Product Rule for $mathbbR^n$ The Next CEO of Stack OverflowCan't see how this function is differentiable Spivak's Calculus on Manifolds Exercise 2-4Proving differentiability at an end point.Proof using the definition of the limit.Frechet Derivatives of normed spacesProve that $F(x)=alpha(x)f(x)$ is differentiable and compute the derivativeComputing differentials of a quotient (product) of differential functionsDifferentiable $fcolon ItomathbbC$ with bounded derivative is Lipschitz continuousProduct rule for quaternionsWhat we can say about $lim_n to infty b_n$?Analysis Inequality proof technique/tool
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Product Rule for $mathbbR^n$
The Next CEO of Stack OverflowCan't see how this function is differentiable Spivak's Calculus on Manifolds Exercise 2-4Proving differentiability at an end point.Proof using the definition of the limit.Frechet Derivatives of normed spacesProve that $F(x)=alpha(x)f(x)$ is differentiable and compute the derivativeComputing differentials of a quotient (product) of differential functionsDifferentiable $fcolon ItomathbbC$ with bounded derivative is Lipschitz continuousProduct rule for quaternionsWhat we can say about $lim_n to infty b_n$?Analysis Inequality proof technique/tool
$begingroup$
Let g: $mathbbR^m$ → $mathbbR^n$ differentiable and h: $mathbbR^m$ → $mathbbR$ differentiable with $xin mathbbR^m$.
How to prove product rule that ($hg$)($x$)= $h'(x)g(x) + h(x)g'(x)$
I got to do with this definition:
$$
lim_k to 0 fraclVert g(x+k) - g(x) - g'(x)(k) rVertlVert k rVert = 0
$$
But I cannot sum and subtract $g(x)h(x+k)$ - $g(x)h(x+k)$ as I do in $mathbbR$ because $g(x)h(x+k)$ remains and I can't apply limit again. Is there any trick here?
Thanks!
real-analysis derivatives
asked Mar 19 at 19:53
iaguetiaguet
265
$endgroup$
add a comment |
$begingroup$
Let g: $mathbbR^m$ → $mathbbR^n$ differentiable and h: $mathbbR^m$ → $mathbbR$ differentiable with $xin mathbbR^m$.
How to prove product rule that ($hg$)($x$)= $h'(x)g(x) + h(x)g'(x)$
I got to do with this definition:
$$
lim_k to 0 fraclVert g(x+k) - g(x) - g'(x)(k) rVertlVert k rVert = 0
$$
But I cannot sum and subtract $g(x)h(x+k)$ - $g(x)h(x+k)$ as I do in $mathbbR$ because $g(x)h(x+k)$ remains and I can't apply limit again. Is there any trick here?
Thanks!
real-analysis derivatives
asked Mar 19 at 19:53
iaguetiaguet
265
$endgroup$
$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
$endgroup$
– user647486
Mar 19 at 20:06
$begingroup$
The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
$endgroup$
– John Bentin
Mar 19 at 21:43
add a comment |
$begingroup$
Let g: $mathbbR^m$ → $mathbbR^n$ differentiable and h: $mathbbR^m$ → $mathbbR$ differentiable with $xin mathbbR^m$.
How to prove product rule that ($hg$)($x$)= $h'(x)g(x) + h(x)g'(x)$
I got to do with this definition:
$$
lim_k to 0 fraclVert g(x+k) - g(x) - g'(x)(k) rVertlVert k rVert = 0
$$
But I cannot sum and subtract $g(x)h(x+k)$ - $g(x)h(x+k)$ as I do in $mathbbR$ because $g(x)h(x+k)$ remains and I can't apply limit again. Is there any trick here?
Thanks!
real-analysis derivatives
asked Mar 19 at 19:53
iaguetiaguet
265
$endgroup$
Let g: $mathbbR^m$ → $mathbbR^n$ differentiable and h: $mathbbR^m$ → $mathbbR$ differentiable with $xin mathbbR^m$.
