# Differentiable function - Easy Calc

So I am studying for my calculus midterm and I had the following question:

If $f_{x}(a, b)$ and $f_{y}(a,b)$ both exist then $f$ is differentiable at $(a,b)$

I answered true, but the answer is false. Why is that? I haven't done real analysis yet (dropped it this term) so I won't know what your talking about if its too advanced. If you attempt to use epsilons and deltas then please try to explain it as simple as possible.

Thanks!

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What is the definition of differentiable that you/your book are using? – Andrés E. Caicedo Nov 25 '10 at 3:47

Others have given you some of the standard examples that you can verify are indeed examples. Let me try to give you some intuition of why the partials are not enough.

Intuitively, a function $f\colon\mathbb{R}^2\to\mathbb{R}$ is differentiable at a point $(x_0,y_0)$ if you can find a plane that is tangent to the graph of $f$ at the point $(x_0,y_0,f(x_0,y_0))$. Here, "tangent" is interpreted the same way that a line is said to be tangent to a curve when dealing with $g\colon\mathbb{R}\to\mathbb{R}$:

In the one-variable case, a line $y=mx+b$ is tangent to the graph of $y=g(x)$ at the point $c$ if and only $mc+b=g(c)$ (they go through the same point) and the approximation given by the tangent as $x$ approaches $c$ goes to $0$ faster than $x$ approaches $c$; that is, $$\lim_{x\to c}\frac{g(x)-(mx+b)}{x-c}=0.$$ Note: To see that this certainly will happen for the tangent, remember that the tangent will have $m=g'(c)$ and $b=g(c) - g'(c)c$, so the expression in the limit becomes $$\frac{g(x) - (g'(c)x + g(c)-g'(c)c)}{x-c} = \frac{g(x) - g(c) - g'(c)(x-c)}{x-c} = \frac{g(x)-g(c)}{x-c} - g'(c).$$ But in fact, one can define the tangent as the (unique) line with this property.

We do the same thing with two variables, but with planes instead of lines. Differentiability at $(x_0,y_0)$ for $f(x,y)$ means that there is a plane $z=ax+by+c$, with $ax_0 + by_0 + c = f(x_0,y_0)$, and such that $$\frac{f(x,y) - (ax + by + c)}{||(x,y)-(x_0,y_0)||}$$ goes to $0$ as $(x,y)\to (x_0,y_0)$: the plane approaches the graph, in every direction, faster than the arguments are approaching the point. (Here, $||\cdot||$ is the usual norm in $\mathbb{R}^2$: $||(a,b)-(c,d)||=\sqrt{(a-c)^2+(b-d)^2}$).

But this has to happen along all directions of approach to $(x_0,y_0)$; just like with limits of two variables, approaching along vertical lines alone or along horizontal lines alone is not enough (even approaching along lines is not enough; it really has to be along all directions).

By contrast, the partial derivative with respect to $x$ only tell you that if you "slice" the graph with a plane parallel to the XZ-plane and you look at the "shadow" on that plane (which will look just like the graph of a one-variable function), then you can approach the bit of graph with an appropriate line; that is, you can approach the graph of $z=f(x,y)$ (which is a three-dimensional thing) with a line parallel to the XZ plane; and the partial derivative with respect to $y$ tells you that you can approach the graph with a line parallel to the YZ-plane.

Certainly, if you can approach the graph with a plane appropriately, then you can also approach it with lines (just make sure the lines are contained in the plane). But being able to approach it in the two coordinate directions alone is not enough to make sure you can actually get a whole plane close to the graph in all directions. That is why knowing that the partial derivatives exist does not tell you that the function is differentiable.

One nice thing, as mentioned by Bryan Yocks, is that if if the partial derivatives, viewed as functions of two variables, not only exist but are also continuous, then you can leverage this extra information into showing you can find a plane. Intuitively, what happens here is that the lines do give you enough information because you know that points close to $(x_0,y_0)$ will have similar lines that are very close to the ones you have, and so you can picture in your head perhaps that as you draw all these lines, "locally" it's going to look very much like a plane because the change in either of the two directions is very small. But if all you know is that the partial derivatives exist, this is not enough for the reasons described above.

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$f=\frac{xy}{x^2+y^2}$ when $(x,y)\neq (0,0)$ and $f(0,0)=0$ is a standard counterexample for this. $f_x(0,0)=f_y(0,0)=0$, but $f$ is not differentiable (or even continuous at the origin).

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Need the partial derivatives exist and are $continuous$ in a open ball around $(a, b)$. This is sufficient.

Edit You can check : http://en.wikipedia.org/wiki/Partial_derivative

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Think about when can you say a function is differentiable?

Remember that $f_x$ and $f_y$ gives you the partial derivatives in just 2 directions namely $X$ axis and $Y$ axis respectively.

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An even easier example than Timothy's is $f(x,y)=1$ when $x=0$ or $y=0$, and $f(x,y)=0$ for all other points. Then $f_x(0,0)=0, f_y(0,0)=0$ but $f$ is not differentiable (or even continuous at the origin).

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