Ytukyg

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

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 $$lim_{h to 0}frac{left|f(p+h)-f(h)-L(h)right|}{|h|} = 0$$ 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).

This pertains to the first question. $f_y(0, 0)$ is neither 0 nor 1 in the limit. This is because $f_y(x,y)$ is not continuous at (0, 0).

To see this, approach the point (0, 0) along the line y = x. You get,$$f_y(0, 0)= 1$$

If you approach along y = 2x (put x = t and y = 2t), you get, $$f_y(0, 0) = 28/25$$

Thus, $lim_{(x, y)to(0, 0)} f_y(x, y)$ does not exist.

Now the definition of continuity requires the value at $f_y(0, 0)$ to equal the limit, which does not exist.

Therefore, $f_y(x, y)$ is not continuous at (0, 0).

Definition of partial derivative of $f$ in $(a, b)$ respect to $y$ is
$$
f_y(a, b)=lim_{hrightarrow 0}frac{f(a, b+h)-f(a, b)}{h}
$$
then
$$
f_y(0, 0)=lim_{hrightarrow 0}frac{f(0, h)-f(0, 0)}{h}=lim_{hrightarrow 0}frac{frac{h^3}{0+h^2}-0}{h}=lim_{hrightarrow 0}frac{h}{h}=1
$$

As @Joker said $f_y$ isn't continue in $0$ then that example doesn't contraddict the general principle. $f$ isn't differentiable because if it is then $nabla f(0, 0)=(0, 1)$ (their coordinate must be the partial derivatives) and
$$
lim_{x, y)rightarrow (0, 0)}frac{frac{y^3}{x^2+y^2}-0-0cdot x-1cdot y}{sqrt{x^2+y^2}}=lim_{x, y)rightarrow (0, 0)}frac{-x^2y}{(x^2+y^2)^{frac 32}}
$$
passing to polar coordinates with $x=rhocostheta$, $y=rhosintheta$
$$
frac{-x^2y}{(x^2+y^2)^{frac 32}}=frac{-rho^3cos^2thetasintheta}{rho^3}=cos^2thetasintheta
$$
and doesn't go to $0$ when $rhorightarrow 0$ then limit doesn't exists and $f$ isn't differentiable.