How to prove product rule that ($hg$)($x$)= $h'(x)g(x) + h(x)g'(x)$
I got to do with this definition:
$$
lim_k to 0 fraclVert g(x+k) - g(x) - g'(x)(k) rVertlVert k rVert = 0
$$
But I cannot sum and subtract $g(x)h(x+k)$ - $g(x)h(x+k)$ as I do in $mathbbR$ because $g(x)h(x+k)$ remains and I can't apply limit again. Is there any trick here?
Thanks!
real-analysis derivatives
real-analysis derivatives
asked Mar 19 at 19:53
iaguetiaguet
265
asked Mar 19 at 19:53
iaguetiaguet
265
edited Mar 19 at 19:59
iaguet
asked Mar 19 at 19:53
iaguetiaguet
265
asked Mar 19 at 19:53
iaguetiaguet
265
asked Mar 19 at 19:53
iaguetiaguet
265
265
$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
$endgroup$
– user647486
Mar 19 at 20:06
$begingroup$
The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
$endgroup$
– John Bentin
Mar 19 at 21:43
add a comment |
$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
$endgroup$
– user647486
Mar 19 at 20:06
$begingroup$
The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
$endgroup$
– John Bentin
Mar 19 at 21:43
$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
$endgroup$
– user647486
Mar 19 at 20:06
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
$endgroup$
– user647486
Mar 19 at 20:06
$begingroup$
The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
$endgroup$
– John Bentin
Mar 19 at 21:43
$begingroup$
The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
$endgroup$
– John Bentin
Mar 19 at 21:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The total derivative of the map $x mapsto h(x)g(x) in mathbbR^n$ at $x in mathbbR^m$ is defined as the unique linear transformation $(hg)'(x)$ such that
$$lim_k to 0 fracVert (hg)(x+k)-(hg)(x) -(hg)'(x)(k) VertVert k Vert = 0$$
Where $k in mathbbR^m$. To check that your formula is correct, we consider that for any $x,k in mathbbR^m$
$beginalign
Vert (h & g)(x+k) -(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert (hg)(x+k) - h(x+k)g(x)+h(x+k)g(x)-(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) + (h(x+k)-h(x)-h'(x)(k))g(x) Vert \
& leq Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) Vert + Vert (h(x+k)-h(x)-h'(x)(k))g(x)Vert \
& = Vert h(x+k)(g(x+k)-g(x) - g'(x)(k)) + (h(x+k)-h(x))g'(x)(k) Vert \ & qquad+ vert h(x+k)-h(x)-h'(x)(k) vert Vert g(x) Vert \
& leq |h(x+k)||g(x+k)-g(x)-g'(x)(k)| +|h(x+k)-h(x)||g'(x)(k)| \ & qquad + | h(x+k)-h(x)-h'(x)(k)| | g(x) |
endalign $
So dividing through by $|k|$ and taking the limit as $k to 0$, the first term goes to zero by the definition of $g'(x)$ and continuity of $h$, the second goes to zero by continuity of $h$ and the fact that $fracg'(x)(k)=g'(x) left( frack right)$ is bounded, since linear maps on finite dimensional vector spaces are continuous and the unit sphere is compact, and the third term goes to zero by definition of $h'(x)$, showing that the expression $h'(x)g(x)+h(x)g'(x)$ satisfies the defining equation of $(hg)'(x)$.
