partial dirivative of $v$ when you take derivative to $f$ ? is it suppose to be $\frac{df}{dx} = 2x + \frac{dv(x)}{dx}$ ? $\endgroup$
    – peng yu
    Apr 20, 2020 at 14:58

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