Question #273258

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


(0

,

0) (that is

f

x

(0

,

0) and

f

y

(0

,

0) both exist), but

f

is NOT differentiable at (0

,

0).



1
Expert's answer
2021-12-03T14:24:05-0500

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


f(x,y)={2xyx2+y2(x,y)(0,0)0(x,y)=(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,f(0,y)=0, so limy0f(0,y)=0lim_{y\to 0}f(0,y)=0.

choosing the path x=yx=y we see that f(x,x)=2x22x2=1,f(x,x)=\frac{2x^2}{2x^2}=1, so limx0f(x,x)=1lim_{x\to 0}f(x,x)=1. The two path theorem implies that lim(x,y)(0,0)f(x,y)lim_{(x,y)\to(0, 0)}f(x,y) does not exist


(b) Finding fx(0,0) and fy(0,0)f_x(0,0) \ and\ f_y(0,0)

the partial derivatives are defined at (0,0)

fx(0,0)=limh01h[f(0+h,0)f(0,0)]=limh01h[00]=0f_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


fy(0,0)=limh01h[f(0,0+h)f(0,0)]=limh01h[00]=0f_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 fx(0,0)=fy(0,0)=0f_x(0,0)=f_y(0,0)=0

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