I'd be interested to understand why is the total derivative of a function $f(t,x,y)$, where $x = x(t)$ and $y=y(t)$ defined as:

$$\frac{df}{dt} = \frac{\partial f}{\partial t}\frac{dt}{dt} + \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$$

This formula seems intuitive to me, but if I would need to prove the general case analytically, how would I do it? :)

Thank you for any help! :)


In computing the partial derivative $\frac{\partial f}{\partial x_i}$ of a function $f(x_1,\dots,x_n)$ w.r.t. to a variable $x_i$, one assumes that the other variables do not vary in the computational process. This is a consequence of the very definition of partial derivative. More generally, if the variables $x_2,\dots,x_n$ are functions of the variable $x_1$, then the partial derivative $\frac{\partial f}{\partial x_1}$ does not catch the overall change of $f$ while $x_1$ varies.

One needs to introduce another measure of such change, i.e. the total derivative

$$\frac{df}{dx_1}:=\frac{\partial f}{\partial x_1}+\sum_{i=2}^n \frac{\partial f}{\partial x_i}\frac{d x_i}{d x_1}.$$

From its definition (this is the point: I take it as a definition, although you can prove it using the chain rule on $f(x_1,x_2(x_1),\dots,x_n(x_1)))$ it is clear that the total derivative takes into account of all changes when the variable $x_1$ varies.

To arrive at a geometrical interpretation of the total derivative, let us multiply both sides of the above expression with the infinitesimal increment "$dx_1$", arriving at

$$df(x):=\frac{df}{dx_1}dx_1=\frac{\partial f}{\partial x_1}dx_1+\sum_{i=2}^n \frac{\partial f}{\partial x_i} d x_i~~(*).$$

We can interpret $df(x)$ in $(*)$ as the total differential of $f$ at $x=(x_1,\dots,x_n)$ as

$$df(x)=\frac{\partial f}{\partial x_1}dx_1+\sum_{i=2}^n \frac{\partial f}{\partial x_i} d x_i=\sum_{i=1}^n \frac{\partial f}{\partial x_i} d x_i=\langle \nabla f(x), dx \rangle $$

where $dx=(dx_1,\dots,dx_n)$ represents an infinitesimal increment.

| cite | improve this answer | |
  • $\begingroup$ +1 Thank you @Avitus really nice answer :) I get it now :) $\endgroup$ – jjepsuomi Jul 8 '13 at 11:38

Try starting with f(x+Δx ,y+Δy) - f(x,y) = f(x+Δx, y+∆y) - f(x+Δx, y) + f(x+Δx, y) - f(x, y).

Reference https://www.physicsforums.com/threads/proof-of-the-total-differential-of-f-x-y.467858/

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.