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June 28th, 2017, 08:39 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! 
June 28th, 2017, 08:55 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,829 Thanks: 753  Quote:
Quote:
 
June 28th, 2017, 09:13 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 11:56 AM. 
June 28th, 2017, 09:21 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,164 Thanks: 1423 
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). 
June 28th, 2017, 09:43 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! 
June 28th, 2017, 10:15 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  
June 28th, 2017, 10:29 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!

June 28th, 2017, 10:38 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,175 Thanks: 90  
June 28th, 2017, 10:50 AM  #9 
Newbie Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 
but isnt df/dx=1 since df/dg*dg/dx=1 ?

June 28th, 2017, 10:59 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 11:57 AM. 

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