January 11th, 2017, 11:23 AM  #21 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra 
I think that's more vague handwaving in an attempt to avoid a circular definition. They clearly aren't ordinary variables as highlighted by the OP.
Last edited by v8archie; January 11th, 2017 at 12:00 PM. 
January 11th, 2017, 11:51 AM  #22  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68  Quote:
Sorry you didn't understand some of my other posts which specifically addressed the OP, where I distinguish between the differential of u(x,y,z), and the differential of u(x,y,z) with two of the variables held fixed, in direct response to OP question why isn't 3$\displaystyle \partial u/\partial u$=3, noting that \partial u is not a differential.  
January 11th, 2017, 12:05 PM  #23 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra  The fact that you don't understand what a precise mathematical definition is makes most of what you write at best unreliable and at worst unintelligible. As I said before, I'd encourage the reader to focus on what other people write. 
January 11th, 2017, 06:55 PM  #24  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68  Quote:
If u =u(x,y), $\displaystyle du=\frac{\partial u}{\partial x}dx + \frac{\partial u} {\partial y}dy$ If you divide by du, $\displaystyle \frac{\mathrm{d} u}{\mathrm{d}u }= \frac{\partial u}{\partial x}\frac{\mathrm{d} x}{\mathrm{d}u } + \frac{\partial u} {\partial y}\frac{\mathrm{d} y}{\mathrm{d}u }$ Which is just another relation between differentials. Just like a linear equation z=Ax+By can be divided by z to give: z/z=Ax/z + By/z. When in doubt, write the independent variables the dependent variables are a function of. $\displaystyle du(x,y)=\frac{\partial u(x,y)}{\partial x}dx + \frac{\partial u(x,y)} {\partial y}dy$ You can't do it for the OP. $\displaystyle \partial u$ and $\displaystyle \partial x$ are not differentials, so there is no cancellation. Last edited by zylo; January 11th, 2017 at 07:04 PM. Reason: typ  
January 11th, 2017, 08:40 PM  #25  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68  Quote:
However, strangely enough, $\displaystyle \frac{\partial du}{\partial dx}$ makes sense because an equation in differentials, $\displaystyle du=u_{x}dx+u_{y}dy$ is a linear relation between the variables du, dx and dy. So if du=z, dx=v, and dy=w, then $\displaystyle z=u_{x}v+u_{y}w$ and $\displaystyle \frac{\partial du}{\partial dx}= \frac{\partial z}{\partial v} = u_{x}$ Last edited by zylo; January 11th, 2017 at 08:46 PM.  
January 11th, 2017, 08:47 PM  #26 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra 
It was supposed to be $\partial u$ and $\partial x$  they don't exist.

January 11th, 2017, 09:17 PM  #27  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 861 Thanks: 68  Quote:
Nevertheless your typo led to an interesting point see my previous post. But what is the point of alleging something doesn't exist that nobody said existed.  
January 12th, 2017, 03:37 PM  #28 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra  I'm pretty certain that you have written more nonsense. And you were the first to talk about $\partial u$ and $\partial x$. But, as you may have guessed fron the way I deleted a post above before you had replied to it, I'm getting tired of this subject. 
January 15th, 2017, 03:03 PM  #29 
Newbie Joined: Jan 2017 From: Canada Posts: 6 Thanks: 1 
Thanks to everyone for taking the time to respond to this post. It seems to have ignited a fiery discussion! To answer the question about "What are x,y,and z functions of?" the following should help: u = u(x, y, z) x = x(u, v, w) v = v(x, y, z) y = y(u, v, w) w = w(x, y, z) z = z(u, v, w) So x, y, z, u, v, and w are very much overloaded symbols that represent: 1) the axis of coordinate systems, 2) the value along the coordinates, and 3) a function defining that value. x, y, z represent a Cartesian coordinate set and u, v, w represent another, possible curvilinear nonorthogonal, coordinate set; the two are related by the six functions given above. At the risk of using jargon without learning basic principles these are all a leadin to defining the covariant and contravariant base vectors of a curvilinear coordinate set. I wasn't looking for the definition of partial derivatives, that I have, I was hoping for some inside help to solving that original equality ... and the answer is you can't, "you have to have come across that type of equation before"! For details on the introduction to tensors that this is leading to, and for your reading pleasure, refer to the bottom of page 17 and page 18 of: https://www.grc.nasa.gov/www/k12/Nu...2002211716.pdf Last edited by photogo01; January 15th, 2017 at 03:06 PM. 
January 15th, 2017, 03:57 PM  #30  
Newbie Joined: Jan 2017 From: Canada Posts: 6 Thanks: 1 
Also, as a follow up ... Quote:
Examples of differentials are ds, dx, and dy which can be viewed as infinitesimal lengths. I erred in seeing âˆ‚x as dx and thinking "differentials". A great introduction to differentials starts on page 93 of: https://ocw.mit.edu/ans7870/resource...s/Calculus.pdf  

Tags 
cancelling, differentials 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Justification for cancelling terms in limits?  Mr Davis 97  Calculus  7  August 14th, 2015 12:58 PM 
How to justify the cancelling of variables in a rational expression?  Mr Davis 97  Calculus  3  February 19th, 2015 06:16 AM 
Differentials  delphine cormier  Calculus  1  August 17th, 2014 08:04 AM 
Cancelling factors.  zengjinlian  Probability and Statistics  2  April 4th, 2014 02:35 PM 
Differentials cos(57°)  math999  Calculus  1  May 5th, 2013 09:04 PM 