Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App