# Differentiablility of a multivariable function

I have the function: $$f(x,y) = xy^{1/2}$$

I want to check what the parital derivatives are at the origin and whether it is differentiable at the origin.

For the partial derivatives at the origin I obtained:

$$f_x = y^{1/2} = 0$$

$$f_y = \frac{1}{2}x y^{-1/2}=0$$

Please let me know if these are correct.

Where I'm struggling is showing whether the function is differentiable or not. One idea I have is to say that since $$xy^{1/2}$$ is a composition of two differentiable functions, it too must be differentiable. However, I'm not sure if this is correct in this case and if it is enough to prove differentiability.

Any help would be appreciated, thanks in advance.

• Your function is not defined any neighborhood of the origin. Jul 23, 2021 at 11:54
• Differentiabilty (and even continuity) requires that the function is defined in some neighborhood of $(0,0)$. Jul 23, 2021 at 12:04
• @CharlieP The partial derivatives were not both $0$ - see my answer Jul 23, 2021 at 12:10
• @CharlieP If the partial derivatives were both actually zero then that would be fine - it would indicate that the surface of the function would be flat at the origin Jul 23, 2021 at 12:11
• Yes, they are saying that a function cannot be differentiable on the boundary of its domain. It can only be differentiable in the interior of its domain.
– Joe
Jul 23, 2021 at 12:19

Be careful and remember the negative power reciprocal laws:

$$f_y=\frac{x}{2\sqrt{y}}=\frac{0}{0}, (x,y)=(0,0).$$

This is potentially resolvable using L'Hopital's rule, since both $$x,y$$ go to zero I will treat them as one variable, $$u$$ - this examines one of infinitely many paths of approach to the origin, but by showing that this path fails I show all paths fail (i.e if the limit does not exist for one way, then by definition it cannot exist at all):

$$\frac{d}{du}\frac{u}{2\sqrt{u}}=\frac{d}{du}\frac{\sqrt{u}}{2}=-\frac{1}{2\sqrt{u}}$$

Which is undefined at zero and further applications of the law will not help. This shows that there is one path of approaching the origin for which the partial derivative w.r.t $$y$$ does not exist, showing that it does not exist at the origin in general. For $$f$$ to be differentiable, we must have that all partial derivatives exist.

I conclude that it is not differentiable at the origin.

Look at these $$3$$D sketches from Wolfram Alpha of the partial derivative with respect to $$y$$:

See how choppy and not well defined they are at $$(0,0)$$?

• The function is not defined when $y<0$. So the question of differentiabilty or even continuity does not arise. Jul 23, 2021 at 12:03
• @Joe I showed that one path was undefined; this shows the derivative could not exist, if one path of approach did not exist Jul 23, 2021 at 12:04
• Which is what I showed in my post - perhaps I should have been clearer, @KaviRamaMurthy Jul 23, 2021 at 12:07
• You calculated it fine - it's just not possible to evaluate it at $0$. You wrote $\frac{1}{2}xy^{-1/2}$ and I wrote $x/(2\sqrt{y})$ - these are equivalent statements, I was just trying to emphasise the "division by zero" aspect Jul 23, 2021 at 12:24
• @CharlieP, so not only can you not take the derivative at the origin, since that is on the boundary of the domain, but the other error you had was evaluating $xy^{-1/2} = 0$ at $(0,0)$. A variable to a non-positive power isn't defined at zero.
– Joe
Jul 23, 2021 at 12:31