Xrfgtjtk

I'm really not clear with the subtlety between derivative and differential. Let $f:mathbb R^2to mathbb R$, then the derivative is $nabla f(x,y)=(partial _x(x,y) f,partial _y f(x,y))$ where as the differential is $df=partial f_xdx+partial _y dy$. What these notation mean ? Also, I always thought that differential and derivative where almost the same, but with thise definitions, I'm completely confuse.

I'm going to try to answer this:

If you consider a function $f:mathbb{R}^n to mathbb{R}$, then the vector of its partial derivatives, $nabla f(x_1,.....,x_n)=(frac{partial f}{partial x_1},....,frac{partial f}{partial x_n})$ is called the gradient and it is a vector in $mathbb{R}^n$. It is a generalisation of the derivative in a one-dimensional setting.

The total derivative (what you called differential), in the case where the codomain is one-dimensional, is simply a scalar and is the result of multiplying the gradient $nabla f(x_1,.....,x_n)=(frac{partial f}{partial x_1},....,frac{partial f}{partial x_n})$ with a vector $vec{v} in mathbb{R}^n $ (vetor multiplication). It represents the linear approximation of the variation of f when the the coordinates vary by $vec{v}$.

Example
$f(x,y) = x^2+ y^2$, then $nabla f(x,y) = (2x,2y)$. If you consider the point (1,1), then $nabla f(1,1) = (2,2)$ . If you have a variation of the coordinates by the vector $vec{v}= (3,3)$, so you go from $(1,1)$ to $(4,4)$, then your linear approximation of the variation of the function is $ (2,2)*(3,3) = <(2,2),(3,3)> = 12$. This means that the differential (linear approximation of the variation) is of 12.

So if you compute $f(1,1) + nabla f(1,1)*(3,3) = 14$, you obtain a linear approximation of $f(4,4) = 32$. Now this is a bad approximation because we only took the first-order derivative in consideration (look at Taylor Series for more detail on it). If instead we had $f(x,y) = 2x + 2y$, you would see that the linear approximation would give you the exact value.