derivative of a modulus of a gradient of a function with respect to the same function
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I've tryied to calculate the following derivative using the chain rule, but in vain. Could you please help me with it? So I have a absolute value of a gradient of a function f, and I need to calculate the derivative of it with respect to the same function f.
$$fracnabla fdf$$
thanks!
To correct my question.
I came to the $$fracnabla fdf$$ derivative mistakenly.
The expression that wanted to evalualte was $$nabla e^f(x)$$, and thanks to @mfl now I know how to do it.
calculus vector-analysis
asked Aug 9 at 18:42
user582956
62
 |Â
show 1 more comment
up vote
1
down vote
favorite
I've tryied to calculate the following derivative using the chain rule, but in vain. Could you please help me with it? So I have a absolute value of a gradient of a function f, and I need to calculate the derivative of it with respect to the same function f.
$$fracnabla fdf$$
thanks!
To correct my question.
I came to the $$fracnabla fdf$$ derivative mistakenly.
The expression that wanted to evalualte was $$nabla e^f(x)$$, and thanks to @mfl now I know how to do it.
calculus vector-analysis
asked Aug 9 at 18:42
user582956
62
Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've tryied to calculate the following derivative using the chain rule, but in vain. Could you please help me with it? So I have a absolute value of a gradient of a function f, and I need to calculate the derivative of it with respect to the same function f.
$$fracnabla fdf$$
thanks!
To correct my question.
I came to the $$fracnabla fdf$$ derivative mistakenly.
The expression that wanted to evalualte was $$nabla e^f(x)$$, and thanks to @mfl now I know how to do it.
calculus vector-analysis
asked Aug 9 at 18:42
user582956
62
I've tryied to calculate the following derivative using the chain rule, but in vain. Could you please help me with it? So I have a absolute value of a gradient of a function f, and I need to calculate the derivative of it with respect to the same function f.
$$fracnabla fdf$$
thanks!
To correct my question.
I came to the $$fracnabla fdf$$ derivative mistakenly.
The expression that wanted to evalualte was $$nabla e^f(x)$$, and thanks to @mfl now I know how to do it.
calculus vector-analysis
asked Aug 9 at 18:42
user582956
62
edited Aug 14 at 11:03
asked Aug 9 at 18:42
user582956
62
asked Aug 9 at 18:42
user582956
62
asked Aug 9 at 18:42
user582956
62
62
Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35
 |Â
show 1 more comment
Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35
Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35
 |Â
show 1 more comment
1 Answer
1
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up vote
0
down vote
Partial answer
We have that
beginalign
dfracpartialpartial x_ie^f(x)&=e^f(x)dfracpartialpartial x_i(f(x)|nabla f(x)|) \&=e^f(x)left(|nabla f(x)|dfracpartialpartial x_if(x)+f(x)dfracpartialpartial x_i|nabla f(x)|right)
endalign
Now, it is
$$|nabla f|=sqrtsum_i=j^nleft(dfracpartial fpartial x_jright)^2$$
and thus
$$dfracpartialpartial x_i|nabla f(x)|=dfracdisplaystyledfracpartialpartial x_isum_j=1^nleft(dfracpartial fpartial x_jright)^2displaystyle2sqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2=dfracdisplaystylesum_j=1^ndfracpartial^2 fpartial x_ipartial x_jdisplaystylesqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2.$$
And now I hope you can get the answer.
answered Aug 9 at 19:45
mfl
24.6k12040
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Partial answer
We have that
beginalign
dfracpartialpartial x_ie^f(x)&=e^f(x)dfracpartialpartial x_i(f(x)|nabla f(x)|) \&=e^f(x)left(|nabla f(x)|dfracpartialpartial x_if(x)+f(x)dfracpartialpartial x_i|nabla f(x)|right)
endalign
Now, it is
$$|nabla f|=sqrtsum_i=j^nleft(dfracpartial fpartial x_jright)^2$$
and thus
$$dfracpartialpartial x_i|nabla f(x)|=dfracdisplaystyledfracpartialpartial x_isum_j=1^nleft(dfracpartial fpartial x_jright)^2displaystyle2sqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2=dfracdisplaystylesum_j=1^ndfracpartial^2 fpartial x_ipartial x_jdisplaystylesqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2.$$
And now I hope you can get the answer.
