# Question about the differentiability of a function of more than one variable

## Main Question or Discussion Point

If a function of two or more variables is differentiable at some point, does this imply that all its partial derivatives are continuous at that point?

It's probably not really correct to talk about differentiability of a function in that sense. Instead, we'd say that it is differentiable with respect to a particular variable. For instance, the function:

f(x, y) = y*sgn(x)

where "sgn" is the sign function (+1 for positive values, -1 for negative values, and sgn(0)=0). This function is differentiable at 0 with respect to y but not with respect to x. To say that it's just "differentiable" might be a useful shorthand for saying that it's "differentiable in both variables," but I've actually never heard that, and it's a bit ambiguous.

One other note: a function can be "differentiable," and not have a continuous derivative. A function that is "continuously differentiable" has a continuous derivative.

It sounds like you may have the implication backwards. In general, differentiable implies continuous, not the other way around. For example, since f(x, y) = y*sgn(x) is differentiable with respect to y along the slice x=0, we can say that slice is continuous, or that f(y)=f(0,y)=y*0 is continuous.

I'm not sure who you're talking to, but I don't believe I or the original poster ever assumed that continuous implies differentiable. What are you referring to exactly?

In hindsight, however, I picked a crappy example, since when either argument is 0, the function is constant and identically 0.

f(x,y) = y + sgn(x)

illustrates the property I intended better. At the point (0,0), the function is differentiable with respect to y, but not with respect to x.

I'm not sure who you're talking to, but I don't believe I or the original poster ever assumed that continuous implies differentiable. What are you referring to exactly?
If a function of two or more variables is differentiable at some point, does this imply that all its partial derivatives are/is continuous at that point?
It sounds like it could have been incorrectly generalized from the converse, differentiable -> continuous. Of course, I could be completely wrong.

Oh I see what you're saying, and I think it's a different mistake from that. It doesn't appear that he's going from f differentiable to f continuous, but from f differentiable to f' continuous, which is also incorrect, but slightly different.

It's probably not really correct to talk about differentiability of a function in that sense. Instead, we'd say that it is differentiable with respect to a particular variable. For instance, the function:

f(x, y) = y*sgn(x)

where "sgn" is the sign function (+1 for positive values, -1 for negative values, and sgn(0)=0). This function is differentiable at 0 with respect to y but not with respect to x. To say that it's just "differentiable" might be a useful shorthand for saying that it's "differentiable in both variables," but I've actually never heard that, and it's a bit ambiguous.

One other note: a function can be "differentiable," and not have a continuous derivative. A function that is "continuously differentiable" has a continuous derivative.
Saying that a function is differentiable is standard terminology really and has little to do with the partial derivatives. In general a function $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is differentiable in a if there exists a linear function $L:\mathbb{R}^n\rightarrow \mathbb{R}^m$ such that

$$\lim_{h\rightarrow 0}{\frac{f(a+h)-f(h)-L(h)}{\|h\|}}=0$$

Im confused now... According to my book, a function of two or more variables is differentiable at a point P if a linear function can approximate it "Well enough" at other points "Close to P". I believe there was also another one that involved delta z. (Where z=f(x,y)) If z is differentiable at some point, then $\Delta$z=f$_{x}$$\Delta$x+f$_{y}$$\Delta$y+$\epsilon$1$\Delta$x+$\epsilon$2$\Delta$y

Where both $\epsilon$1 and $\epsilon$2 approach zero as delta x and delta y approach zero.

If $\Delta$z can indeed be expressed in that way at some point (a,b), then the function is differentiable at the point.

Now, there is an easier way of finding out if a function of two or more variables is differentiable at some point. If all the partial derivatives of a function are continuos at some point, then f(x,y) is differentiable at that same point.

This is where my question comes in... does the differentiability of a function (The way I defined it above) imply that its partial derivatives are all continuos at that point? or does it merely imply that they exist?

