# 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

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.