answered Aug 9 at 19:45
mfl
24.6k12040
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
add a comment |Â
up vote
0
down vote
Partial answer
We have that
beginalign
dfracpartialpartial x_ie^f(x)&=e^f(x)dfracpartialpartial x_i(f(x)|nabla f(x)|) \&=e^f(x)left(|nabla f(x)|dfracpartialpartial x_if(x)+f(x)dfracpartialpartial x_i|nabla f(x)|right)
endalign
Now, it is
$$|nabla f|=sqrtsum_i=j^nleft(dfracpartial fpartial x_jright)^2$$
and thus
$$dfracpartialpartial x_i|nabla f(x)|=dfracdisplaystyledfracpartialpartial x_isum_j=1^nleft(dfracpartial fpartial x_jright)^2displaystyle2sqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2=dfracdisplaystylesum_j=1^ndfracpartial^2 fpartial x_ipartial x_jdisplaystylesqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2.$$
And now I hope you can get the answer.
answered Aug 9 at 19:45
mfl
24.6k12040
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Partial answer
We have that
beginalign
dfracpartialpartial x_ie^f(x)&=e^f(x)dfracpartialpartial x_i(f(x)|nabla f(x)|) \&=e^f(x)left(|nabla f(x)|dfracpartialpartial x_if(x)+f(x)dfracpartialpartial x_i|nabla f(x)|right)
endalign
Now, it is
$$|nabla f|=sqrtsum_i=j^nleft(dfracpartial fpartial x_jright)^2$$
and thus
$$dfracpartialpartial x_i|nabla f(x)|=dfracdisplaystyledfracpartialpartial x_isum_j=1^nleft(dfracpartial fpartial x_jright)^2displaystyle2sqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2=dfracdisplaystylesum_j=1^ndfracpartial^2 fpartial x_ipartial x_jdisplaystylesqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2.$$
And now I hope you can get the answer.
answered Aug 9 at 19:45
mfl
24.6k12040
Partial answer
We have that
beginalign
dfracpartialpartial x_ie^f(x)&=e^f(x)dfracpartialpartial x_i(f(x)|nabla f(x)|) \&=e^f(x)left(|nabla f(x)|dfracpartialpartial x_if(x)+f(x)dfracpartialpartial x_i|nabla f(x)|right)
endalign
Now, it is
$$|nabla f|=sqrtsum_i=j^nleft(dfracpartial fpartial x_jright)^2$$
and thus
$$dfracpartialpartial x_i|nabla f(x)|=dfracdisplaystyledfracpartialpartial x_isum_j=1^nleft(dfracpartial fpartial x_jright)^2displaystyle2sqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2=dfracdisplaystylesum_j=1^ndfracpartial^2 fpartial x_ipartial x_jdisplaystylesqrtsum_j=1^nleft(dfracpartial fpartial x_jright)^2.$$
And now I hope you can get the answer.
answered Aug 9 at 19:45
mfl
24.6k12040
answered Aug 9 at 19:45
mfl
24.6k12040
answered Aug 9 at 19:45
mfl
24.6k12040
answered Aug 9 at 19:45
mfl
24.6k12040
24.6k12040
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
add a comment |Â
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
thanks so much, now I see my mistake. That derivative of the magnitude appeared because I used incorrectly the chain rule for a gradient of a composite function.
– user582956
Aug 9 at 21:26
add a comment |Â
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Look again. A gradient is a vector. Absolute values are for scalars. So those vertical bars mean what?
– mvw
Aug 9 at 18:44
@mvw It is $|nabla f|=sqrtnabla fcdot nabla f.$ In any case it is not clear to me what derivative you want to have. Is it possible that you are looking for the derivative of $|nabla f|$ in the direction of $nabla f?$
– mfl
Aug 9 at 18:57
Don't tell me.There seems to be something missing or wrong.
– mvw
Aug 9 at 19:05
@mfl So I have the magnitude of the gradient of a function $f$ with respect to the same function $f$. It came when I was calculating the following: $nabla (exp(f(textbfr)|nabla f(textbfr)|))$
– user582956
Aug 9 at 19:30
So, you want to get $nabla e^f(x).$ Is it correct?
– mfl
Aug 9 at 19:35