How to differentiate the function $f(x,y)=(y,x)$ at $ a=(a_1,a_2)$?
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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
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up vote
1
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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
1,21810
1,21810
asked Nov 20 at 2:09
AColoredReptile
1178
1178
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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
21.5k32457
21.5k32457
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
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