Answer to Question #108914 in Differential Equations for TUHIN SUBHRA DAS

Question

find fx(0,0) and fx (x,y), where (x,y)≠(0,0) for the following function f(x,y)= {xy^3/(x^2+y^2), (x,y) ≠(0,0) 0, (x,y)=(0,0) is fx continuous at (0,0)?

Answer 1

$$f_x(x,y)=y^3\cdot{x^2+y^2-2x^2 \over (x^2+y^2)^2}=y^3{y^2-x^2 \over (x^2+y^2)^2},\ (x,y)\not=(0,0)$$

$$f_y(x,y)=x\cdot{3y^2(x^2+y^2)-2y^4\over (x^2+y^2)^2}=xy^2{3x^2+y^2\over (x^2+y^2)^2},\ (x,y)\not=(0,0)$$

The partial derivatives are defined at (0, 0).

$$f_x(0,0)=\lim\limits_{h\to0}{1\over h}(f(0+h,0)-f(0,0))=\lim\limits_{h\to0}(0-0)=0$$

$$f_y(0,0)=\lim\limits_{h\to0}{1\over h}(f(0,0+h)-f(0,0))=\lim\limits_{h\to0}(0-0)=0$$

Therefore

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

Let $x=rcos\theta, y=r\sin\theta$

$$\lim\limits_{(x,y)\to(0,0)}f_x(x,y)=\lim\limits_{r\to0}\bigg[r^3\sin^3\theta{r^2\sin^2\theta-r^2\cos^2\theta \over (r^2\sin^2\theta+r^2\cos^2\theta)^2}\bigg]=$$ $$=\lim\limits_{r\to0}\big[r^3\sin^3\theta(-\cos(2\theta))\big]=0$$

$$\lim\limits_{(x,y)\to(0,0)}f_x(x,y)=0=f_x(0,0)$$

$f_x(x,y)$ is continuous at (0, 0)