# In general, how can we make a multivariable function differentiable at a specific point?

Let's start with an example:

Determine the value of the positive integer $$p$$ for which the function

$$f(x,y) = \begin{cases} \frac{(x-y)^{p}}{x^2+y^2} & \text{if (x,y) \neq (0,0)} \\ 0 & \text{if (x,y)= (0,0)} \end{cases}$$ is differentiable at $$(0,0)$$

I understand that the definition of differentiability of multivariable function begins with the notion of the equation of a tangent plane (or hyperplane):

$$h(a,b) = f(a,b) + f_x(a,b)(x-a)+f_y(a,b)(y-b)$$

And the function is said to be differentiable at $$(0,0)$$ if $$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = 0$$ holds. But for the question in the example, I do not even know if the partial derivative exists. I then try to assume that it exists and evaluate the equation of the hyperplane by expressing the partial derivatives in terms of $$x, y$$ and then producing the equation of the tangent plane. Then:

$$h(x,y) = \frac{(x-y)^{p}(p-2)}{x^2+y^2}$$

And if $$(x,y)=(0,0)$$, the equation seems to be not well-defined. In the case of not even having a candidate tangent plane to start with, how should I chosse $$p$$ to make the function differentiable at $$(0,0)$$? And more importantly, what are some general methods for this type of question?

• $f_x(0,0)=\lim_{x\to 0}\frac{x^p}{x^2}=\lim_{x\to 0} x^{p-2}$ exists if and only if $p$ ... – mfl Oct 1 '18 at 13:12
• @mfl Maybe I am misunderstanding something, but I think it exists regardless of the values of p? If p=2, then the limit is 1; but for other p isn't the limit 0? – hephaes Oct 1 '18 at 13:42
• What happens if $p=1?$ – mfl Oct 1 '18 at 13:53
• @mfl I see. For p>3, the partial derivative is 0. But should I prove the partial derivative is continuous first? – hephaes Oct 1 '18 at 14:22

## 1 Answer

We have that

$$f_x(0,0)=\lim_{x\to 0}\frac{x^p}{x^2}=\lim_{x\to 0} x^{p-2}$$

and

$$f_y(0,0)=\lim_{y\to 0}\frac{y^p}{y^2}=\lim_{y\to 0} y^{p-2}.$$

If $$p=1$$ the partial derivatives don't exist; if $$p=2$$ we have $$f_x(0,0)=f_(0,0)=1$$ and if $$p\ge 3$$ then $$f_x(0,0)=f_y(0,0)=0.$$

Assume $$p=2:$$

$$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \dfrac{\dfrac{(x-y)^2}{x^2+y^2}-x-y}{\sqrt{x^2+y^2}}.$$ But

$$\lim_{x\to 0} \frac{f(x,0)-h(x,0)}{\sqrt{x^2}} = \lim_{x\to 0} \dfrac{\dfrac{x^2}{x^2}-x}{\sqrt{x^2}}$$ doesn't exist. So $$f$$ is not differentible at $$(0,0).$$

Assume $$p\ge 3:$$

$$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-h(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \dfrac{\dfrac{(x-y)^p}{x^2+y^2}}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)} \dfrac{(x-y)^p}{(x^2+y^2)^{3/2}}.$$

If $$p=3$$ the limit doesn't exist (Consider the limit along the paths $$(x,kx)$$). So $$f$$ is not differentible at $$(0,0).$$

If $$p\ge 4$$ the limit is $$0.$$ So $$f$$ is differentible at $$(0,0).$$