# Confused about differentiability/continuity/partial derivative existence

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

• Oh I understand now. The theorem says it has to be continuous at (a,b) for the PARTIAL derivatives, not the function itself. Okay. Still wondering about my first and second questions though? – Gary Choi Dec 7 '13 at 1:14
• Are you sure those results are not a mixed partial, like $f_{xy}$? – Amzoti Dec 7 '13 at 1:43

## 3 Answers

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).

• You haven't explained how this relates to the questions the OP asked in their post. – Michael Albanese Jan 19 '16 at 1:31

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

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 $$\lim_{x, y)\rightarrow (0, 0)}\frac{\frac{y^3}{x^2+y^2}-0-0\cdot x-1\cdot y}{\sqrt{x^2+y^2}}=\lim_{x, y)\rightarrow (0, 0)}\frac{-x^2y}{(x^2+y^2)^{\frac 32}}$$ passing to polar coordinates with $$x=\rho\cos\theta$$, $$y=\rho\sin\theta$$ $$\frac{-x^2y}{(x^2+y^2)^{\frac 32}}=\frac{-\rho^3\cos^2\theta\sin\theta}{\rho^3}=\cos^2\theta\sin\theta$$ and doesn't go to $$0$$ when $$\rho\rightarrow 0$$ then limit doesn't exists and $$f$$ isn't differentiable.