Partial Derivative chain rule help. “Y is temporarily held constant” [closed]
up vote
-1
down vote
favorite
Can someone explain to me how they know the partial of Y with regards to X is 0. Why is this the case and when would it be different? (Refer to Image example)
[1][1]: https://i.stack.imgur.com/7NtwM.png
multivariable-calculus partial-derivative
asked Nov 15 at 23:27
Drake Wilde
11
closed as unclear what you're asking by TheGeekGreek, Tom-Tom, José Carlos Santos, Kelvin Lois, user10354138 Nov 16 at 14:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
Can someone explain to me how they know the partial of Y with regards to X is 0. Why is this the case and when would it be different? (Refer to Image example)
[1][1]: https://i.stack.imgur.com/7NtwM.png
multivariable-calculus partial-derivative
asked Nov 15 at 23:27
Drake Wilde
11
closed as unclear what you're asking by TheGeekGreek, Tom-Tom, José Carlos Santos, Kelvin Lois, user10354138 Nov 16 at 14:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Can someone explain to me how they know the partial of Y with regards to X is 0. Why is this the case and when would it be different? (Refer to Image example)
[1][1]: https://i.stack.imgur.com/7NtwM.png
multivariable-calculus partial-derivative
asked Nov 15 at 23:27
Drake Wilde
11
Can someone explain to me how they know the partial of Y with regards to X is 0. Why is this the case and when would it be different? (Refer to Image example)
[1][1]: https://i.stack.imgur.com/7NtwM.png
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
asked Nov 15 at 23:27
Drake Wilde
11
asked Nov 15 at 23:27
Drake Wilde
11
edited Nov 20 at 7:56
asked Nov 15 at 23:27
Drake Wilde
11
asked Nov 15 at 23:27
Drake Wilde
11
asked Nov 15 at 23:27
Drake Wilde
11
11
closed as unclear what you're asking by TheGeekGreek, Tom-Tom, José Carlos Santos, Kelvin Lois, user10354138 Nov 16 at 14:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by TheGeekGreek, Tom-Tom, José Carlos Santos, Kelvin Lois, user10354138 Nov 16 at 14:01
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53
add a comment |
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
The explanation is bad. $x$ and $y$ are independent variables, i.e. changing one does not necessarily change the other. Thus, the change in $y$ with respect to a change in $x$ is $0$.
– AlexanderJ93
Nov 15 at 23:32
I agree with @AlexanderJ93. I think the written explanation is quite lousy. When it's written $f(x,y,z)$ it should be read as $f(x,y,h(x,y))$. The chain rule that is, I think, much clearer in this case.
– Tom-Tom
Nov 16 at 8:53