
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
February 16th, 2017, 01:47 AM  #1 
Newbie Joined: Jan 2016 From: United Kingdom Posts: 29 Thanks: 0  Multivariable Chain rule proof : is it correct?
Hello, I've had a go at proving the chain rule for a composition that maps from R^2 to R to R. There are two images attached. I've used the "arithmetic of limits" rules throughout, that if two limits exist, their product limit and sum limit exist etc. Any thoughts / scrutiny would be appreciated, since I can't find a proof elsewhere. James 
February 16th, 2017, 03:49 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 Math Focus: Dynamical systems, analytic function theory, numerics 
2 problems with your proof: 1. Look over your computations on page 1 (about halfway down). Consider if $f$ is constant and notice that none of these statements you have made are true. Namely, the RHS of (*) is not true when $f$ is constant. Unfortunately, this computation is the crux of your proof and the rest of the computation falls through. The difficult part of proving the chain theorem occurs precisely when $f$ has zero derivative on some sequence converging to $(a,b)$. 2. On page 2, you state from one line to the next that $$ \lim_{h \rightarrow 0 } \frac{g(f(a+h,b))g(f(a,b))}{f(a+h,b)  f(a,b)} = g'(f(a,b))$$ but this is precisely the statement that requires proof. Implicitly, in the line above you have interchanged the limit for the variable you call $t$ with the variable $h$. In this case that can be justified but this isn't always the case and any proof should argue why this is allowed here. As a hint toward a proof I suggest you consider the following expression $$ F(h) = \frac{g(f(a+h,b))  g(f(a,b))}{h}  g'(f(a,b)) \qquad h \neq 0 \\ F(0) = 0. $$ Now try to prove that $F$ is continuous at $h= 0$. Hope this helps. 
February 16th, 2017, 10:16 AM  #3 
Newbie Joined: Jan 2016 From: United Kingdom Posts: 29 Thanks: 0 
SDK , Thanks for the reply, I've attached an image to (try) to build on what you said about the constant f function. As for the rest, I'm going to assume the function is non constant and proceed with your idea. James 
February 16th, 2017, 10:28 AM  #4 
Newbie Joined: Jan 2016 From: United Kingdom Posts: 29 Thanks: 0 
Also, am I right in saying that to prove the continuity, one would have to use the fact that g is differentiable f(a,b)? Then manipulate the limit?


Tags 
chain, correct, multivariable, proof, rule 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof of chain rule questions  Christhanksalot  Calculus  6  June 19th, 2015 10:09 AM 
Real differentiable function, proof of chain rule  king.oslo  Complex Analysis  0  September 29th, 2014 09:07 AM 
chain rule constant rule  ungeheuer  Calculus  1  July 30th, 2013 05:10 PM 
proof of the chain rule  one thing I don't understand  NeuroFuzzy  Calculus  1  September 21st, 2011 03:37 AM 
how do yo solve this one using chain rule with quotient rule  Peter1107  Calculus  1  September 8th, 2011 10:25 AM 