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Tensor chain rule reference request

Generalization of chain rule to tensorsMath Major: How to read textbooks in better style or method ? And how to select best books?Show that the sequence $Omega^0Bbb{R}^2 longrightarrow Omega^1Bbb{R}^2 longrightarrow Omega^2Bbb{R}^2$ is exact.Differentiation of scalar fields using tensor notationGetting the most general form of Mayer-Vietoris from the axioms of homologyMultivariable Taylor SeriesReference request for tensor analysisProof of this fairly obscure differentiation trick?Reference request: alternative references on tensor product of modulesFind the gradient and hessian of $f(Ax+b)$ for real value $f$ and matrix $A$$frac{d}{d x_1} int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3overset{!}{=}-v_1(x_1,x_2,F(x_1,x_2))frac{partial}{partial x_1}F(x_1,x_2)$

0

$begingroup$

I am a Maths major student.

Question: Given a function $f:mathbb{R}^2tomathbb{R},$ $g:mathbb{R}^2tomathbb{R}^2$ and $(a_1,a_2)in mathbb{N}^2.$
Assume that $f$ and $g$ are infinitely differentiable.
Is there a formula for
$$frac{partial^{a_1+a_2}} {partial x_1^{a_1} partial x_2^{a_2}} f(g(x_1,x_2))?$$

We need to use chain rule for multivariable version.
I think it has something to do with tensor, as suggested by this post.
But I am unable to pintpoint some references (books or lecture notes) to build my knowledge on tensor.

calculus reference-request vector-analysis tensors

share|cite|improve this question

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

$endgroup$

add a comment | 

0

$begingroup$

I am a Maths major student.

Question: Given a function $f:mathbb{R}^2tomathbb{R},$ $g:mathbb{R}^2tomathbb{R}^2$ and $(a_1,a_2)in mathbb{N}^2.$
Assume that $f$ and $g$ are infinitely differentiable.
Is there a formula for
$$frac{partial^{a_1+a_2}} {partial x_1^{a_1} partial x_2^{a_2}} f(g(x_1,x_2))?$$

We need to use chain rule for multivariable version.
I think it has something to do with tensor, as suggested by this post.
But I am unable to pintpoint some references (books or lecture notes) to build my knowledge on tensor.

calculus reference-request vector-analysis tensors

share|cite|improve this question

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

$endgroup$

add a comment | 

$begingroup$

I am a Maths major student.

Question: Given a function $f:mathbb{R}^2tomathbb{R},$ $g:mathbb{R}^2tomathbb{R}^2$ and $(a_1,a_2)in mathbb{N}^2.$
Assume that $f$ and $g$ are infinitely differentiable.
Is there a formula for
$$frac{partial^{a_1+a_2}} {partial x_1^{a_1} partial x_2^{a_2}} f(g(x_1,x_2))?$$

We need to use chain rule for multivariable version.
I think it has something to do with tensor, as suggested by this post.
But I am unable to pintpoint some references (books or lecture notes) to build my knowledge on tensor.

calculus reference-request vector-analysis tensors

share|cite|improve this question

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

$endgroup$

share|cite|improve this question

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

share|cite|improve this question

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

asked Mar 15 at 8:59

IdonknowIdonknow

2,576850117

1 Answer
1

active

oldest

votes

1

$begingroup$

The Wikipedia page on the chain rule has some info on this.

It will help to write $g=(g_1,g_2)$. Then, we have
$$partial_1(f(g_1(x_1,x_2),g_2(x_1,x_2))) = partial_1f(g(x))partial_1g_1(x) + partial_2f(g(x))partial_1g_2(x) = partial_jf(g(x))partial_1g_j(x).$$
Then, you just iterate this process and determine the general structure (find a pattern). Admittedly, these multivariable chain rule problems can get quite messy when computing higher derivatives.

