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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! :)

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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.

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

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