Cancellation rules for partial derivatives
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3
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Is it possible to do some kind of simplifications on an expression like
$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$
So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.
I would say no, that's not allowed, but I'm just wondering.
Thank you.
partial-derivative
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up vote
3
down vote
favorite
Is it possible to do some kind of simplifications on an expression like
$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$
So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.
I would say no, that's not allowed, but I'm just wondering.
Thank you.
partial-derivative
I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is it possible to do some kind of simplifications on an expression like
$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$
So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.
I would say no, that's not allowed, but I'm just wondering.
Thank you.
partial-derivative
Is it possible to do some kind of simplifications on an expression like
$$
f : x, y to mathbb{R} \
frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}} = frac{partial f}{partial y}
$$
So that $frac{partial f}{partial y}$ in the denominator reduces $frac{partial^2 f}{partial y^2}$ to $frac{partial f}{partial y}$ like $frac{a^2}{a}=a$, without knowing the function $f$.
I would say no, that's not allowed, but I'm just wondering.
Thank you.
partial-derivative
partial-derivative
asked Nov 26 at 22:56
WolfgangP
1625
1625
I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00
add a comment |
I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00
I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00
add a comment |
4 Answers
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3
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accepted
$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$
just as
$$ frac{y^{primeprime}}{y^prime}ne y^prime $$
add a comment |
up vote
2
down vote
That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
$$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
add a comment |
up vote
1
down vote
No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.
add a comment |
up vote
0
down vote
No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$
just as
$$ frac{y^{primeprime}}{y^prime}ne y^prime $$
add a comment |
up vote
3
down vote
accepted
$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$
just as
$$ frac{y^{primeprime}}{y^prime}ne y^prime $$
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$
just as
$$ frac{y^{primeprime}}{y^prime}ne y^prime $$
$$frac{frac{partial^2 f}{partial y^2}}{frac{partial f}{partial y}}=frac{frac{partial}{partial y}left(frac{partial f}{partial y}right)}{frac{partial f}{partial y}}nefrac{partial f}{partial y}$$
just as
$$ frac{y^{primeprime}}{y^prime}ne y^prime $$
answered Nov 26 at 23:17
John Wayland Bales
13.8k21137
13.8k21137
add a comment |
add a comment |
up vote
2
down vote
That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
$$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
add a comment |
up vote
2
down vote
That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
$$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
add a comment |
up vote
2
down vote
up vote
2
down vote
That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
$$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$
That is indeed not allowed, even for single-variable derivatives. Take for example $f(x) = x^2$. Then
$$ frac{f''}{f'} = frac{2}{2x} = frac{1}{x} not= 2x = f' $$
answered Nov 26 at 23:00
MisterRiemann
5,0731623
5,0731623
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
add a comment |
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
1
1
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
Actually the case in the question is single-variable, as $x$ plays no role.
– Yves Daoust
Nov 26 at 23:01
1
1
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
@YvesDaoust Indeed. I just wanted to emphasize this because of the way that the question was asked.
– MisterRiemann
Nov 26 at 23:02
add a comment |
up vote
1
down vote
No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.
add a comment |
up vote
1
down vote
No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.
add a comment |
up vote
1
down vote
up vote
1
down vote
No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.
No. Remember the second derivative is just the derivative of the first derivative, so using $g$ for the first derivative, your equation is equivalent to $frac{partial g}{partial y}=g^2$, which of course can't always be true.
answered Nov 26 at 23:00
Y. Forman
11.4k423
11.4k423
add a comment |
add a comment |
up vote
0
down vote
No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.
add a comment |
up vote
0
down vote
No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.
add a comment |
up vote
0
down vote
up vote
0
down vote
No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.
No, usually these kinds of analogies don't work. Take for example $f(x,y)=y$.
answered Nov 26 at 23:01
Scientifica
6,13141332
6,13141332
add a comment |
add a comment |
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I think that the closest you have to it is the chain rule: $$ frac{partial f}{partial u} = frac{partial x}{partial u} frac{partial f}{partial x} $$
– rafa11111
Nov 26 at 22:59
No, a second derivative has nothing to do with the square of the first derivative. This quotient cannot be simplified.
– Yves Daoust
Nov 26 at 23:00