Difference of total derivative and partial derivative

1. Aug 16, 2014

mikengan

many books only tell the operation of total derivative and partial derivative,

so i now confuse the application of these two.

when doing problem, when should i use total derivative and when should i use partial derivative.
such a difference is detrimental when doing Physics problem, so i help someone can help me explain the meaning of these two derivatives.

2. Aug 21, 2014

Maharshi Roy

It depends upon the problem. Please provide an example so that I can explain it better. However refer below for a little help:-

Q.1) Find the rate of change of ideal gas pressure with respect to volume?
Soln:
P=nRT/V
We require rate of change of pressure only with respect to volume, hence we're gonna ignore the other parameters like temp., moles, and treat them as constants. Then comes partial differentiation.
(∂P/∂V)=(-nRT)/(V^2)

Q. 2) Calculate the differential change in ideal gas pressure?
Soln:
Now total differential is required.
dP=(nR/V)dT + (RT/V)dn - (nRT/V^2)dV

3. Aug 21, 2014

Blazejr

Let $f$ be function of three independent variables: $f=f(x,y,z)$. Partial derivative with respect to one of them is just rate of change of $f$ with respect to that variable.
As an example, take our function to be potential energy which depends only on a position of a particle: $V=V(x,y,z)$. If you wanted to know by how much potential energy of your particle changes if you move it just a tiny little bit along the x axis, you would compute $\frac{\partial V}{\partial x} \Delta x$. Since we decided that V only depends on position, we also have $\frac{\partial V}{\partial t}=0$, because if you changed $t$ a little bit and !held other variables fixed!, nothing would change. That's what we mean when we say that $V$ doesn't depend on time explicitly.
Now, if we are solving problem in mechanics, let's say, the coordinates $x,y,z$ will in fact be functions of $t$, and finding them will often be your task. Now you can, in a way think of $V$ as implicit function of t. What does it mean? Since coordinates $x,y,z$ are functions of time, you can "plug them into" $V$ function, and get that $V(t)=V(x(t),y(t),z(t))$. Now as time goes by in mechanical system, the particle moves and the coordinates change. If we now treat $V$ as a function of single variable $t$ in a way I presented, we can compute ordinary derivative known from calculus 1. It is radically different object though - we are not assuming that other variables are held fixed now! That is what some people would call "total derivative" with respect to time.

4. Aug 21, 2014

HallsofIvy

Staff Emeritus
Are you sure the phrase is "total derivative" rather than "total differential"? I think the term "total differential is more common than "total derivative" although I have seen the latter used occasionally (with a meaning different from "total derivative").

If f(x, y, z) is a function of the three variables, x, y, and z, then the partial derivatives are, of course, $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, and $\frac{\partial f}{\partial z}$. If, in addition, x, y, and z are themselves all functions of some other variable, t, we could replace each of x, y, and z with its expression as a function of t, reducing f to a function of the single variable t, which then has derivative $\frac{df}{dt}$.

By the "chain rule" for several variables we have
$$\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}+ \frac{\partial f}{\partial z}\frac{dz}{dt}$$

That is what is called the "total derivative" though, as I said, the "total differential", which would be
$$df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz$$
is more often used. Notice that there is no dependence on an additional variable, t, so this is much more general.