logo

Ask Questions, Get Answers

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

Varify $\large\frac{\partial^{2} y}{\partial x\partial y}=\frac{\partial^{2} y}{\partial y\partial x}$ for the following function;$\;u=\tan^{-1}\large(\frac{x}{y})$

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
Toolbox:
  • 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=\tan^{-1} \bigg(\large\frac{x}{y}\bigg)$
Step 1:
$\large\frac{\partial u}{\partial y}=\large\frac{1}{1+(\Large\frac{x}{y})^2}.\frac{-x}{y^2}=\frac{-x}{\Large\frac{y^2(y^2+x^2)}{y^2}}=\frac{-x}{x^2+y^2}$
$\large\frac{\partial^2 u}{\partial x \partial y}=\frac{(x^2+y^2)(-1)-(-x)(2x)}{(x^2+y^2)^2}$
$\qquad=\large\frac{-x^2-y^2+2x^2}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}$
Step 2:
$\large\frac{\partial u}{\partial x}=\large\frac{1}{1+(\Large\frac{x}{y})^2}.\frac{1}{y}=\frac{1}{\Large\frac{y(x^2+y^2)}{y^2}}=\frac{y}{x^2+y^2}$
$\large\frac{\partial^2 u}{\partial y \partial x}=\frac{(x^2+y^2)(1)-y(2y)}{(x^2+y^2)^2}$
$\qquad=\large\frac{x^2-y^2}{(x^2+y^2)^2}$
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...