answered Mar 19 at 20:56
jawheelejawheele
40729
$endgroup$
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The total derivative of the map $x mapsto h(x)g(x) in mathbbR^n$ at $x in mathbbR^m$ is defined as the unique linear transformation $(hg)'(x)$ such that
$$lim_k to 0 fracVert (hg)(x+k)-(hg)(x) -(hg)'(x)(k) VertVert k Vert = 0$$
Where $k in mathbbR^m$. To check that your formula is correct, we consider that for any $x,k in mathbbR^m$
$beginalign
Vert (h & g)(x+k) -(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert (hg)(x+k) - h(x+k)g(x)+h(x+k)g(x)-(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) + (h(x+k)-h(x)-h'(x)(k))g(x) Vert \
& leq Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) Vert + Vert (h(x+k)-h(x)-h'(x)(k))g(x)Vert \
& = Vert h(x+k)(g(x+k)-g(x) - g'(x)(k)) + (h(x+k)-h(x))g'(x)(k) Vert \ & qquad+ vert h(x+k)-h(x)-h'(x)(k) vert Vert g(x) Vert \
& leq |h(x+k)||g(x+k)-g(x)-g'(x)(k)| +|h(x+k)-h(x)||g'(x)(k)| \ & qquad + | h(x+k)-h(x)-h'(x)(k)| | g(x) |
endalign $
So dividing through by $|k|$ and taking the limit as $k to 0$, the first term goes to zero by the definition of $g'(x)$ and continuity of $h$, the second goes to zero by continuity of $h$ and the fact that $fracg'(x)(k)=g'(x) left( frack right)$ is bounded, since linear maps on finite dimensional vector spaces are continuous and the unit sphere is compact, and the third term goes to zero by definition of $h'(x)$, showing that the expression $h'(x)g(x)+h(x)g'(x)$ satisfies the defining equation of $(hg)'(x)$.
answered Mar 19 at 20:56
jawheelejawheele
40729
$endgroup$
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
add a comment |
$begingroup$
The total derivative of the map $x mapsto h(x)g(x) in mathbbR^n$ at $x in mathbbR^m$ is defined as the unique linear transformation $(hg)'(x)$ such that
$$lim_k to 0 fracVert (hg)(x+k)-(hg)(x) -(hg)'(x)(k) VertVert k Vert = 0$$
Where $k in mathbbR^m$. To check that your formula is correct, we consider that for any $x,k in mathbbR^m$
$beginalign
Vert (h & g)(x+k) -(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert (hg)(x+k) - h(x+k)g(x)+h(x+k)g(x)-(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) + (h(x+k)-h(x)-h'(x)(k))g(x) Vert \
& leq Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) Vert + Vert (h(x+k)-h(x)-h'(x)(k))g(x)Vert \
& = Vert h(x+k)(g(x+k)-g(x) - g'(x)(k)) + (h(x+k)-h(x))g'(x)(k) Vert \ & qquad+ vert h(x+k)-h(x)-h'(x)(k) vert Vert g(x) Vert \
& leq |h(x+k)||g(x+k)-g(x)-g'(x)(k)| +|h(x+k)-h(x)||g'(x)(k)| \ & qquad + | h(x+k)-h(x)-h'(x)(k)| | g(x) |
endalign $
So dividing through by $|k|$ and taking the limit as $k to 0$, the first term goes to zero by the definition of $g'(x)$ and continuity of $h$, the second goes to zero by continuity of $h$ and the fact that $fracg'(x)(k)=g'(x) left( frack right)$ is bounded, since linear maps on finite dimensional vector spaces are continuous and the unit sphere is compact, and the third term goes to zero by definition of $h'(x)$, showing that the expression $h'(x)g(x)+h(x)g'(x)$ satisfies the defining equation of $(hg)'(x)$.
answered Mar 19 at 20:56
jawheelejawheele
40729
$endgroup$
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
add a comment |
$begingroup$
The total derivative of the map $x mapsto h(x)g(x) in mathbbR^n$ at $x in mathbbR^m$ is defined as the unique linear transformation $(hg)'(x)$ such that
$$lim_k to 0 fracVert (hg)(x+k)-(hg)(x) -(hg)'(x)(k) VertVert k Vert = 0$$
Where $k in mathbbR^m$. To check that your formula is correct, we consider that for any $x,k in mathbbR^m$
$beginalign
Vert (h & g)(x+k) -(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert (hg)(x+k) - h(x+k)g(x)+h(x+k)g(x)-(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) + (h(x+k)-h(x)-h'(x)(k))g(x) Vert \
& leq Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) Vert + Vert (h(x+k)-h(x)-h'(x)(k))g(x)Vert \
& = Vert h(x+k)(g(x+k)-g(x) - g'(x)(k)) + (h(x+k)-h(x))g'(x)(k) Vert \ & qquad+ vert h(x+k)-h(x)-h'(x)(k) vert Vert g(x) Vert \
& leq |h(x+k)||g(x+k)-g(x)-g'(x)(k)| +|h(x+k)-h(x)||g'(x)(k)| \ & qquad + | h(x+k)-h(x)-h'(x)(k)| | g(x) |
endalign $
So dividing through by $|k|$ and taking the limit as $k to 0$, the first term goes to zero by the definition of $g'(x)$ and continuity of $h$, the second goes to zero by continuity of $h$ and the fact that $fracg'(x)(k)=g'(x) left( frack right)$ is bounded, since linear maps on finite dimensional vector spaces are continuous and the unit sphere is compact, and the third term goes to zero by definition of $h'(x)$, showing that the expression $h'(x)g(x)+h(x)g'(x)$ satisfies the defining equation of $(hg)'(x)$.
