# Partial derivatives must exist and be continuous on all defined points of $f$ for $f$ to be differentiable?

Today my professor explained that $$f(x,y)=\frac{2xy}{(x^2+y^2)^2}$$ is differentiable even though for $$(x,y)=(0,0)$$ $$f$$ is not defined.

The partial derivatives are $$\frac{\partial f}{\partial x}=\frac{2y^3-6x^2y}{(x^2+y^2)^3} \text{ and } \frac{\partial f}{\partial y}=\frac{2x^3-6y^2x}{(x^2+y^2)^3}$$. So, they exist.

Doesn't differentiability require the partial derivaties to be continuous? My professor said that we ignore $$(0,0)$$ on the partial derivatives because it isn't defined for $$f$$. I was hoping someone here could fully explain this.

Also could someone explain the difference between being in $$C^1$$ and just being differentiable?

First of all, you should care about the domain of the functions. Usually, the functions are defined by formulas, just as you wrote $f(x,y)=\frac{2xy}{(x^2+y^2)^2}$, and we understand its domain to be the largest in which the formula makes sense. In this example, the domain is $U=\mathbb{R}^2\setminus\left\{(0,0)\right\}$. This function cannot be extended to $\mathbb{R}^2$ to a continuous function (try to prove this: show that the limit $\lim_{t,s\rightarrow 0}f(s,t)$ does not exist).

Now, forgetting some technicalities, the partial derivative $\frac{\partial f}{\partial x}$ is defined to be the function given by $\frac{\partial f}{\partial x}(x,y)=\lim_{t\rightarrow 0}\frac{f(x+t,y)-f(x,y)}{t}$, as you know, an its domain is the set of points $(x,y)$ for which such limit exists, so $\frac{\partial f}{\partial x}:U\rightarrow\mathbb{R}$ is a function, and the most correct way to write it in this case is $\frac{\partial f}{\partial x}(x,y)=\frac{2y^3-6x^2y}{(x^2+y^2)^3}$ for all $(x,y)\in U$.

The proper definition of a "differentiable function" may be a bit confusing for you, but let's say that a function is differentiable if all its partial derivatives exists (this is not the correct meaning). However, a function is said to be of class $C^1$ iff all its partial derivatives exist in the whole domain and they are all continuous functions. Obviously, $C^1$ functions are differentiable (simply from the definition of $C^1$), but not all differentiable functions are $C^1$: For example, take $g:\mathbb{R}^2\rightarrow\mathbb{R}$ given by $g(x,y)=x^2\sin(1/x)$ if $x\neq 0$ and $g(0,y)=0$. Then $g$ is differentiable and, for $(x,y)\in\mathbb{R}^2$, $$\frac{\partial g}{\partial y}(x,y)=0;$$ $$\frac{\partial g}{\partial x}(x,y)=\begin{cases}2x\in(1/x)-\cos(1/x)&\text{, if }x\neq 0\\0&\text{, if }x=0\end{cases}$$ so $g$ is not $C^1$, since $\partial g/\partial x$ is discontinuous at the points $(0,y)$, $y\in\mathbb{R}$.

• So $f$ in this case is $C^1$ provided we specify that $(x,y)=(0,0)$ ∉ the domain of $f$? – Bobby Lee Feb 12 '14 at 2:24
• Exactly. Maybe your professor won't really bother saying all this about domains, but I believe it is important that you know what you're doing. – Luiz Cordeiro Feb 12 '14 at 2:45

If a function is differentiable at a point $p$, then not only is it defined at $p$ it is also continuous at $p$. This is a classical result following almost immediately from the definition of differentiability.

Functions in $C^1$ are those functions with a continuous first derivative. Not all differentiablefunctions have continuous first derivatives, so not every differentiable function is in $C^1$.

• Why do we ignore the discontinuity at $(x,y)=(0,0)$ and still call $f$, in this case, differentiable? – Bobby Lee Feb 12 '14 at 1:56
• We don't. We first extend $f$ to include $(0,0)$ in its domain making sure the function is continuous. – Ittay Weiss Feb 12 '14 at 1:58