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 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! June 28th, 2017, 07:55 AM   #2
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 Originally Posted by mgho 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! 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. June 28th, 2017, 08:21 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 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 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! June 28th, 2017, 09:15 AM   #6
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 Originally Posted by mgho 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 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! June 28th, 2017, 09:38 AM   #8
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 Originally Posted by mgho 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 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 ? 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. Tags chain, dfx or dx, dgx or dx, fgx, finding, rule Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Dmath Calculus 4 June 14th, 2014 03:39 PM ungeheuer Calculus 1 July 30th, 2013 05:10 PM unwisetome3 Calculus 4 October 19th, 2012 01:21 PM Peter1107 Calculus 1 September 8th, 2011 10:25 AM nyluappe Calculus 2 February 6th, 2009 09:34 PM

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