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 $\partial$?
Asked
Active
Viewed 1.7k times
22
-
3Total derivative vs partial derivative. – Aug 16 '12 at 18:28
-
@Jen, maybe elaborate a bit more in an answer? :) – J. M. ain't a mathematician Aug 16 '12 at 18:33
-
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 $\frac{df}{dx}$? – iblue Aug 16 '12 at 18:35
-
@iblue, the first one treats all the other independent variables as if they were constants. The second one doesn't. – J. M. ain't a mathematician Aug 16 '12 at 18:40
-
@iblue Read what Jennifer linked to, it will help. – Pedro Aug 16 '12 at 18:40
-
She edited her comment, and added the link to the wikipedia article on total deriviates. Now I see... – iblue Aug 16 '12 at 18:41
2 Answers
19
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 $x.$)
15
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 $$ \frac{d\phi}{ds}= \sum_m \frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds} $$
one too many
- 196
- 4
-
1Why it's partial dirivative of $v$ when you take derivative to $f$ ? is it suppose to be $\frac{df}{dx} = 2x + \frac{dv(x)}{dx}$ ? – peng yu Apr 20 '20 at 14:58