Question

se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y2 , (x, y) ≠ (0, 0) 0, (x, y) = (0, 0)

Answer 1

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

• To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$    exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$   exists.

• To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).