Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

If $u=\log\begin{pmatrix}\large\frac{x^{2}+y^{2}}{xy}\end{pmatrix}$ then $x\large\frac{\partial u}{\partial x}+$$y\large\frac{\partial u}{\partial y}$ is

\[\begin{array}{1 1}(1)0&(2)u\\(3)2u&(4)u^{-1}\end{array}\]

1 Answer

Comment
A)
$u= \log \bigg( \large\frac{x^2+y^2} {xy}\bigg)$
$e^u=-e^{\log} \bigg( \large\frac{x^2+y^2} {xy}\bigg)$
$e= \large\frac{x^2+y^2} {xy} $
$e^u $ is a homogeneous function of degree 0.
by Eulers theorem,
$x.\large\frac{\partial}{\partial x} (e^u)+ y . \large\frac{\partial }{\partial y} (e^u)=0.e^u$
$x.e^u \large\frac{\partial u}{\partial x} + y .e^u.\large\frac{\partial u}{\partial y} =0$
$e^u \bigg( x \large\frac{\partial u}{\partial x} + y .e^u.\large\frac{\partial u}{\partial y} \bigg)=0$
$ x \large\frac{\partial u}{\partial x} + \large\frac{\partial u}{\partial y} =0$
Hence 1 is the correct answer.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...