Background:
I am studying Calculus from Stewart which is not rigorous. So, Stewart does not talk about this problem. Also, multivariable calculus is dealt mostly in context of three dimensions which is alright for me. And, I would like the answer in the context of 3 dimensions.
Definitions:
Differentiable: A function is differentiable at a point iff the tangents with respect to every direction (a,b) lies on one plane. This definition is just an intuition. I may be wrong and it may not be equivalent to the standard definition. But the problem still stands even if the two definitions are not equivalent.
Motivation:
Problem: If the partial derivatives are continuous at a point, then the function is differentiable at the point. This problem that if the partial derivatives are continuous at a point, then all the tangents lie on a single plane has been bugging me for a while.
Sidenote: I saw in one forum post that existence of both partial derivatives at the point, and continuity of one is enough for the above problem to be true.
Trials and thoughts:
I don't seem to have a mental picture of of how this theorem is true and how the argument works. I have thought about proving the inverse: If all the tangents lie on a plane, then one of the partial derivatives might not be continuous. But, I haven;t succeeded.
All I have done is set up the problem using the respective definitions. First I found the equation of the plane which contains the the two partial tangents and the point. Then I found the slope of the line which lies on the plane and is directed in a particular direction (a,b). Then, I tried to show that this ratio is equal to the directional derivative in the direction (a,b). The proof of this will illuminate why this theorem works the way it works.
Thanks for the answers.
Best Answer
Suppose that $f$ has continuous partial derivatives. Then it is differentiable. Formally, this means that $$ \frac{f(x+h,y+k)-f(x,y)-h\,\frac{\partial f}{\partial x}(x,y)-k\,\frac{\partial f}{\partial y}(x,y)}{\sqrt{h^2+k^2}}\to0,\ \ \ \text{ as} (h,k)\to0. $$ In any case this implies that the directional derivatives exist: for $(a,b)$ with $a^2+b^2=1$, $$ D_{\vec{(a,b})}f(x,y)=\lim_{t\to0}\frac{f(x+at,y+bt)-f(x,y)}t $$ exists at it is equal to $$ D_{\vec{(a,b})}f(x,y)=(a,b)\cdot\nabla f(x,y). $$ So the tangent vectors at $(x,y,f(x,y))$ are given by $$(a,b,(a,b)\cdot\nabla f(x,y))=(a,b,a\,\frac{\partial f}{\partial x}+b\,\frac{\partial f}{\partial y}),$$ for any choice of $a,b$ with $a^2+b^2=1$.
We can explicitly see that the vector $(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1)$ is perpendicular to all such vectors, and this shows that they are in the same plane.