Partial derviative of prediction (sigmoid applied) with respect to weightGradient Descent Step for word2vec negative samplingImplementing RMSProp, but finding differences between reference versionsWhy is vanishing gradient a problem?SGD learning gets stuck when using a max pooling layer (but it works fine with just conv + fc)Why are optimization algorithms slower at critical points?Why Gradient methods work in finding the parameters in Neural Networks?Is gradient descent slower for finite differences?Creating a convolutional layer with weight normalization?
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Partial derviative of prediction (sigmoid applied) with respect to weight
Gradient Descent Step for word2vec negative samplingImplementing RMSProp, but finding differences between reference versionsWhy is vanishing gradient a problem?SGD learning gets stuck when using a max pooling layer (but it works fine with just conv + fc)Why are optimization algorithms slower at critical points?Why Gradient methods work in finding the parameters in Neural Networks?Is gradient descent slower for finite differences?Creating a convolutional layer with weight normalization?
$begingroup$
I am very confused as to where a seemingly "extra" term is included in the above mentioned calculation in my Udacity course.
The above is taking the derivative of a sigmoid so why isn't it just
$$=sigma(Wx+b)(1-sigma(Wx+b)$$
but rather has $fracpartialpartial w_j(Wx+b)$ tacked on the tail?
gradient-descent
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am very confused as to where a seemingly "extra" term is included in the above mentioned calculation in my Udacity course.
The above is taking the derivative of a sigmoid so why isn't it just
$$=sigma(Wx+b)(1-sigma(Wx+b)$$
but rather has $fracpartialpartial w_j(Wx+b)$ tacked on the tail?
gradient-descent
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am very confused as to where a seemingly "extra" term is included in the above mentioned calculation in my Udacity course.
The above is taking the derivative of a sigmoid so why isn't it just
$$=sigma(Wx+b)(1-sigma(Wx+b)$$
but rather has $fracpartialpartial w_j(Wx+b)$ tacked on the tail?
gradient-descent
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am very confused as to where a seemingly "extra" term is included in the above mentioned calculation in my Udacity course.
The above is taking the derivative of a sigmoid so why isn't it just
$$=sigma(Wx+b)(1-sigma(Wx+b)$$
but rather has $fracpartialpartial w_j(Wx+b)$ tacked on the tail?
gradient-descent
gradient-descent
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
flexitarian33
asked yesterday
flexitarian33flexitarian33
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
flexitarian33flexitarian33
256
asked yesterday
flexitarian33flexitarian33
256
256
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
flexitarian33 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
Recall that for chain rule, we have $$fracddwh(g(w))=h'(g(w))g'(w)$$
For the context of your question, $h(t)=sigma(t)$ and $g(w)=Wx+b$,
hence that is why we have one more term.
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
$endgroup$
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Recall that for chain rule, we have $$fracddwh(g(w))=h'(g(w))g'(w)$$
For the context of your question, $h(t)=sigma(t)$ and $g(w)=Wx+b$,
hence that is why we have one more term.
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
$endgroup$
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
add a comment |
$begingroup$
Recall that for chain rule, we have $$fracddwh(g(w))=h'(g(w))g'(w)$$
For the context of your question, $h(t)=sigma(t)$ and $g(w)=Wx+b$,
hence that is why we have one more term.
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
$endgroup$
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
add a comment |
$begingroup$
Recall that for chain rule, we have $$fracddwh(g(w))=h'(g(w))g'(w)$$
For the context of your question, $h(t)=sigma(t)$ and $g(w)=Wx+b$,
hence that is why we have one more term.
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
$endgroup$
Recall that for chain rule, we have $$fracddwh(g(w))=h'(g(w))g'(w)$$
For the context of your question, $h(t)=sigma(t)$ and $g(w)=Wx+b$,
hence that is why we have one more term.
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
answered yesterday
Siong Thye GohSiong Thye Goh
1,302418
1,302418
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
add a comment |
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
$begingroup$
I can see now that it's a composite function since sigmoid itself is a function :)
$endgroup$
– flexitarian33
yesterday
add a comment |
flexitarian33 is a new contributor. Be nice, and check out our Code of Conduct.
flexitarian33 is a new contributor. Be nice, and check out our Code of Conduct.
flexitarian33 is a new contributor. Be nice, and check out our Code of Conduct.
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