# 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? Dec 7 '13 at 1:14
• Are you sure those results are not a mixed partial, like $f_{xy}$? Dec 7 '13 at 1:43

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).

• Pardon me for asking after almost 6 years but shouldn't the limit equation be equal to $\lim_{h \to 0}\frac{\left|f(p+h)-f(p)-L(h)\right|}{|h|} = 0$ instead of $\lim_{h \to 0}\frac{\left|f(p+h)-f(h)-L(h)\right|}{|h|} = 0$? May 24 '19 at 9:03
• Yup, that is a typo! Correcting. May 29 '19 at 11:14

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. 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.