Question

Give an example of a function of two variables whose first order partial derivatives exist at

(0, 0) (that is fx(0, 0) and fy(0, 0) both exist), but f is NOT differentiable at (0, 0).

Explain also why the function is NOT differentiable at (0, 0).

Answer 1

Example

(a)Here is discontinuous function at (0,0) having partial derivatives at (0,0)

$f(x,y)= \begin{cases} \frac{2xy}{x^2+y^2} &\text{(x,y)} \neq (0,0)\\ 0 &\text{(x,y)} =(0,0) \end{cases}$

showing f is not continuous at (0,0)

choosing the path x = 0 we see that $f(0,y)=0,$ so $lim_{y\to 0}f(0,y)=0$.

choosing the path $x=y$ we see that $f(x,x)=\frac{2x^2}{2x^2}=1,$ so $lim_{x\to 0}f(x,x)=1$. The two path theorem implies that $lim_{(x,y)\to(0, 0)}f(x,y)$ does not exist

(b) Finding $f_x(0,0) \ and\ f_y(0,0)$

the partial derivatives are defined at (0,0)

$f_x(0,0)=lim_{h\to 0}\frac{1}{h}[f(0+h,0)-f(0,0)]=lim_{h\to 0}\frac{1}{h}[0-0]=0$

$f_y(0,0)=lim_{h\to 0}\frac{1}{h}[f(0,0+h)-f(0,0)]=lim_{h\to 0}\frac{1}{h}[0-0]=0$

Therefore $f_x(0,0)=f_y(0,0)=0$