Proof regarding homogeneous functions of degree $n$
$begingroup$
If $f$ is homogeneous of degree $n$, show that $f_{x}(tx,ty) = t^{n-1}f_{x}(x,y)$.
My attempt at a solution:
Let $u = tx$ and $v = ty$.$$frac{partial f(u,v)}{partial x} = frac{partial f(u,v)}{partial u}frac{partial u}{partial x} + frac{partial f(u,v)}{partial v}frac{partial v}{partial x}.$$ But $frac{partial u}{partial x} = t$ and $frac{partial v}{partial x} = 0.$ Hence,
$$tfrac{partial f(u,v)}{partial u} = t^{n}f_{x}(x,y)$$ or $$frac{partial f(u,v)}{partial u} = t^{n-1}f_{x}(x,y).$$ How do I get from the term on the left hand side to $frac{partial f(u,v)}{partial x}$? Thank you in advance.
calculus multivariable-calculus derivatives
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
$endgroup$
add a comment |
$begingroup$
If $f$ is homogeneous of degree $n$, show that $f_{x}(tx,ty) = t^{n-1}f_{x}(x,y)$.
My attempt at a solution:
Let $u = tx$ and $v = ty$.$$frac{partial f(u,v)}{partial x} = frac{partial f(u,v)}{partial u}frac{partial u}{partial x} + frac{partial f(u,v)}{partial v}frac{partial v}{partial x}.$$ But $frac{partial u}{partial x} = t$ and $frac{partial v}{partial x} = 0.$ Hence,
$$tfrac{partial f(u,v)}{partial u} = t^{n}f_{x}(x,y)$$ or $$frac{partial f(u,v)}{partial u} = t^{n-1}f_{x}(x,y).$$ How do I get from the term on the left hand side to $frac{partial f(u,v)}{partial x}$? Thank you in advance.
calculus multivariable-calculus derivatives
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
$endgroup$
$begingroup$
math.stackexchange.com/questions/977677/…
$endgroup$
– Rhaldryn
Dec 23 '18 at 5:08
add a comment |
$begingroup$
If $f$ is homogeneous of degree $n$, show that $f_{x}(tx,ty) = t^{n-1}f_{x}(x,y)$.
My attempt at a solution:
Let $u = tx$ and $v = ty$.$$frac{partial f(u,v)}{partial x} = frac{partial f(u,v)}{partial u}frac{partial u}{partial x} + frac{partial f(u,v)}{partial v}frac{partial v}{partial x}.$$ But $frac{partial u}{partial x} = t$ and $frac{partial v}{partial x} = 0.$ Hence,
$$tfrac{partial f(u,v)}{partial u} = t^{n}f_{x}(x,y)$$ or $$frac{partial f(u,v)}{partial u} = t^{n-1}f_{x}(x,y).$$ How do I get from the term on the left hand side to $frac{partial f(u,v)}{partial x}$? Thank you in advance.
calculus multivariable-calculus derivatives
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
$endgroup$
If $f$ is homogeneous of degree $n$, show that $f_{x}(tx,ty) = t^{n-1}f_{x}(x,y)$.
My attempt at a solution:
Let $u = tx$ and $v = ty$.$$frac{partial f(u,v)}{partial x} = frac{partial f(u,v)}{partial u}frac{partial u}{partial x} + frac{partial f(u,v)}{partial v}frac{partial v}{partial x}.$$ But $frac{partial u}{partial x} = t$ and $frac{partial v}{partial x} = 0.$ Hence,
$$tfrac{partial f(u,v)}{partial u} = t^{n}f_{x}(x,y)$$ or $$frac{partial f(u,v)}{partial u} = t^{n-1}f_{x}(x,y).$$ How do I get from the term on the left hand side to $frac{partial f(u,v)}{partial x}$? Thank you in advance.
calculus multivariable-calculus derivatives
calculus multivariable-calculus derivatives
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
edited Nov 8 '15 at 17:54
Davide Giraudo
126k16150261
126k16150261
asked Oct 31 '15 at 20:30
saadsaad
1507
asked Oct 31 '15 at 20:30
saadsaad
1507
asked Oct 31 '15 at 20:30
saadsaad
1507
1507
$begingroup$
math.stackexchange.com/questions/977677/…
$endgroup$
– Rhaldryn
Dec 23 '18 at 5:08
add a comment |
$begingroup$
math.stackexchange.com/questions/977677/…
$endgroup$
– Rhaldryn
Dec 23 '18 at 5:08
$begingroup$
math.stackexchange.com/questions/977677/…
$endgroup$
– Rhaldryn
Dec 23 '18 at 5:08
$begingroup$
math.stackexchange.com/questions/977677/…
$endgroup$
– Rhaldryn
Dec 23 '18 at 5:08
add a comment |
1 Answer
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$begingroup$
$frac {partial f(u,v)}{partial u}$ means partial derivative with respect to the first variable, which is also denoted by $f_x(u,v)$. Here the subscript $x$ denotes that the partial derivative is with respect to the first variable $u=tx$ and not $x$ itself. So in our case $frac {partial f(u,v)}{partial u}=f_x(u,v)=f_x(tx,ty)=t^{n-1}f_x(x,y)$ and you are already done.
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
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1 Answer
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1 Answer
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$begingroup$
$frac {partial f(u,v)}{partial u}$ means partial derivative with respect to the first variable, which is also denoted by $f_x(u,v)$. Here the subscript $x$ denotes that the partial derivative is with respect to the first variable $u=tx$ and not $x$ itself. So in our case $frac {partial f(u,v)}{partial u}=f_x(u,v)=f_x(tx,ty)=t^{n-1}f_x(x,y)$ and you are already done.
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
$endgroup$
add a comment |
$begingroup$
$frac {partial f(u,v)}{partial u}$ means partial derivative with respect to the first variable, which is also denoted by $f_x(u,v)$. Here the subscript $x$ denotes that the partial derivative is with respect to the first variable $u=tx$ and not $x$ itself. So in our case $frac {partial f(u,v)}{partial u}=f_x(u,v)=f_x(tx,ty)=t^{n-1}f_x(x,y)$ and you are already done.
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
$endgroup$
add a comment |
$begingroup$
$frac {partial f(u,v)}{partial u}$ means partial derivative with respect to the first variable, which is also denoted by $f_x(u,v)$. Here the subscript $x$ denotes that the partial derivative is with respect to the first variable $u=tx$ and not $x$ itself. So in our case $frac {partial f(u,v)}{partial u}=f_x(u,v)=f_x(tx,ty)=t^{n-1}f_x(x,y)$ and you are already done.
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
$endgroup$
$frac {partial f(u,v)}{partial u}$ means partial derivative with respect to the first variable, which is also denoted by $f_x(u,v)$. Here the subscript $x$ denotes that the partial derivative is with respect to the first variable $u=tx$ and not $x$ itself. So in our case $frac {partial f(u,v)}{partial u}=f_x(u,v)=f_x(tx,ty)=t^{n-1}f_x(x,y)$ and you are already done.
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
answered Dec 23 '18 at 5:03
RhaldrynRhaldryn
361414
361414
add a comment |
add a comment |
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– Rhaldryn
Dec 23 '18 at 5:08