Doubt about Implicit differentiation

  • Thread starter vineethbs
  • Start date
  • #1
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Hi all,

I was reading a paper in which implicit differentiation was used as follows

[tex] x \in R, \lambda \in R [/tex]
Given [tex] G(x,\lambda) = 0 [/tex]
[tex] \frac{\partial G(x,\lambda)}{\partial x} \frac{\partial x}{\partial \lambda} + \frac{\partial G(x,\lambda)}{\partial \lambda} = 0 [/tex]

My doubt is related to whether it is possible to do this even if x is say a function from R to R and G is therefore a functional. Is there a implicit differentiation rule for functionals ?

Thanks for your time !
 

Answers and Replies

  • #2
726
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I am a little unsure of your question.

x is a function from R to R since it is a function of lamda.
What do you mean by functional?
 
  • #3
lurflurf
Homework Helper
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That is the most simple example of implicit differentiation possible. If anything is wrong with i, all implicit differentiation is wrong. There is an analogous calculus for funtionals,, but it is not needed in this case. A functional is a function that takes another function as its argument. So true that
y|->F(y(x),x)
2x|->sin(2x)
x^2|->x^2+1
f(x)|->3*f(5)
f(x)|->0
f|->g(f(x))
are functionals they only use information at a single point thus no complications arise.

The chain rule
[g(f(x))]'=g'(f(x))f'(x)
is another example since g is a functional.
 
  • #4
8
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I am a little unsure of your question.

x is a function from R to R since it is a function of lamda.
What do you mean by functional?
Sorry about the confusion. Yes, x is an implicit function of \lamba , but the idea is that x(\lambda) is a family of functions paramterized by \lambda
 
  • #5
8
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If anything is wrong with i, all implicit differentiation is wrong.

Sorry but what is "i" ? I did not understand what this means.

There is an analogous calculus for funtionals,, but it is not needed in this case.
Why is it not needed if x is a function ? And could you please point me to some easy references on calculus for functionals ?

Thank you !
 

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