# Is there such a thing as a total partial derivative?

1. Jun 25, 2013

### nayanm

Is there such a thing as a total "partial" derivative?

Total Derivative as I've Been Taught
From my understanding, if we have a function s = f(x, y) where the two arguments x and y are related by another function y = g(x), then there is a great deal of difference between ds/dx and ∂s/∂x.

∂s/∂x is simply a partial derivative and can be calculated by treating y as a constant and differentiating f(x, y) with respect to x.

On the other hand, the "total derivative" ds/dx takes the y = g(x) relationship into account and, by the Chain Rule, gives:
$\frac{ds}{dx}$ = $\frac{∂s}{∂x}$$\frac{dx}{dx}$ + $\frac{∂s}{∂y}$$\frac{dy}{dx}$​

This approach is very well explained in Wikipedia:
http://en.wikipedia.org/wiki/Total_derivative#Differentiation_with_indirect_dependencies

A Different Case
However, what happens if we have a function s = f(x, y, z) and only two of the arguments are related, as through y = g(x).

As before, ∂s/∂x can still be calculated by differentiating f(x, y, z) and treating y and z as constants, but what of the total derivative in terms of x?

Such a total "partial" derivative would take the form:
(total partial derivative in terms of x) = $\frac{∂s}{∂x}$$\frac{dx}{dx}$ + $\frac{∂s}{∂y}$$\frac{dy}{dx}$​

But, clearly, we can't notate this as ds/dx since s is also a function of z. Neither can we call is ∂s/∂x since that notation is reserved for the regular partial derivative.

So my question: is there such a concept as a total "partial" derivative"? I haven't been able to find any discussion on such a concept and was curious about whether something like this even exists.

Any replies are appreciated, and thank you in advance!

Last edited: Jun 25, 2013
2. Jun 25, 2013

### JJacquelin

$\frac{ds}{dx}$ = $\frac{∂s}{∂x}$ + $\frac{∂s}{∂y}$$\frac{dy}{dx}$+ $\frac{∂s}{∂z}$$\frac{dz}{dx}$
If $\frac{dz}{dx}$=0 then $\frac{ds}{dx}$ = $\frac{∂s}{∂x}$ + $\frac{∂s}{∂y}$$\frac{dy}{dx}$

3. Jun 25, 2013

### nayanm

If $\frac{dz}{dx}$=0, why would we refer to the quantity on the right as $\frac{ds}{dx}$?

This is STILL only a partial derivative of s since s depends on z as well.

4. Jun 25, 2013

### JJacquelin

I think you are kidding !

5. Jun 25, 2013

### nayanm

Pardon? Forgive me if I'm missing something.

s depends on x, y, and z.
x and y are related; z is independent of the two.

Thus, considering the relation between x and y, we can think of s as a function of x and z only.
s then has two PARTIAL derivatives ∂s/∂x and ∂s/∂z.