Respuesta :
Answer:
It is proved that [tex]f_x, f_y[/tex] exixts at (0,0) but not differentiable there.
Step-by-step explanation:
Given function is,
[tex]f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)[/tex]
- To show exixtance of [tex]f_x(0,0), f_y(0,0)[/tex] we take,
[tex]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[/tex] exists.
And,
[tex]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}[/tex] exists.
- To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,
[tex]\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}[/tex] 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).
SInce [tex]\frac{\delta^2 f}{\delta x \delta y} \neq\frac{\delta^2 f}{\delta y \delta y}[/tex], hence the function is not differentiable at (0, 0)
For the given function to be differentiable at ()0, 0), then the following must be true, hence the function is not differentiable.
- [tex]\frac{\delta^2 y}{\delta x \delta y} =\frac{\delta^2 y}{\delta y \delta y}[/tex]
[tex]\frac{\delta^2 f}{\delta y \delta x} = \frac{\delta f}{\delta y}(\frac{\delta f}{\delta x} ) = \frac{\delta }{\delta y}(18x^2y+4x^3)=36xy[/tex]
Similarly for the other expression;
[tex]\frac{\delta^2 f}{\delta x\delta y} = \frac{\delta f}{\delta x} (\frac{\delta f}{\delta y} ) = \frac{\delta }{\delta x}(9x^2+2y)=18x[/tex]
SInce [tex]\frac{\delta^2 f}{\delta x \delta y} \neq\frac{\delta^2 f}{\delta y \delta y}[/tex], hence the function is not differentiable at (0, 0)
Learn more on partial derivative here: https://brainly.com/question/2293382