answered Mar 19 at 20:56
jawheelejawheele
40729
$endgroup$
The total derivative of the map $x mapsto h(x)g(x) in mathbbR^n$ at $x in mathbbR^m$ is defined as the unique linear transformation $(hg)'(x)$ such that
$$lim_k to 0 fracVert (hg)(x+k)-(hg)(x) -(hg)'(x)(k) VertVert k Vert = 0$$
Where $k in mathbbR^m$. To check that your formula is correct, we consider that for any $x,k in mathbbR^m$
$beginalign
Vert (h & g)(x+k) -(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert (hg)(x+k) - h(x+k)g(x)+h(x+k)g(x)-(hg)(x)-h'(x)(k)g(x)-h(x)g'(x)(k) Vert \
& = Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) + (h(x+k)-h(x)-h'(x)(k))g(x) Vert \
& leq Vert h(x+k)(g(x+k)-g(x)) -h(x)g'(x)(k) Vert + Vert (h(x+k)-h(x)-h'(x)(k))g(x)Vert \
& = Vert h(x+k)(g(x+k)-g(x) - g'(x)(k)) + (h(x+k)-h(x))g'(x)(k) Vert \ & qquad+ vert h(x+k)-h(x)-h'(x)(k) vert Vert g(x) Vert \
& leq |h(x+k)||g(x+k)-g(x)-g'(x)(k)| +|h(x+k)-h(x)||g'(x)(k)| \ & qquad + | h(x+k)-h(x)-h'(x)(k)| | g(x) |
endalign $
So dividing through by $|k|$ and taking the limit as $k to 0$, the first term goes to zero by the definition of $g'(x)$ and continuity of $h$, the second goes to zero by continuity of $h$ and the fact that $fracg'(x)(k)=g'(x) left( frack right)$ is bounded, since linear maps on finite dimensional vector spaces are continuous and the unit sphere is compact, and the third term goes to zero by definition of $h'(x)$, showing that the expression $h'(x)g(x)+h(x)g'(x)$ satisfies the defining equation of $(hg)'(x)$.
answered Mar 19 at 20:56
jawheelejawheele
40729
edited Mar 19 at 22:00
answered Mar 19 at 20:56
jawheelejawheele
40729
answered Mar 19 at 20:56
jawheelejawheele
40729
answered Mar 19 at 20:56
jawheelejawheele
40729
40729
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
add a comment |
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
1
1
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
$begingroup$
The main formula on the third line is missing "$=0$".
$endgroup$
– John Bentin
Mar 19 at 21:48
add a comment |
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$begingroup$
x is a vector in Rn or Rm? And h:Rm to R , or to Rn?
$endgroup$
– Ameryr
Mar 19 at 19:58
$begingroup$
@Ameryr it's Rm...I'm sorry, edited it
$endgroup$
– iaguet
Mar 19 at 19:59
$begingroup$
Compute the differential of the function $p(x,y)=x+y$, use/prove the chain rule and apply it to $p(h,g)$.
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– user647486
Mar 19 at 20:06
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The product rule is $$(hg)'(x)=g(x)h'(x)+h(x)g'(x).$$ Also, "$g'(x)(k)$" should be written $g'(x)k.$
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– John Bentin
Mar 19 at 21:43