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

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 $\partial$?

• – user2468
Aug 16, 2012 at 18:28
• @Jen, maybe elaborate a bit more in an answer? :) Aug 16, 2012 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 $\frac{df}{dx}$? Aug 16, 2012 at 18:35
• @iblue, the first one treats all the other independent variables as if they were constants. The second one doesn't. Aug 16, 2012 at 18:40
• @iblue Read what Jennifer linked to, it will help.
– Pedro
Aug 16, 2012 at 18:40

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 $x.$)
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 $$\frac{d\phi}{ds}= \sum_m \frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds}$$
• Why it's partial dirivative of $v$ when you take derivative to $f$ ? is it suppose to be $\frac{df}{dx} = 2x + \frac{dv(x)}{dx}$ ? Apr 20, 2020 at 14:58