Determine if $f(x,y)=(1+x+2y)^2$ is differentiable in $(1,-1).$

Question

To show this, I can use the definition for differentiability for two variables:

A function $f(x,y)$ of two variables is differentiable in $(x,y)=(a,b)$ iff $ \ \exists \ $ constants $A,B:$

$$f(1+h,1+k)-f(a,b)=Ah+Bk+\sqrt{h^2+k^2} \ \rho(h,k).$$

Here I go:

$$\begin{array}{lcl} f(x,y) & = & (1+x+2y)^2 \\ f(1,-1) & = & (1+1+2(-2))^2=0 \\ f(1+h,-1+k) & = & (1+1+h+2(-1+k))^2=4k^2+4kh+h^2. \end{array}$$

$$\implies f(1+h,-1+k)-f(1,-1)=4k^2+4kh+h^2=Ah+Bk+\sqrt{h^2+k^2}\rho(h,k) \quad (1).$$

Clearly, there exists no constants $A,B$ such that $(1)$ holds, thus, the function is not differentiable in $(1,-1).$

Questions:

1. Will my answer do?
2. How can I show whether this function is $C^1$ or not?

Answer 1

Your $f$ is a polynomial in $x$ and $y$. If polynomials were not differentiable we would have given up differential calculus long ago. Your problem is that you don't yet have fully grasped the notion of differentiability in the multivariate case.

A simple calculation gives $$f(1+h,-1+k)-f(1,-1)=(h+2k)^2=0h +0k +h^2+4hk+4k^2\ .$$ Now $(2|h|-|k|)^2\geq0$ and therefore $4|hk|\leq 4h^2+k^2$ It follows that $${|f(1+h,-1+k)-f(1,-1)|\over\sqrt{h^2+k^2}}\leq5\sqrt{h^2+k^2}\to0\qquad\bigl((h,k)\to(0,0)\bigr)\ .$$ This shows that $df(1,-1)=0\ .$

Of course this is all obvious using the standard rules of this game.

Answer 2

Note that in this case you can simply observethat by Differentiability theorem since all the partial derivatives exist and are continuous in a neighborhood of $(0,0)$ then (i.e. sufficient condition) the function is differentiable at $(0,0)$.

Note that when partial derivatives are not continuos at (0,0) you can't yet conclude anything about differentiability. You need to check directly differentiability by definition that:

$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$