My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree3Thanks
  • 1 Post By Country Boy
  • 1 Post By skipjack
  • 1 Post By zylo
Reply
 
LinkBack Thread Tools Display Modes
June 28th, 2017, 07:39 AM   #1
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

using chain rule for finding dg(x)/dx given f(g(x)) and df(x)/dx

I have a problem that says:

f(x) and g(x) are differentiable, given that f(g(x))=x and df(x)/dx = 1+[f(x)]^2
show that dg(x)/dx= 1/(1+x^2)!

Am I correct to say since f(g(x))=x, then df(x)/dg(x)=1 ?
then I wrote df(x)/dx = 1+[f(x)]^2= df(x)/dg(x) * dg(x)/dx =dg(x)/dx!

I don't know what to do next, or even what I was doing was correct or not..!!

I appreciate your help in advance!
mgho is offline  
 
June 28th, 2017, 07:55 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,524
Thanks: 643

Quote:
Originally Posted by mgho View Post
I have a problem that says:

f(x) and g(x) are differentiable, given that f(g(x))=x and df(x)/dx = 1+[f(x)]^2
show that dg(x)/dx= 1/(1+x^2)!

Am I correct to say since f(g(x))=x, then df(x)/dg(x)=1 ?
No, you are not. By the chain rule d(f(g(x))/dx= (df/dg)(dg/dx)= 1 so that df/dg= 1/(dg/dx).

Quote:
then I wrote df(x)/dx = 1+[f(x)]^2= df(x)/dg(x) * dg(x)/dx =dg(x)/dx!

I don't know what to do next, or even what I was doing was correct or not..!!

I appreciate your help in advance!
Thanks from mgho
Country Boy is offline  
June 28th, 2017, 08:13 AM   #3
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

Okay thanks, I got it! But then I get df(x)/dx = df(x)/dg(x)*dg(x)/dx= 1/(dg/dx)* dg/dx=1 which doesn't make sense since the problem is saying df(x)/dx=1+[f(x)]^2...

Can you show me what I should do next to show that dg(x)/dx= 1/(1+x^2)?

Last edited by skipjack; June 28th, 2017 at 10:56 AM.
mgho is offline  
June 28th, 2017, 08:21 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 17,533
Thanks: 1322

As f$\,^\prime\!$(x) = 1 + (f(x))², f$\,^\prime\!$(g(x)) = 1 + (f(g(x))².

As f(g(x)) = x, f$\,^\prime\!$(g(x))g$\,^\prime\!$(x) = 1,
and so g$\,^\prime\!$(x) = 1/f$\,^\prime\!$(g(x)) = 1/(1 + (f(g(x))²) = 1/(1 + x²).
Thanks from mgho
skipjack is offline  
June 28th, 2017, 08:43 AM   #5
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

I understood your answer; thank you skipjack..

Can you tell me what I am doing wrong here though?

from f(g(x))=x, CountryBoy showed me that dg/dx=1/(df/dg), which is what you used also..

Next, can I say as df/dx=1+f^2, then df/dg*dg/dx=1+f^2 ? but then the left side of this equation is equal to 1 since dg/dx=1/(df/dg) resulting in 1=1+f^2 ->f=0 ???!!

Can you help me, I don't understand where I am messing it up..
thanx!
mgho is offline  
June 28th, 2017, 09:15 AM   #6
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by mgho View Post
Am I correct to say since f(g(x))=x, then df(x)/dg(x)=1 ?
No. You either differentiate wrt g or x.
df/dg=dx/dg or (df/dg)(dg/dx)=1
Thanks from mgho
zylo is offline  
June 28th, 2017, 09:29 AM   #7
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

I understood that and corrected it ) Can you check my reply on the 5th post and tell me where I am going wrong? Thanx Zylo!
mgho is offline  
June 28th, 2017, 09:38 AM   #8
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,079
Thanks: 87

Quote:
Originally Posted by mgho View Post
I
... since dg/dx=1/(df/dg) resulting in 1=1+f^2 ->f=0 ???!!

Can you help me, I don't understand where I am messing it up..
thanx!
(dg/dx)(df/dg)=df/dx
zylo is offline  
June 28th, 2017, 09:50 AM   #9
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

but isnt df/dx=1 since df/dg*dg/dx=1 ?
mgho is offline  
June 28th, 2017, 09:59 AM   #10
Newbie
 
Joined: Jun 2017
From: Long Beach

Posts: 20
Thanks: 0

Thank you everybody! I got it: I think my mistake was not understanding that df(x)/dx is different from df(g(x))/dx !

Last edited by skipjack; June 28th, 2017 at 10:57 AM.
mgho is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
chain, dfx or dx, dgx or dx, fgx, finding, rule



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
chain rule Dmath Calculus 4 June 14th, 2014 03:39 PM
chain rule- constant rule ungeheuer Calculus 1 July 30th, 2013 05:10 PM
Product rule into chain rule unwisetome3 Calculus 4 October 19th, 2012 01:21 PM
how do yo solve this one using chain rule with quotient rule Peter1107 Calculus 1 September 8th, 2011 10:25 AM
Problem involving Chain Rule, Finding Maximum/Minimum point nyluappe Calculus 2 February 6th, 2009 09:34 PM





Copyright © 2017 My Math Forum. All rights reserved.