Answer to Question #273258 in Electrical Engineering for zohaib

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

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