# First order approximation of multivariable function

Let $O$ be an open subset of $\mathbb R^2$ and suppose the function $f:O\to\mathbb R$ is continuous at the point $(x_0,y_0)$ in $O$.

Define tangent plane as $\phi(x,y)=a+b(x-x_0)+c(y-y_0)$ where $a,b,c$ are real numbers, which has the property that

$$\lim_{(x,y)\to(x_0,y_0)}\frac{f(x,y)-\phi(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0$$

clearly that $\phi$ is a first order approximation of $f$.

I know that if $f$ is continuously differentiable, then the tangent plane at the point $(x_0,y_0)$ is defined by $$\phi(x,y)=f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)$$

However I don't know how to prove the converse theorem:

If $\phi$ is a tangent plane of a function $f:O\to\mathbb R$ at the point $(x_0,y_0,f(x_0,y_0))$ (which at the point is continuous?), then $f$ has first order partial derivatives at $(x_0,y_0)$ and $a=f(x_0,y_0)$, $b=\frac{\partial f}{\partial x}(x_0,y_0)$, $c=\frac{\partial f}{\partial y}(x_0,y_0)$ such that the tangent plane is unique. Also $f$ has directional derivatives in all direction at the point $(x_0,y_0)$

The bracket (which at the point is continuous?) means that I don't know whether it is a necessary condition for the theorem to hold, since I do not know whether tangent plane exist implies continuous at the point.

If you can draw a tangent plane to a multivaraite function $f$ at $(x_0,y_0)$, then this function is surely differentiable at $(x_0,y_0)$. If $f$ is differentiable at $(x_0,y_0)$, then it is also continuous at that point.
You can find all the directional derivatives by the inner product of the vector $\left( \frac{\partial f(x_0,y_0)}{\partial x}, \frac{\partial f(x_0,y_0)}{\partial y} \right)$ and the direction vector $(v_1,v_2)$ (please normalize the direction vector first).