Ok, so in my notes it says
Prop 1: If a function is differentiable, it will be continuous AND it will also have partial derivatives.
Prop 2: If a function is continuous, or has partial derivatives, or has both, it does not guarantee the function is differentiable.
And the example to follow for prop 2 is:
$f(x,y)=\frac{y^3}{x^2+y^2}$ if $(x,y )\ne (0,0)$
$f(x,y)=0 $ if $(x,y)=(0,0)$
$f_x(0,0)=0$
$f_y(0,0)=1$ (how????)
The function is also continuous at $(0,0)$ since $\lim f(x,y)=0$ (using squeeze theorem)
So it says the partial derivatives exist.
My first question is, why is $f_y(0,0)=1$? shouldn't it be $0$? Not that it makes a difference. The partials will exist regardless.
My second question is, it says that this function is not differentiable. How do they know that?
My third question: It says in the calculus textbook, one of the theorems (theorem 8 of chapter 14.4 for stewert's book): If the partial derivatives $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$ then $f$ is differentiable at $(a,b)$. How does this make sense? The example in my notes just said a function can be continuous and have partials, but still not be differentiable
Best Answer
A function $f:\mathbb{R}^2 \to \mathbb{R}$ is differentiable at a point $p$ if there is a linear map $L:\mathbb{R}^2 \to \mathbb{R}$ such that $$\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$$ If the function is differentiable, then $L$ is the function $L(x,y) = \dfrac{\partial f }{\partial x}\big|_p x +\dfrac{\partial f}{\partial y}\big|_p y$.
If all you know is that the partial derivatives exist, you do not even know that the function is continuous, let alone is well approximated by a tangent plane. For instance $f(x,y) = 0$ if $x=0$ or $y=0$ and $f(x,y) = 1$ otherwise. The partial derivatives of $f$ at $(0,0)$ are all $0$, but the tangent plane is a really crappy approximation to $f$ off of the coordinate axes.
The example you were looking at is a little harder to visualize, but think about what happens to $f$ on lines other than horizontal and vertical ones.
In other words, morally, for a function to be differentiable at $(a,b)$ we need that $f(a+\Delta x,b+\Delta y) \approx f(a,b) + \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$. This talks about approximating $f$ in all directions around $(a,b)$, whereas existence of the partial derivatives only means you have good approximations in two directions (the coordinate directions).