Im sorry if my initial post was a bit unclear... also, I apologize if my equation seems messy, its my first shot at using latex

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Im confused now... According to my book, a function of two or more variables is differentiable at a point P if a linear function can approximate it "Well enough" at other points "Close to P". I believe there was also another one that involved delta z. (Where z=f(x,y)) If z is differentiable at some point, then $\Delta$z=f$_{x}$$\Delta$x+f$_{y}$$\Delta$y+$\epsilon$1$\Delta$x+$\epsilon$2$\Delta$y

Where both $\epsilon$1 and $\epsilon$2 approach zero as delta x and delta y approach zero.

If $\Delta$z can indeed be expressed in that way at some point (a,b), then the function is differentiable at the point.

Now, there is an easier way of finding out if a function of two or more variables is differentiable at some point. If all the partial derivatives of a function are continuos at some point, then f(x,y) is differentiable at that same point.

This is where my question comes in... does the differentiability of a function (The way I defined it above) imply that its partial derivatives are all continuos at that point? or does it merely imply that they exist?

Im sorry if my initial post was a bit unclear... also, I apologize if my equation seems messy, its my first shot at using latex
To determine if f(x,y) is differentiable at (x0, y0), you see if both the limits, [TEX]\lim_{h \to 0} \frac{f(x_0+h,y_0)-f(x_0,y_0)}{h}[/TEX] and [TEX]\lim_{h \to 0} \frac{f(x_0,y_0+h)-f(x_0,y_0)}{h}[/TEX]
exist.

Hopefully, knowing the definition in 1D and 2D, you can see how it generalizes to higher dimensions, because I don't feel like trying to make a general formula. :)

HallsofIvy
Homework Helper
No, that is not sufficient. That only shows that the two partial derivatives exist and, as has been said before in this thread, partial derivatives existing at a point is not sufficient to say that the function is differentiable there.

As an example, take f(x,y)= 0 is xy= 0, 1 otherwise.

For all h, f(h, 0)= 0 so f(h, 0)- f(0, 0)= 0 and f(0, h)- (0, 0)= h so the limits you give exist and are 0 but the function is not even continuous at (0, 0).

Indeed, only seeing that the partial derivatives exist is not enough. But if the partial derivatives exist AND they are continuous, then the function is differentiable. Maybe use that?

HallsofIvy
Homework Helper
Yes, if a function has continuous partial derivatives in some neighborhood of a point (not just continuous at the point) then the function is differentiable at that point.

Yes, if a function has continuous partial derivatives in some neighborhood of a point (not just continuous at the point) then the function is differentiable at that point.
Can a function be continuous at a point? Wouldn't the fact that $\lim_{x \to c}f(x)=f(c)$ imply that it is continuous in some neighborhood of c?

Can a function be continuous at a point? Wouldn't the fact that $\lim_{x \to c}f(x)=f(c)$ imply that it is continuous in some neighborhood of c?
Yes, a function can be continuous in one point only. The example are perhaps a little farfetched, but they exist anyways. If you're interested:

$$f:[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if x is rational}\\ 0~\text{if x is irrational}\end{array}\right.$$

is continuous only in 0.

Yes, a function can be continuous in one point only. The example are perhaps a little farfetched, but they exist anyways. If you're interested:

$$f:[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if x is rational}\\ 0~\text{if x is irrational}\end{array}\right.$$

is continuous only in 0.
Does $\lim_{x \to 0} f(x)$ even exist? (Out of curiosity, I would like to see a proof of that.)

I'm not disagreeing that it does. I don't have the knowledge to know for sure either way.

Does $\lim_{x \to 0} f(x)$ even exist? (Out of curiosity, I would like to see a proof of that.)

I'm not disagreeing that it does. I don't have the knowledge to know for sure either way.
Yes, the limit is 0. The easiest way by seeing this is by using the squeeze theorem:

$$0\leq f(x)\leq x$$

so in 0, the limit of 0 and x is 0. So the limit of f(x) is 0.

We can also do it by epsilon-delta. Indeed, if $|x|<\delta$, then $|f(x)|\leq |x|<\epsilon$. So by the epsilon-delta definition of limits, the limit is 0.

Analogously, every sequence $(x_n)_n$ which converges to 0 will have that [itex]f(x_n)\rightarrow 0. this is very easily seen. So the limit indeed exists!!