Edit: To extend this to higher derivatives gets very involved beyond the second or third derivatives. You can look up Faà di Bruno's formula (there is a Wikipedia page on it) to see the generalized expression. I'm virtually sure you can find a proof online, so I won't reproduce it here. Unless you really can't find one, in which case I'll have a cup of coffee and give it a whirl.

share|cite|improve this answer

edited Mar 15 at 13:37

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to compute lower derivative. Question is that I am not able to see a pattern if I have higher derivative.
  $endgroup$
  – Idonknow
  Mar 15 at 12:39
  

  
  
  
  

add a comment | 

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1

$begingroup$

The Wikipedia page on the chain rule has some info on this.

It will help to write $g=(g_1,g_2)$. Then, we have
$$partial_1(f(g_1(x_1,x_2),g_2(x_1,x_2))) = partial_1f(g(x))partial_1g_1(x) + partial_2f(g(x))partial_1g_2(x) = partial_jf(g(x))partial_1g_j(x).$$
Then, you just iterate this process and determine the general structure (find a pattern). Admittedly, these multivariable chain rule problems can get quite messy when computing higher derivatives.

Edit: To extend this to higher derivatives gets very involved beyond the second or third derivatives. You can look up Faà di Bruno's formula (there is a Wikipedia page on it) to see the generalized expression. I'm virtually sure you can find a proof online, so I won't reproduce it here. Unless you really can't find one, in which case I'll have a cup of coffee and give it a whirl.

share|cite|improve this answer

edited Mar 15 at 13:37

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to compute lower derivative. Question is that I am not able to see a pattern if I have higher derivative.
  $endgroup$
  – Idonknow
  Mar 15 at 12:39
  

  
  
  
  

add a comment | 

1

$begingroup$

The Wikipedia page on the chain rule has some info on this.

It will help to write $g=(g_1,g_2)$. Then, we have
$$partial_1(f(g_1(x_1,x_2),g_2(x_1,x_2))) = partial_1f(g(x))partial_1g_1(x) + partial_2f(g(x))partial_1g_2(x) = partial_jf(g(x))partial_1g_j(x).$$
Then, you just iterate this process and determine the general structure (find a pattern). Admittedly, these multivariable chain rule problems can get quite messy when computing higher derivatives.

Edit: To extend this to higher derivatives gets very involved beyond the second or third derivatives. You can look up Faà di Bruno's formula (there is a Wikipedia page on it) to see the generalized expression. I'm virtually sure you can find a proof online, so I won't reproduce it here. Unless you really can't find one, in which case I'll have a cup of coffee and give it a whirl.

share|cite|improve this answer

edited Mar 15 at 13:37

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to compute lower derivative. Question is that I am not able to see a pattern if I have higher derivative.
  $endgroup$
  – Idonknow
  Mar 15 at 12:39
  

  
  
  
  

add a comment | 

$begingroup$

The Wikipedia page on the chain rule has some info on this.

It will help to write $g=(g_1,g_2)$. Then, we have
$$partial_1(f(g_1(x_1,x_2),g_2(x_1,x_2))) = partial_1f(g(x))partial_1g_1(x) + partial_2f(g(x))partial_1g_2(x) = partial_jf(g(x))partial_1g_j(x).$$
Then, you just iterate this process and determine the general structure (find a pattern). Admittedly, these multivariable chain rule problems can get quite messy when computing higher derivatives.

Edit: To extend this to higher derivatives gets very involved beyond the second or third derivatives. You can look up Faà di Bruno's formula (there is a Wikipedia page on it) to see the generalized expression. I'm virtually sure you can find a proof online, so I won't reproduce it here. Unless you really can't find one, in which case I'll have a cup of coffee and give it a whirl.

share|cite|improve this answer

edited Mar 15 at 13:37

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

$endgroup$

share|cite|improve this answer

edited Mar 15 at 13:37

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

answered Mar 15 at 10:32

Gary MoonGary Moon

77616

answered Mar 15 at 10:32

Gary MoonGary Moon

77616