I found this definition of total variation of a function $F(x, y)$ \begin{eqnarray} dF(x, y) &\equiv& F(x+dx, y+dy) -F(x, y)\\ & = & [F(x+dx, y+dy) - F(x, y+dy)] + [F(x, y+dy) - F(x, y)] \\ & = & \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy \end{eqnarray}
But I did not understand why $[F(x+dx, y+dy) - F(x, y+dy)]$ and $[F(x, y+dy) - F(x, y)]$ are equals to $\frac{\partial F}{\partial x}dx$ and $\frac{\partial F}{\partial y}dy$ respectively. Can anyone help me?
$\endgroup$ 11 Answer
$\begingroup$By the definition of partial derivative, it is the rate of change of the function with respect to some variable, while keeping all others fixed.
Therefore, at any x
$$\frac{\partial F}{\partial y}=\frac{F(x, y+dy) - F(x, y)}{dy}$$
Also, at any y $$\frac{\partial F}{\partial x}=\frac{F(x+dx, y) - F(x, y)}{dx}$$
Now in the second expression we can replace $y$ with $y+dy$ everywhere, and it will still remain a partial derivative, but now evaluated at $y+dy$ instead of $y$, which is the same thing at this order. Thus $$\frac{\partial F}{\partial x}=\frac{F(x+dx, y+dy) - F(x, y+dy)}{dx}$$
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