How to calculate the partial derivative of a vector

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Let suppose we have the following vector

$ mathbfb= [b_1, b_2, b_3]$

$:R_jleft(b_jright)=dfracb_j Q^2sum _i=1^3left(b_iright):-dfracb_j Qsum_i=1^3left(b_iright): $

$ dfracpartial R_jleft(mathbfbright)partial b_j:=?: $

How to differentiate?

differential-equations derivatives partial-derivative vector-analysis

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edited Jul 20 at 4:33

asked Jul 19 at 10:59

jehan

106

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up vote
0
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favorite

Let suppose we have the following vector

$ mathbfb= [b_1, b_2, b_3]$

$:R_jleft(b_jright)=dfracb_j Q^2sum _i=1^3left(b_iright):-dfracb_j Qsum_i=1^3left(b_iright): $

$ dfracpartial R_jleft(mathbfbright)partial b_j:=?: $

How to differentiate?

differential-equations derivatives partial-derivative vector-analysis

share|cite|improve this question

edited Jul 20 at 4:33

asked Jul 19 at 10:59

jehan

106

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let suppose we have the following vector

$ mathbfb= [b_1, b_2, b_3]$

$:R_jleft(b_jright)=dfracb_j Q^2sum _i=1^3left(b_iright):-dfracb_j Qsum_i=1^3left(b_iright): $

$ dfracpartial R_jleft(mathbfbright)partial b_j:=?: $

How to differentiate?

differential-equations derivatives partial-derivative vector-analysis

share|cite|improve this question

edited Jul 20 at 4:33

asked Jul 19 at 10:59

jehan

106

Let suppose we have the following vector

$ mathbfb= [b_1, b_2, b_3]$

$:R_jleft(b_jright)=dfracb_j Q^2sum _i=1^3left(b_iright):-dfracb_j Qsum_i=1^3left(b_iright): $

$ dfracpartial R_jleft(mathbfbright)partial b_j:=?: $

How to differentiate?

differential-equations derivatives partial-derivative vector-analysis

share|cite|improve this question

edited Jul 20 at 4:33

asked Jul 19 at 10:59

jehan

106

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 4:33

edited Jul 20 at 4:33

edited Jul 20 at 4:33

asked Jul 19 at 10:59

jehan

106

asked Jul 19 at 10:59

jehan

106

asked Jul 19 at 10:59

jehan

106

106

add a comment | 

add a comment | 

1 Answer
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active

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votes

up vote
0
down vote



accepted

There is no problem at all; you just have to set up things correctly. You have the three functions
$$R_j(bf b):=b_joversum_i b_i>(Q^2-Q)qquad(1leq jleq3) ,$$
whereby $Q$ is assumed constant. Then
$$partial R_joverpartial b_j=1cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=sum_ine j b_ioverleft(sum_i b_iright)^2>(Q^2-Q) .$$
When $kne j$ we similarly have
$$partial R_joverpartial b_k=0cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=-b_joverleft(sum_i b_iright)^2>(Q^2-Q)qquad(kne j) .$$

share|cite|improve this answer

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
  – jehan
  Jul 21 at 3:59
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

There is no problem at all; you just have to set up things correctly. You have the three functions
$$R_j(bf b):=b_joversum_i b_i>(Q^2-Q)qquad(1leq jleq3) ,$$
whereby $Q$ is assumed constant. Then
$$partial R_joverpartial b_j=1cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=sum_ine j b_ioverleft(sum_i b_iright)^2>(Q^2-Q) .$$
When $kne j$ we similarly have
$$partial R_joverpartial b_k=0cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=-b_joverleft(sum_i b_iright)^2>(Q^2-Q)qquad(kne j) .$$

share|cite|improve this answer

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
  – jehan
  Jul 21 at 3:59
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

There is no problem at all; you just have to set up things correctly. You have the three functions
$$R_j(bf b):=b_joversum_i b_i>(Q^2-Q)qquad(1leq jleq3) ,$$
whereby $Q$ is assumed constant. Then
$$partial R_joverpartial b_j=1cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=sum_ine j b_ioverleft(sum_i b_iright)^2>(Q^2-Q) .$$
When $kne j$ we similarly have
$$partial R_joverpartial b_k=0cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=-b_joverleft(sum_i b_iright)^2>(Q^2-Q)qquad(kne j) .$$

share|cite|improve this answer

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
  – jehan
  Jul 21 at 3:59
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

There is no problem at all; you just have to set up things correctly. You have the three functions
$$R_j(bf b):=b_joversum_i b_i>(Q^2-Q)qquad(1leq jleq3) ,$$
whereby $Q$ is assumed constant. Then
$$partial R_joverpartial b_j=1cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=sum_ine j b_ioverleft(sum_i b_iright)^2>(Q^2-Q) .$$
When $kne j$ we similarly have
$$partial R_joverpartial b_k=0cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=-b_joverleft(sum_i b_iright)^2>(Q^2-Q)qquad(kne j) .$$

share|cite|improve this answer

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

There is no problem at all; you just have to set up things correctly. You have the three functions
$$R_j(bf b):=b_joversum_i b_i>(Q^2-Q)qquad(1leq jleq3) ,$$
whereby $Q$ is assumed constant. Then
$$partial R_joverpartial b_j=1cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=sum_ine j b_ioverleft(sum_i b_iright)^2>(Q^2-Q) .$$
When $kne j$ we similarly have
$$partial R_joverpartial b_k=0cdotsum_i b_i-b_jcdot 1overleft(sum_i b_iright)^2>(Q^2-Q)=-b_joverleft(sum_i b_iright)^2>(Q^2-Q)qquad(kne j) .$$

share|cite|improve this answer

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 8:18

edited Jul 21 at 8:18

edited Jul 21 at 8:18

answered Jul 20 at 8:33

Christian Blatter

164k7107306

answered Jul 20 at 8:33

Christian Blatter

164k7107306

answered Jul 20 at 8:33

Christian Blatter

164k7107306

164k7107306

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
  – jehan
  Jul 21 at 3:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
  – jehan
  Jul 21 at 3:59
  

  
  
  
  

Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
– jehan
Jul 21 at 3:59

Awesome, its mean that I should ignore other bids and consider the "j-th" only. Then yes I can calculate the same way as you described. Thank you
– jehan
Jul 21 at 3:59

add a comment | 

 

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