What is the complete Partial Derivatives approximation formula to calculate the value of a function? We know the approximation formula using partial derivatives to calculate the value of a function with some variables $x$ and $y$, which can be extended to any number of variables to be 
\begin{equation}
f(x, y) \approx \frac{\partial f(x, y)}{\partial x} \Delta x + \frac{\partial f(x, y)}{\partial y} \Delta y
\end{equation}
What are the higher order terms involving second and $n^{th}$ order derivatives if you want the approximation to be better? 
I recently learnt the formula and I don't know who derived it first or have any leads on what papers to read to get the full formula or proof so any help is appreciated.
Edit:
Assuming it is the Taylor's theorem, is the second order term for this approximation
\begin{equation}
\frac{1}{2!} (\frac{\partial^2 f(x, y)}{\partial x^2} {\Delta x}^2 + \frac{\partial^2 f(x, y)}{\partial y^2} {\Delta y}^2 + \frac{2\partial^2 f(x,y)}{\partial x \partial y}\Delta x \Delta y )
\end{equation}
and there was a constant for $0^{th}$ order term $f(x_{1}, y_{1})$ where you plug in the value of $x$ and $y$ which I am unsure of, so the approximation up until the second order term should be 
\begin{equation}
f(x, y) \approx f(x_{1}, y_{1}) + \frac{\partial f(x, y)}{\partial x} \Delta x + \frac{\partial f(x, y)}{\partial y} \Delta y + 
\frac{1}{2!} (\frac{\partial^2 f(x, y)}{\partial x^2} {\Delta x}^2 + \frac{\partial^2 f(x, y)}{\partial y^2} {\Delta y}^2  \\ +  \frac{2\partial^2 f(x,y)}{\partial x \partial y}\Delta x \Delta y )
\end{equation}
where 
$${\Delta x} = x - x_{1}, \Delta y = y - y_{1} $$ and you plug in $$x = x_{1}, y = y_{1}$$ after partial differentiation except in $$\Delta x, \Delta y$$ and the $0^{th}$ order term is a bit unclear, so I'm unsure if there's a mistake understanding the realization of $x$ and $y$ in there somewhere, or what the notation means when they define it in 2 variables/dimensions here with respect to $f(\alpha)$, perhaps they meant $f(x_{1},y_{1})$ as the $0^{th}$ order term? So we can use it to calculate properties from the equation at some point $x_{1}$, $y_{1}$ on $f(x,y)$, like closeness to some tangent plane or asymptotic analysis to get approximation with predetermined error value with higher order terms.
 A: With the generalized Taylor series :
$$f(x,y)= \sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\left(\frac{\partial^{j+k} f}{\partial x^j\partial y^k}\right)_{(x_1,y_1)} \frac{(x-x_1)^j(y-y_1)^k}{j! k!}$$
See the section "Taylor series in several variables" in :
https://en.wikipedia.org/wiki/Taylor_series#Generalization
The first term with $(j=0,k=0)$ is $f(x_1,y_1)$.
The term with $(j=1,k=0)$ is $\left(\frac{\partial f}{\partial x}\right)_{(x_1,y_1)} \frac{(x-x_1)}{1}$.
The term with $(j=0,k=1)$ is $\left(\frac{\partial f}{\partial y}\right)_{(x_1,y_1)} \frac{(y-y_1)}{1}$.
The term with $(j=1,k=1)$ is $\left(\frac{\partial^{2} f}{\partial x\partial y}\right)_{(x_1,y_1)} \frac{(x-x_1)(y-y_1)}{1}$
The term with $(j=2,k=0)$ is $\left(\frac{\partial^2 f}{\partial x^2}\right)_{(x_1,y_1)} \frac{(x-x_1)^2}{2}$.
The term with $(j=0,k=2)$ is $\left(\frac{\partial^2 f}{\partial y^2}\right)_{(x_1,y_1)} \frac{(y-y_1)^2}{2}$.
and so on …
For example :
The term with $(j=3,k=2)$ is $\left(\frac{\partial^5 f}{\partial x^3\partial y^2}\right)_{(x_1,y_1)} \frac{(x-x_1)^3(y-y_1)^2}{12}$.
For an approximate, takes only a limited number of terms $j\leq J$ and $k\leq K$ instead of $\infty$.
