Partial derivative of a function defined in a rectangular area
I noticed in a book that it considers a function $u(x,y)$ in the rectangular area : $xin (0, 1)$ ,$yin (0, 1]$, and takes there the $partial_y$. As I am used to only to take partial derivatives in open sets, I would appreciate a clarification of the value of $partial_y(x, 1)$. Is it $$partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$ or something else? Thanks in advance.
calculus multivariable-calculus derivatives
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dmtri
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I noticed in a book that it considers a function $u(x,y)$ in the rectangular area : $xin (0, 1)$ ,$yin (0, 1]$, and takes there the $partial_y$. As I am used to only to take partial derivatives in open sets, I would appreciate a clarification of the value of $partial_y(x, 1)$. Is it $$partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$ or something else? Thanks in advance.
calculus multivariable-calculus derivatives
asked yesterday
dmtri
1,4321521
add a comment |
I noticed in a book that it considers a function $u(x,y)$ in the rectangular area : $xin (0, 1)$ ,$yin (0, 1]$, and takes there the $partial_y$. As I am used to only to take partial derivatives in open sets, I would appreciate a clarification of the value of $partial_y(x, 1)$. Is it $$partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$ or something else? Thanks in advance.
calculus multivariable-calculus derivatives
asked yesterday
dmtri
1,4321521
I noticed in a book that it considers a function $u(x,y)$ in the rectangular area : $xin (0, 1)$ ,$yin (0, 1]$, and takes there the $partial_y$. As I am used to only to take partial derivatives in open sets, I would appreciate a clarification of the value of $partial_y(x, 1)$. Is it $$partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$ or something else? Thanks in advance.
calculus multivariable-calculus derivatives
calculus multivariable-calculus derivatives
asked yesterday
dmtri
1,4321521
asked yesterday
dmtri
1,4321521
edited yesterday
asked yesterday
dmtri
1,4321521
asked yesterday
dmtri
1,4321521
asked yesterday
dmtri
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1,4321521
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1 Answer
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Yes, your definition of partial derivative with respect to $y$ at $(x,1)$
$$ partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$
is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.
answered yesterday
Mohammad Riazi-Kermani
41k42059
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes, your definition of partial derivative with respect to $y$ at $(x,1)$
$$ partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$
is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.
answered yesterday
Mohammad Riazi-Kermani
41k42059
add a comment |
Yes, your definition of partial derivative with respect to $y$ at $(x,1)$
$$ partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$
is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.
answered yesterday
Mohammad Riazi-Kermani
41k42059
add a comment |
Yes, your definition of partial derivative with respect to $y$ at $(x,1)$
$$ partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$
is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.
answered yesterday
Mohammad Riazi-Kermani
41k42059
Yes, your definition of partial derivative with respect to $y$ at $(x,1)$
$$ partial_y{u(x, 1)}=lim_{tto 1^{-}}frac{u(x, t) -u(x, 1)}{t-1}$$
is correct. Of course it is one sided because we do not have $y>1$ as part of the domain.
answered yesterday
Mohammad Riazi-Kermani
41k42059
answered yesterday
Mohammad Riazi-Kermani
41k42059
answered yesterday
Mohammad Riazi-Kermani
41k42059
answered yesterday
Mohammad Riazi-Kermani
41k42059
41k42059
add a comment |
add a comment |
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