Circular working out with partial derivatives The Next CEO of Stack OverflowGalilean...
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Circular working out with partial derivatives
The Next CEO of Stack OverflowGalilean transformation and differentiationA question about chain rule for partial derivativesCan you use the chain rule when only one partial derivative is continuous?How to show that this equation involving partial derivatives is true? (Change of variables)Changing variables: partial derivatives of a tensorFinding Second Order DerivativesFind $frac{partial^2f}{partial rpartial s}$ for $f(x(r,s),y(r,s))$.Partial derivatives vs. Total Derivatives for chain rule.chain rule , partial derivativeFactorising functions out of partial derivativesApply the chain rule with partial derivatives
$begingroup$
My question is the related to the example below:
Why don't we use chain rule when differentiating this:
Example: Suppose $x,y$ are functions of $u,v$ and $z = x^2+y$, where
begin{cases} x=e^u cos(v) \ y=e^u sin(v) end{cases}
Required to find: $$ frac{partial z}{partial x} $$
If I'm differentiating $z=x^2+y$ with respect to $ x $, my first impulse would be to say
$$ frac{partial z}{partial x} = 2x$$
But when I think about it, I don't understand why we don't use chain rule (when differentiating y) again because $ y $ and $x $ are related so isn't $y$ also a function of $ x$. $$ y = x tan(v) $$
So why isn't the answer as below?:
$$frac{partial z}{partial x} = 2x+ frac{dy}{dx} = 2x+tan(v) $$
Can someone please clarify for me when chain rule is required and when it is not and please explain what is wrong with the logic/reasoning above. Thanks!
multivariable-calculus partial-derivative chain-rule
asked Mar 18 at 2:30
user523384user523384
177
$endgroup$
|
show 3 more comments
$begingroup$
My question is the related to the example below:
Why don't we use chain rule when differentiating this:
Example: Suppose $x,y$ are functions of $u,v$ and $z = x^2+y$, where
begin{cases} x=e^u cos(v) \ y=e^u sin(v) end{cases}
Required to find: $$ frac{partial z}{partial x} $$
If I'm differentiating $z=x^2+y$ with respect to $ x $, my first impulse would be to say
$$ frac{partial z}{partial x} = 2x$$
But when I think about it, I don't understand why we don't use chain rule (when differentiating y) again because $ y $ and $x $ are related so isn't $y$ also a function of $ x$. $$ y = x tan(v) $$
So why isn't the answer as below?:
$$frac{partial z}{partial x} = 2x+ frac{dy}{dx} = 2x+tan(v) $$
Can someone please clarify for me when chain rule is required and when it is not and please explain what is wrong with the logic/reasoning above. Thanks!
multivariable-calculus partial-derivative chain-rule
asked Mar 18 at 2:30
user523384user523384
177
$endgroup$
$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05
|
show 3 more comments
$begingroup$
My question is the related to the example below:
Why don't we use chain rule when differentiating this:
Example: Suppose $x,y$ are functions of $u,v$ and $z = x^2+y$, where
begin{cases} x=e^u cos(v) \ y=e^u sin(v) end{cases}
Required to find: $$ frac{partial z}{partial x} $$
If I'm differentiating $z=x^2+y$ with respect to $ x $, my first impulse would be to say
$$ frac{partial z}{partial x} = 2x$$
But when I think about it, I don't understand why we don't use chain rule (when differentiating y) again because $ y $ and $x $ are related so isn't $y$ also a function of $ x$. $$ y = x tan(v) $$
So why isn't the answer as below?:
$$frac{partial z}{partial x} = 2x+ frac{dy}{dx} = 2x+tan(v) $$
Can someone please clarify for me when chain rule is required and when it is not and please explain what is wrong with the logic/reasoning above. Thanks!
multivariable-calculus partial-derivative chain-rule
asked Mar 18 at 2:30
user523384user523384
177
$endgroup$
My question is the related to the example below:
Why don't we use chain rule when differentiating this:
Example: Suppose $x,y$ are functions of $u,v$ and $z = x^2+y$, where
begin{cases} x=e^u cos(v) \ y=e^u sin(v) end{cases}
Required to find: $$ frac{partial z}{partial x} $$
If I'm differentiating $z=x^2+y$ with respect to $ x $, my first impulse would be to say
$$ frac{partial z}{partial x} = 2x$$
But when I think about it, I don't understand why we don't use chain rule (when differentiating y) again because $ y $ and $x $ are related so isn't $y$ also a function of $ x$. $$ y = x tan(v) $$
So why isn't the answer as below?:
$$frac{partial z}{partial x} = 2x+ frac{dy}{dx} = 2x+tan(v) $$
Can someone please clarify for me when chain rule is required and when it is not and please explain what is wrong with the logic/reasoning above. Thanks!
multivariable-calculus partial-derivative chain-rule
multivariable-calculus partial-derivative chain-rule
asked Mar 18 at 2:30
user523384user523384
177
asked Mar 18 at 2:30
user523384user523384
177
edited Mar 18 at 4:44
user523384
asked Mar 18 at 2:30
user523384user523384
177
asked Mar 18 at 2:30
user523384user523384
177
asked Mar 18 at 2:30
user523384user523384
177
177
$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05
|
show 3 more comments
$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05
$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05
|
show 3 more comments
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$begingroup$
Here $y$ is treated as constant when the derivative is taken with respect to $x.$
$endgroup$
– Sean Roberson
Mar 18 at 2:47
$begingroup$
I think you mean $dfrac{partial z}{partial x} = 2x$ ("my first impulse would be to say…"). Or do you indeed mean $dfrac{partial y}{partial x} = y$. That would be a weird conclusion, even if you thought $y$ depended on $x$ (for that would imply $y = ke^x$).
$endgroup$
– M. Vinay
Mar 18 at 2:51
$begingroup$
Yes I did, apologies edited
$endgroup$
– user523384
Mar 18 at 3:54
$begingroup$
@Sean Robertson but this would not be the case if $y$ was explicitly a function of $x$, which would require chain rule. Isn't It possible to write it this way by eliminating the parameters?
$endgroup$
– user523384
Mar 18 at 3:58
$begingroup$
You would use the chain rule if you wanted to calculate $frac{partial z}{partial u}$ or $frac{partial z}{partial v}$.
$endgroup$
– John Douma
Mar 18 at 4:05