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

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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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Quote:
 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: 20,370 Thanks: 2007 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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Quote:
 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.

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