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Verify $\large\frac{\partial ^{2} y}{\partial x\partial y}=\frac{\partial ^{2} y}{\partial y\partial x}$ for the following function;$\;u=\large\frac{x}{y^{2}}-\frac{y}{x^{2}}$

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1 Answer

  • If $u=f(x,y)$ is the function of two independent variables then
  • $\large\frac{\partial u}{\partial y (x_0,y_0)}=\frac{d}{dy}$$f(x_0,y)\;and \;\large\frac{\partial u}{\partial x(x_0,y_0)}=\large\frac{d}{dy}$$ f(x,y_0)$ Provided they exist
  • The Second order parral derivaties are $\large\frac{\partial}{\partial x} \frac{\partial u}{\partial y}=\frac{\partial ^2 u}{\partial x \partial y},\frac{\partial}{\partial y} \frac{\partial u}{\partial y}=\frac{\partial ^2 u}{\partial y^2},\frac{\partial}{\partial y} \frac{\partial u}{\partial x}=\frac{\partial ^2 u}{\partial y \partial x},$$\;and\;\large\frac{\partial}{\partial x} \frac{\partial u}{\partial x}=\frac{\partial ^2 u}{\partial x ^2}$ Partial derivatives of functions of more similarly defined
Given $u=\large\frac{x}{y^2}-\frac{y}{x^2}$
Step 1:
$\large\frac{\partial u}{\partial y}$$=\large\frac{-2x}{y^3}-\frac{1}{x^2}$
$\large\frac{\partial^2 u}{\partial x \partial y}=\frac{-2}{y^3}+\frac{2}{x^3}$
Step 2:
$\large\frac{\partial u}{\partial x}=\large\frac{1}{y^2}+\frac{2y}{x^3}$
$\large\frac{\partial^2 u}{\partial y \partial x}=\frac{-2}{y^3}+\frac{2}{x^3}$
Step 3:
$\large\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$
answered Aug 12, 2013 by meena.p

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