How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
up vote
1
down vote
favorite
How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
$$begin{align}
limlimits_{h to 0}frac{Vert f(a+h)-f(a)-Df(a)hVert}{Vert hVert}
&=limlimits_{h to 0}frac{Vert (h_2+a_2,h_1+a_1)
-(a_2,a_1)-(0,a_1h_1)-(a_2h_2,0)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2,h_1)-(a_2h_2,a_1h_1)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2(1-a_2),h_1(1-a_1))vertvert}{Vert hVert}
end{align}
$$
At this point I don't think I can do anything because I don't know what $a_1$ and $a_2$ are. If I knew they were between $[-1,1]$, I could use squeeze theorem but I'm not sure what to do.
multivariable-calculus derivatives
add a comment |
up vote
1
down vote
favorite
How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
$$begin{align}
limlimits_{h to 0}frac{Vert f(a+h)-f(a)-Df(a)hVert}{Vert hVert}
&=limlimits_{h to 0}frac{Vert (h_2+a_2,h_1+a_1)
-(a_2,a_1)-(0,a_1h_1)-(a_2h_2,0)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2,h_1)-(a_2h_2,a_1h_1)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2(1-a_2),h_1(1-a_1))vertvert}{Vert hVert}
end{align}
$$
At this point I don't think I can do anything because I don't know what $a_1$ and $a_2$ are. If I knew they were between $[-1,1]$, I could use squeeze theorem but I'm not sure what to do.
multivariable-calculus derivatives
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
$$begin{align}
limlimits_{h to 0}frac{Vert f(a+h)-f(a)-Df(a)hVert}{Vert hVert}
&=limlimits_{h to 0}frac{Vert (h_2+a_2,h_1+a_1)
-(a_2,a_1)-(0,a_1h_1)-(a_2h_2,0)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2,h_1)-(a_2h_2,a_1h_1)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2(1-a_2),h_1(1-a_1))vertvert}{Vert hVert}
end{align}
$$
At this point I don't think I can do anything because I don't know what $a_1$ and $a_2$ are. If I knew they were between $[-1,1]$, I could use squeeze theorem but I'm not sure what to do.
multivariable-calculus derivatives
How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
$$begin{align}
limlimits_{h to 0}frac{Vert f(a+h)-f(a)-Df(a)hVert}{Vert hVert}
&=limlimits_{h to 0}frac{Vert (h_2+a_2,h_1+a_1)
-(a_2,a_1)-(0,a_1h_1)-(a_2h_2,0)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2,h_1)-(a_2h_2,a_1h_1)Vert}{Vert hVert}\
&=limlimits_{h to 0}frac{Vert (h_2(1-a_2),h_1(1-a_1))vertvert}{Vert hVert}
end{align}
$$
At this point I don't think I can do anything because I don't know what $a_1$ and $a_2$ are. If I knew they were between $[-1,1]$, I could use squeeze theorem but I'm not sure what to do.
multivariable-calculus derivatives
multivariable-calculus derivatives
edited Nov 20 at 2:52
user587192
1,295110
1,295110
asked Nov 20 at 2:09
AColoredReptile
1678
1678
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Hint: You are trying to linearly approximate the function
$$
f(x,y)=(y,x)=begin{pmatrix} 0&1\1&0 end{pmatrix}begin{pmatrix}x\y end{pmatrix}
$$
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: You are trying to linearly approximate the function
$$
f(x,y)=(y,x)=begin{pmatrix} 0&1\1&0 end{pmatrix}begin{pmatrix}x\y end{pmatrix}
$$
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
|
show 1 more comment
up vote
1
down vote
accepted
Hint: You are trying to linearly approximate the function
$$
f(x,y)=(y,x)=begin{pmatrix} 0&1\1&0 end{pmatrix}begin{pmatrix}x\y end{pmatrix}
$$
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
|
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: You are trying to linearly approximate the function
$$
f(x,y)=(y,x)=begin{pmatrix} 0&1\1&0 end{pmatrix}begin{pmatrix}x\y end{pmatrix}
$$
Hint: You are trying to linearly approximate the function
$$
f(x,y)=(y,x)=begin{pmatrix} 0&1\1&0 end{pmatrix}begin{pmatrix}x\y end{pmatrix}
$$
answered Nov 20 at 2:16
qbert
21.6k32459
21.6k32459
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
|
show 1 more comment
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
Should $Df(a)h=begin{pmatrix} 0&ah_1\bh_2&0 end{pmatrix}$?
– AColoredReptile
Nov 20 at 2:26
1
1
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Why does your candidate for the derivative depend on the variable in the limit?
– qbert
Nov 20 at 2:28
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
Sorry I mean $Df(a)h$
– AColoredReptile
Nov 20 at 2:29
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
still no, I think you should check your computation of $Df$
– qbert
Nov 20 at 2:41
2
2
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
Oh $Df=begin{pmatrix} D_1f_1&D_2f_1\D_1f_2&D_1f_2 end{pmatrix}=begin{pmatrix} 0&1\1&0 end{pmatrix}$? and $Df(a)h=begin{pmatrix} h_2\h_1 end{pmatrix}$
– AColoredReptile
Nov 20 at 2:46
|
show 1 more 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3005836%2fhow-to-differentiate-the-function-fx-y-y-x-at-a-a-1-a-2%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