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I'm really not clear with the subtlety between derivative and differential. Let $f:\mathbb R^2\to \mathbb R$, 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.

user623855
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    The question is not completely clear. A derivative (or gradient) is a vector. A differential is a linear map. The differential of $f$ at $(a,b)$ is the best approximation of $f$ by a linear map. – Surb Jan 12 '19 at 15:51

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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.

jffi
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