$frac{partial{z}}{partial{y}}$ if $z$ is given implicitly
Suppose $z$ is given implicitly as:
$$e^z-x^2y-y^2z = 0$$
Find $$frac{partial{z}}{partial{y}}$$.
I let $F(x,y) = e^z-x^2y-y^2z$. Then,
$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$
$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$
Therefore,
$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$
However, the exercise cites the answer as
$$frac{x^2+2yz}{e^z-y^2}$$
For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.
multivariable-calculus chain-rule
asked Dec 10 '18 at 5:09
Art
3127
add a comment |
Suppose $z$ is given implicitly as:
$$e^z-x^2y-y^2z = 0$$
Find $$frac{partial{z}}{partial{y}}$$.
I let $F(x,y) = e^z-x^2y-y^2z$. Then,
$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$
$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$
Therefore,
$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$
However, the exercise cites the answer as
$$frac{x^2+2yz}{e^z-y^2}$$
For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.
multivariable-calculus chain-rule
asked Dec 10 '18 at 5:09
Art
3127
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21
add a comment |
Suppose $z$ is given implicitly as:
$$e^z-x^2y-y^2z = 0$$
Find $$frac{partial{z}}{partial{y}}$$.
I let $F(x,y) = e^z-x^2y-y^2z$. Then,
$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$
$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$
Therefore,
$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$
However, the exercise cites the answer as
$$frac{x^2+2yz}{e^z-y^2}$$
For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.
multivariable-calculus chain-rule
asked Dec 10 '18 at 5:09
Art
3127
Suppose $z$ is given implicitly as:
$$e^z-x^2y-y^2z = 0$$
Find $$frac{partial{z}}{partial{y}}$$.
I let $F(x,y) = e^z-x^2y-y^2z$. Then,
$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$
$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$
Therefore,
$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$
However, the exercise cites the answer as
$$frac{x^2+2yz}{e^z-y^2}$$
For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.
multivariable-calculus chain-rule
multivariable-calculus chain-rule
asked Dec 10 '18 at 5:09
Art
3127
asked Dec 10 '18 at 5:09
Art
3127
asked Dec 10 '18 at 5:09
Art
3127
asked Dec 10 '18 at 5:09
Art
3127
asked Dec 10 '18 at 5:09
Art
3127
3127
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21
add a comment |
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21
add a comment |
2 Answers
2
active
oldest
votes
Looking through my copy of Stewart's calculus we have
$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$
and similarly
$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$
So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.
answered Dec 10 '18 at 5:43
carsandpulsars
34117
add a comment |
Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.
$e^z-x^2y-y^2z = 0$
$z_ye^z-x^2-2yz-y^2z_y=0$
$z_y(e^z-y^2)=x^2+2yz$
Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$
Added
This proves the minus sign in the formula you used.
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Looking through my copy of Stewart's calculus we have
$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$
and similarly
$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$
So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.
answered Dec 10 '18 at 5:43
carsandpulsars
34117
add a comment |
Looking through my copy of Stewart's calculus we have
$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$
and similarly
$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$
So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.
answered Dec 10 '18 at 5:43
carsandpulsars
34117
add a comment |
Looking through my copy of Stewart's calculus we have
$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$
and similarly
$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$
So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.
answered Dec 10 '18 at 5:43
carsandpulsars
34117
Looking through my copy of Stewart's calculus we have
$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$
and similarly
$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$
So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.
answered Dec 10 '18 at 5:43
carsandpulsars
34117
answered Dec 10 '18 at 5:43
carsandpulsars
34117
answered Dec 10 '18 at 5:43
carsandpulsars
34117
answered Dec 10 '18 at 5:43
carsandpulsars
34117
34117
add a comment |
add a comment |
Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.
$e^z-x^2y-y^2z = 0$
$z_ye^z-x^2-2yz-y^2z_y=0$
$z_y(e^z-y^2)=x^2+2yz$
Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$
Added
This proves the minus sign in the formula you used.
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
add a comment |
Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.
$e^z-x^2y-y^2z = 0$
$z_ye^z-x^2-2yz-y^2z_y=0$
$z_y(e^z-y^2)=x^2+2yz$
Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$
Added
This proves the minus sign in the formula you used.
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
add a comment |
Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.
$e^z-x^2y-y^2z = 0$
$z_ye^z-x^2-2yz-y^2z_y=0$
$z_y(e^z-y^2)=x^2+2yz$
Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$
Added
This proves the minus sign in the formula you used.
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.
$e^z-x^2y-y^2z = 0$
$z_ye^z-x^2-2yz-y^2z_y=0$
$z_y(e^z-y^2)=x^2+2yz$
Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$
Added
This proves the minus sign in the formula you used.
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
answered Dec 10 '18 at 5:44
Rafa Budría
5,5551825
5,5551825
add a comment |
add a comment |
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I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
– Dair
Dec 10 '18 at 5:17
I think so too. But, is there some reasoning as to why the negative just cancels out?
– Art
Dec 10 '18 at 5:21