Inverse function theorem for partial derivatives of a vector function
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
add a comment |
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
add a comment |
$begingroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
$endgroup$
I have a simple vector function:
$$mathbf{y} = amathbf{x}qquad ainmathbb{R},;mathbf{x},mathbf{y}inmathbb{R}^n$$
The inverse is obviously:
$$mathbf{y}^{-1} = {1over a}mathbf{x}$$
The inverse function theorem (at least for scalars) states that:
$$left[f^{-1}(y)right]'=frac{1}{f'(f^{-1}(y))}$$
But the partial derivatives of $mathbf{y}$ and $mathbf{y}^{-1}$ with respect to $a$ are not related that way:
$$begin{align}(mathbf{y})'_a &= mathbf{x} \ (mathbf{y}^{-1})'_a &= -{1over a^2}mathbf{x} end{align}$$
What is the relation between the above derivatives, then?
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
edited Feb 9 '18 at 17:22
user99914
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
asked Feb 25 '15 at 2:19
LiborLibor
8781822
8781822
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
$endgroup$
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1164273%2finverse-function-theorem-for-partial-derivatives-of-a-vector-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
$endgroup$
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
$endgroup$
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
$endgroup$
The derivative of a function $F:mathbb{R}^nto mathbb{R}^n$ at point x, is (= can be seen as) the jacobian matrix
$$J_F(x) = begin{pmatrix}
dfrac{partial F_1}{partial x_1}(x) & cdots & dfrac{partial F_1}{partial x_n}(x)\
vdots & ddots & vdots\
dfrac{partial F_n}{partial x_1}(x) & cdots & dfrac{partial F_n}{partial x_n}(x) end{pmatrix}$$
And the Jacobian of $F^{-1}$ at point x is the inverse of the Jacobian of $F$ at point x :
$$J_{F^{-1}}(x) = left(J_F(x)right)^{-1}$$
You can read this page for a little more information : http://en.wikipedia.org/wiki/Inverse_function_theorem
edited Feb 25 '15 at 16:03
Libor
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
edited Feb 25 '15 at 16:03
Libor
8781822
edited Feb 25 '15 at 16:03
Libor
8781822
edited Feb 25 '15 at 16:03
Libor
8781822
8781822
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
answered Feb 25 '15 at 2:29
TryssTryss
13k1129
13k1129
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
I read that page but it does not speak about partial derivatives. The partial derivative w.r.t. to $a$ of that vector function is a vector, not matrix. But vectors are not invertible.
$endgroup$
– Libor
Feb 25 '15 at 2:34
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
The Jacobian gives you the partial derivatives...
$endgroup$
– Tryss
Feb 25 '15 at 2:37
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
But the function is something like $mathbf{y}=amathbf{x}=begin{pmatrix}ax_1 \ ax_2end{pmatrix}$. The only variable I care about is $a$ ($mathbf{x}$ is known and fixed), but using the above jacobian, there are three columns $a,x_1,x_2$ and only two rows...
$endgroup$
– Libor
Feb 25 '15 at 2:40
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Oh, you mean your function is from $mathbb{R}times mathbb{R}^2 to mathbb{R}^2$? If it's the case, it's clearly not invertible : $$fleft(1,begin{pmatrix} 0 \ 0 end{pmatrix} right) = 1begin{pmatrix} 0 \ 0 end{pmatrix}= begin{pmatrix} 0 \ 0 end{pmatrix} = 2begin{pmatrix} 0 \ 0 end{pmatrix} = fleft(2,begin{pmatrix} 0 \ 0 end{pmatrix}right)$$
$endgroup$
– Tryss
Feb 25 '15 at 2:43
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
$begingroup$
Yes. The problem is that I already know values of $F$, $F'_{a}$ and $F^{-1}$ and only need to somehow obtain $F'_a(F^{-1})$ using the reciprocal rule...
$endgroup$
– Libor
Feb 25 '15 at 2:48
|
show 1 more comment
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
$endgroup$
The derivative w.r.t. $a$ is
$${mathbb{d}mathbf{y}over mathbb{d}a} = mathbf{x}$$
and the derivative of inverse is the reciproval of above, i.e.
$${mathbb{d}aover mathbb{d}mathbf{y}} = {1over mathbf{x}}$$
The notation caused confusion about what is the inverse function. The inverse is function of $a$.
answered Feb 25 '15 at 17:55
LiborLibor
8781822
edited Feb 25 '15 at 18:29
answered Feb 25 '15 at 17:55
LiborLibor
8781822
answered Feb 25 '15 at 17:55
LiborLibor
8781822
answered Feb 25 '15 at 17:55
LiborLibor
8781822
8781822
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
$begingroup$
What is $1/mathbf x$ when $mathbf x$ is a vector?
$endgroup$
– M. Winter
Jan 10 '18 at 13:17
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1164273%2finverse-function-theorem-for-partial-derivatives-of-a-vector-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
