Question

For the following functions find the f_x, and f_y and show that f_xy = f_yx 

f(x, y) = ${\displaystyle {\tan }^{-1}\left(\frac{x}{y}\right)}$

Answer 1

${\displaystyle \frac{\partial \text{f}}{\partial x}=\frac{1}{1+\frac{x^2}{y^2}}\left(\frac{1}{y}\right)=\frac{y}{x^2+y^2}}$

${\displaystyle \frac{\partial \text{f}}{\partial y}=\frac{1}{1+\frac{x^2}{y^2}}\left(\frac{-x}{y^2}\right)=\frac{-x}{x^2+y^2}}$

${\displaystyle \frac{{\partial }^2\text{f}}{\partial x\partial y}=\frac{\partial }{\partial x}\left[\frac{\partial \text{f}}{\partial y}\right]}$

= ${\displaystyle \frac{\partial }{\partial x}\left[\frac{-x}{x^2+y^2}\right]}$

= ${\displaystyle \frac{(x^2+y^2)[-1]-(-x)\left[2x\right]}{{(x^2+y^2)}^2}}$

= ${\displaystyle \frac{x^2-y^2}{{(x^2+y^2)}^2}}$   ........(1)

${\displaystyle \frac{{\partial }^2\text{f}}{\partial y\partial x}=\frac{\partial }{\partial y}\left[\frac{\partial \text{f}}{\partial x}\right]}$

= ${\displaystyle \frac{\partial }{\partial y}\left[\frac{y}{x^2+y^2}\right]}$

= ${\displaystyle \frac{(x^2+y^2)\left[1\right]-y\left[2y\right]}{{(x^2+y^2)}^2}}$

= ${\displaystyle \frac{x^2-y^2}{{(x^2+y^2)}^2}}$   ..........(2)

From (1) and (2)

⇒ ${\displaystyle \frac{{\partial }^2\text{f}}{\partial x\partial y}=\frac{{\partial }^2\text{f}}{\partial y\partial x}}$