Partial derivatives at critical points

Question

If $f(x,y)$ is a continuous and differentiable function then $f_x=f_y=0$ at local maximum, minimim and saddle points.How to prove this?

Answer 1

We will prove

• If $f(x,y)$ is differentiable on an open set containing $(x_0,y_0)$ and $f$ has a local extremum at $(x_0,y_0)$, then $(x_0,y_0)$ is a critical point of $f$.

Define $B_r(x,y)$ as the open ball of radius $r$ centered at $(x,y)$. Suppose $f(x,y)$ is differentiable on an open set containing $(x_0,y_0)$ and $f$ has a local extremum at $(x_0,y_0)$. Let $r>0$ so that $f$ is differentiable $B_r(x_0,y_0)$. Let $g$ be a function of one variable defined by $g(x)=f(x,y_0)$ with $x$ satisfying $(x,y_0)\in B_r(x_0,y_0)$, i.e., $x\in(x_0-r,x_0+r)$.

Now, $g(x)$ is differentiable and $g$ attains its extreme value at $x=x_0$, since $f_x(x,y_0)=g'(x)$. Therefore, $g'(x_0)=0$. Hence, $f_x(x_0,y_0)=0$. A similar argument can be used to show that $f_y(x_0,y_0)=0$.