Answer to Question #170547 in Real Analysis for Prathibha Rose

Question

Consider the function f:R² to R defined by

f(x,y) ={ (x²y²)/(x⁴+y²) for (x,y) not equal to zero

0 , for (x,y) =(0,0)

Prove that,

1. f_x(0,0)=f_y(0,0)=0

2. f_x is continuous at (0,0)

3. f_y is.not continuous at (0,0)

Answer 1

$Given \ f(x,y)=\frac{x^2y^2}{x^4+y^2}\ when (x,y)\ne(0,0) \\ =0 , when \ (x,y)=(0,0) \\ We \ find \ the \ partial \ derivative \ of \ f(x,y) \ w.r.t 'x', \ an \\ arbitrary \ (x,y)\ by \ Using \ \frac{\partial}{\partial x}(\frac{u}{v})=\frac{vu_x -uv_x}{v^2}, we \ get \\ f_x(x,y)=\frac{(x^4+y^2)\frac{\partial}{\partial x}(x^2y^2)\\ -(x^2y^2)\frac{\partial}{\partial x}(x^4+y^2)}{(x^4+y^2)^2} \\ =\frac{(x^4+y^2)(2xy^2)\\ -(x^2y^2)4x^3}{(x^4+y^2)^2} \\ =\frac{(2x^5y^2+2xy^4-4x^5y^2)}{(x^4+y^2)^2} \\ =\frac{2xy^4-2x^5y^2)}{(x^4+y^2)^2} \\ \Rightarrow \ f_x(x,y)= \frac{2xy^4-2x^5y^2)}{(x^4+y^2)^2}\\ We \ find \ the \ partial \ derivative \ of \ f(x,y) \ w.r.t 'y', \ an \\ arbitrary \ (x,y)\ by \ Using \ \frac{\partial}{\partial y}(\frac{u}{v})=\frac{vu_y -uv_y}{v^2}, we \ get \\ f_y(x,y)=\frac{(x^4+y^2)\frac{\partial}{\partial y}(x^2y^2)\\ -(x^2y^2)\frac{\partial}{\partial y}(x^4+y^2)}{(x^4+y^2)^2} \\ =\frac{(x^4+y^2)(2x^2y)\\ -(x^2y^2)(2y)}{(x^4+y^2)^2} \\ =\frac{2x^6y}{(x^4+y^2)^2} \\ \Rightarrow \ f_y(x,y)=\frac{2x^6y}{(x^4+y^2)^2}\\ To \ find \ the \ partial\ derivative \ of \ f\ w.rt. 'x'\ at (0,0)\\ consider \lim_{h\rightarrow 0}\frac{f(0+h,0)-f(0,0)}{h}=\\ lim_{h\rightarrow 0}\frac{\frac{(h^2(0))}{h^4+0}-0}{h}\\ (By \ the \ definition \ of \ f(x,y))\\ =0\\ \Rightarrow f_x(0,0)=0\\ To \ find \ the \ partial\ derivative \ of \ f\ w.rt. 'y'\ at (0,0)\\ consider \lim_{k\rightarrow 0}\frac{f(0,0+k)-f(0,0)}{k}=\\ lim_{k\rightarrow 0}\frac{\frac{(0^2(k^2))}{0^4+k^2}-0}{k}\\ (By \ the \ definition \ of \ f(x,y))\\ =0\\ \Rightarrow f_y(0,0)=0 \\ To\ examine \ the \ continuity \ of \ f_x(x,y) \ at \ (0,0)\\ Consider \ the \ \lim_{x\rightarrow 0,y\rightarrow 0} f_x(x,y)\\ along \ y=x^2\\ \Rightarrow \lim_{x\rightarrow 0} \frac{2x(x^2)^4)-2x^5(x^2)^2)}{(x^4+(x^2)^2)^2}\\ =\lim_{x\rightarrow 0} \frac{0}{4(x^8)}\\ =0\\ \Rightarrow f_x(0,0)=0=\lim_{x\rightarrow 0,y\rightarrow 0} f_x(x,y)\\ \therefore f_x(x,y) \ is \ continuous \ at \ (0,0)\\ To\ examine \ the \ continuity \ of \ f_y(x,y) \ at \ (0,0)\\ Consider \ the \ \lim_{x\rightarrow 0,y\rightarrow 0} f_y(x,y)\\ along \ y=x^2\\ \Rightarrow \lim_{x\rightarrow 0,y\rightarrow 0} \frac{2x^6y}{(x^4+y^2)^2}\\ =\lim_{x\rightarrow 0} \frac{2x^6(x^2)}{(x^4+(x^2)^2)^2)}\\ =\lim_{x\rightarrow 0} \frac{2x^8}{4(x^8)}=1/2\\ \Rightarrow f_y(0,0)=0\ne\lim_{x\rightarrow 0,y\rightarrow 0} f_y(x,y)=\frac{1}{2}\\\ \therefore f_y(x,y) \ is \ not \ continuous \ at \ (0,0)\\$