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)not=to (0,0)
0 (x,y)= to (0,0). }

Is fx continuous at (0,0) ? Justify your answer.

Answer 1

$$f (x,y)=\begin{cases} \frac{xy^3}{x^2+y^2} &\text{if } (x,y)\neq (0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$$

Find $f_x(0,0).$

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

Find $f_x(x,y) \text{ where } (x,y)\neq(0,0).$

$$f_x(x,y) = \frac{y^3(x^2+y^2) - 2x^2y^3}{(x^2+y^2)^2} = \frac{(y^2-x^2)y^3}{(x^2+y^2)^2}$$

Therefore, the partial derivative $f$ with respest to $x$ is

$$f_x (x,y)=\begin{cases} \frac{(y^2-x^2)y^3}{(x^2+y^2)^2}&\text{if } (x,y)\neq (0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$$

To check continuity, find the next limit.

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

Proceed to polar coordinates $x = r\cos\varphi, y=r\sin\varphi.$

$$\lim\limits_{(x,y)\to(0,0)} \frac{(y^2-x^2)y^3}{(x^2+y^2)^2} = \lim\limits_{r\to0} \frac{(r^2\sin^2\varphi - r^2\cos^2\varphi)r^3\sin^3\varphi}{(r^2\cos^2\varphi+r^2\sin^2\varphi)^2} \\ = \lim\limits_{r\to0} \frac{(\sin^2\varphi - \cos^2\varphi)r^5\sin^3\varphi}{r^4} = 0$$

Therefore, $\lim\limits_{(x,y)\to(0,0)} f_x(x,y) = f(0,0)$ and it means that $f_x$ is continuous.