Chain rule for function in $mathbb{R}^{m}$ and $mathbb{R}^{n}$.
In my text book, say that
Let $g:mathbb{R}^{m}tomathbb{R}^{n}$ and $f:mathbb{R}^{n}tomathbb{R}$, the chain rule implies that $$D_{j}(fcirc g)(a)=sum_{i=1}^{n}{D_{i}f(g(a))cdot D_{j}g^{i}(a)}$$
If $ainmathbb{R}^{m}$. I know the chain rule in basic case, but I don't see where from the terms $g^i(a)$. Any hint will be appreciated.! Thanks!
real-analysis multivariable-calculus
add a comment |
In my text book, say that
Let $g:mathbb{R}^{m}tomathbb{R}^{n}$ and $f:mathbb{R}^{n}tomathbb{R}$, the chain rule implies that $$D_{j}(fcirc g)(a)=sum_{i=1}^{n}{D_{i}f(g(a))cdot D_{j}g^{i}(a)}$$
If $ainmathbb{R}^{m}$. I know the chain rule in basic case, but I don't see where from the terms $g^i(a)$. Any hint will be appreciated.! Thanks!
real-analysis multivariable-calculus
Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59
add a comment |
In my text book, say that
Let $g:mathbb{R}^{m}tomathbb{R}^{n}$ and $f:mathbb{R}^{n}tomathbb{R}$, the chain rule implies that $$D_{j}(fcirc g)(a)=sum_{i=1}^{n}{D_{i}f(g(a))cdot D_{j}g^{i}(a)}$$
If $ainmathbb{R}^{m}$. I know the chain rule in basic case, but I don't see where from the terms $g^i(a)$. Any hint will be appreciated.! Thanks!
real-analysis multivariable-calculus
In my text book, say that
Let $g:mathbb{R}^{m}tomathbb{R}^{n}$ and $f:mathbb{R}^{n}tomathbb{R}$, the chain rule implies that $$D_{j}(fcirc g)(a)=sum_{i=1}^{n}{D_{i}f(g(a))cdot D_{j}g^{i}(a)}$$
If $ainmathbb{R}^{m}$. I know the chain rule in basic case, but I don't see where from the terms $g^i(a)$. Any hint will be appreciated.! Thanks!
real-analysis multivariable-calculus
real-analysis multivariable-calculus
asked Dec 26 at 5:36
user570343
Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59
add a comment |
Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59
Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59
add a comment |
1 Answer
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According to the chain rule, you have:
$$D(f circ g)(a) = Df(g(a)) . Dg(a).$$
And by definition, $D_j(f circ g)(a)=D(f circ g)(a)(e_j)$ where $e_j$ is the $j$-th vector of the canonical basis $(e_1, dots ,e_n)$ of $mathbb R^n$.
So
$$D_j(f circ g)(a) = Df(g(a)) . Dg(a)(e_j)$$
while
$Dg(a)(e_j)$ is the vector $(D_j g^1(a), dots, D_j g^n(a))^T$. Multiplying on the left by $Df(g(a))$ according to the chain rule, you’re done.
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
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1 Answer
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1 Answer
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According to the chain rule, you have:
$$D(f circ g)(a) = Df(g(a)) . Dg(a).$$
And by definition, $D_j(f circ g)(a)=D(f circ g)(a)(e_j)$ where $e_j$ is the $j$-th vector of the canonical basis $(e_1, dots ,e_n)$ of $mathbb R^n$.
So
$$D_j(f circ g)(a) = Df(g(a)) . Dg(a)(e_j)$$
while
$Dg(a)(e_j)$ is the vector $(D_j g^1(a), dots, D_j g^n(a))^T$. Multiplying on the left by $Df(g(a))$ according to the chain rule, you’re done.
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
add a comment |
According to the chain rule, you have:
$$D(f circ g)(a) = Df(g(a)) . Dg(a).$$
And by definition, $D_j(f circ g)(a)=D(f circ g)(a)(e_j)$ where $e_j$ is the $j$-th vector of the canonical basis $(e_1, dots ,e_n)$ of $mathbb R^n$.
So
$$D_j(f circ g)(a) = Df(g(a)) . Dg(a)(e_j)$$
while
$Dg(a)(e_j)$ is the vector $(D_j g^1(a), dots, D_j g^n(a))^T$. Multiplying on the left by $Df(g(a))$ according to the chain rule, you’re done.
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
add a comment |
According to the chain rule, you have:
$$D(f circ g)(a) = Df(g(a)) . Dg(a).$$
And by definition, $D_j(f circ g)(a)=D(f circ g)(a)(e_j)$ where $e_j$ is the $j$-th vector of the canonical basis $(e_1, dots ,e_n)$ of $mathbb R^n$.
So
$$D_j(f circ g)(a) = Df(g(a)) . Dg(a)(e_j)$$
while
$Dg(a)(e_j)$ is the vector $(D_j g^1(a), dots, D_j g^n(a))^T$. Multiplying on the left by $Df(g(a))$ according to the chain rule, you’re done.
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
According to the chain rule, you have:
$$D(f circ g)(a) = Df(g(a)) . Dg(a).$$
And by definition, $D_j(f circ g)(a)=D(f circ g)(a)(e_j)$ where $e_j$ is the $j$-th vector of the canonical basis $(e_1, dots ,e_n)$ of $mathbb R^n$.
So
$$D_j(f circ g)(a) = Df(g(a)) . Dg(a)(e_j)$$
while
$Dg(a)(e_j)$ is the vector $(D_j g^1(a), dots, D_j g^n(a))^T$. Multiplying on the left by $Df(g(a))$ according to the chain rule, you’re done.
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
answered Dec 26 at 8:38
mathcounterexamples.net
24.1k21753
24.1k21753
add a comment |
add a comment |
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Start with the case $n=2$.
– IAmNoOne
Dec 26 at 6:23
Interpret the right-hand side as a matrix product.
– amd
Dec 26 at 7:59