Partial chain rule in multi-dimensional case
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I have a problem with a multi-dimensional chain rule. I can't understand what's wrong!
Well, let's consider such partial derivative:
$partial_k(Fcirc G)(x)$, where $F:mathbbR^drightarrowmathbbR$ and $G:mathbbR^drightarrowmathbbR^d$. So, the composition $Fcirc G:mathbbR^drightarrowmathbbR$.
Thus, the partial derivative should be just a real number. However, with the chain rule I somehow get a vector.
Firstly I derive $G$. It's a vector-valued function, like $(G_1, G_2,...,G_d)$, so the partial derivative is a vector $partial_kG=(partial_kG_1, partial_kG_2,...,partial_kG_d)$.
And now I should multiply it on $F$ derivative which is a number, so finally I get a vector.
I'm totally confused, help me please to deal with it! :)
multivariable-calculus partial-derivative chain-rule
asked Aug 28 at 12:33
Alex Goldstein
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favorite
I have a problem with a multi-dimensional chain rule. I can't understand what's wrong!
Well, let's consider such partial derivative:
$partial_k(Fcirc G)(x)$, where $F:mathbbR^drightarrowmathbbR$ and $G:mathbbR^drightarrowmathbbR^d$. So, the composition $Fcirc G:mathbbR^drightarrowmathbbR$.
Thus, the partial derivative should be just a real number. However, with the chain rule I somehow get a vector.
Firstly I derive $G$. It's a vector-valued function, like $(G_1, G_2,...,G_d)$, so the partial derivative is a vector $partial_kG=(partial_kG_1, partial_kG_2,...,partial_kG_d)$.
And now I should multiply it on $F$ derivative which is a number, so finally I get a vector.
I'm totally confused, help me please to deal with it! :)
multivariable-calculus partial-derivative chain-rule
asked Aug 28 at 12:33
Alex Goldstein
1
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a problem with a multi-dimensional chain rule. I can't understand what's wrong!
Well, let's consider such partial derivative:
$partial_k(Fcirc G)(x)$, where $F:mathbbR^drightarrowmathbbR$ and $G:mathbbR^drightarrowmathbbR^d$. So, the composition $Fcirc G:mathbbR^drightarrowmathbbR$.
Thus, the partial derivative should be just a real number. However, with the chain rule I somehow get a vector.
Firstly I derive $G$. It's a vector-valued function, like $(G_1, G_2,...,G_d)$, so the partial derivative is a vector $partial_kG=(partial_kG_1, partial_kG_2,...,partial_kG_d)$.
And now I should multiply it on $F$ derivative which is a number, so finally I get a vector.
I'm totally confused, help me please to deal with it! :)
multivariable-calculus partial-derivative chain-rule
asked Aug 28 at 12:33
Alex Goldstein
1
I have a problem with a multi-dimensional chain rule. I can't understand what's wrong!
Well, let's consider such partial derivative:
$partial_k(Fcirc G)(x)$, where $F:mathbbR^drightarrowmathbbR$ and $G:mathbbR^drightarrowmathbbR^d$. So, the composition $Fcirc G:mathbbR^drightarrowmathbbR$.
Thus, the partial derivative should be just a real number. However, with the chain rule I somehow get a vector.
Firstly I derive $G$. It's a vector-valued function, like $(G_1, G_2,...,G_d)$, so the partial derivative is a vector $partial_kG=(partial_kG_1, partial_kG_2,...,partial_kG_d)$.
And now I should multiply it on $F$ derivative which is a number, so finally I get a vector.
I'm totally confused, help me please to deal with it! :)
multivariable-calculus partial-derivative chain-rule
asked Aug 28 at 12:33
Alex Goldstein
1
asked Aug 28 at 12:33
Alex Goldstein
1
asked Aug 28 at 12:33
Alex Goldstein
1
asked Aug 28 at 12:33
Alex Goldstein
1
1
add a comment |Â
add a comment |Â
1 Answer
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By the chain rule (please look the following notation symbolically) you have:
$$d_x(Fcirc G)(x)=d_xF(G(x))=d_GF(G)cdot d_xG(x)$$
Note that $d_xG$ is a vector and $d_G F(G)$ is a gradient of a scalar function, which is also a vector. Therefore their product gives a scalar.
answered Aug 28 at 13:17
HBR
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
By the chain rule (please look the following notation symbolically) you have:
$$d_x(Fcirc G)(x)=d_xF(G(x))=d_GF(G)cdot d_xG(x)$$
Note that $d_xG$ is a vector and $d_G F(G)$ is a gradient of a scalar function, which is also a vector. Therefore their product gives a scalar.
answered Aug 28 at 13:17
HBR
1,69349
add a comment |Â
up vote
0
down vote
By the chain rule (please look the following notation symbolically) you have:
$$d_x(Fcirc G)(x)=d_xF(G(x))=d_GF(G)cdot d_xG(x)$$
Note that $d_xG$ is a vector and $d_G F(G)$ is a gradient of a scalar function, which is also a vector. Therefore their product gives a scalar.
answered Aug 28 at 13:17
HBR
1,69349
add a comment |Â
up vote
0
down vote
up vote
0
down vote
By the chain rule (please look the following notation symbolically) you have:
$$d_x(Fcirc G)(x)=d_xF(G(x))=d_GF(G)cdot d_xG(x)$$
Note that $d_xG$ is a vector and $d_G F(G)$ is a gradient of a scalar function, which is also a vector. Therefore their product gives a scalar.
answered Aug 28 at 13:17
HBR
1,69349
By the chain rule (please look the following notation symbolically) you have:
$$d_x(Fcirc G)(x)=d_xF(G(x))=d_GF(G)cdot d_xG(x)$$
Note that $d_xG$ is a vector and $d_G F(G)$ is a gradient of a scalar function, which is also a vector. Therefore their product gives a scalar.
answered Aug 28 at 13:17
HBR
1,69349
answered Aug 28 at 13:17
HBR
1,69349
answered Aug 28 at 13:17
HBR
1,69349
answered Aug 28 at 13:17
HBR
1,69349
1,69349
add a comment |Â
add a comment |Â
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