Partial derivative of x - is quotient rule necessary?
$begingroup$
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
$endgroup$
add a comment |
$begingroup$
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
$endgroup$
1
$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
$begingroup$
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
$endgroup$
Let
$$u(x,y)=frac{x}{x^2+y^2}$$
I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.
However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?
multivariable-calculus derivatives partial-derivative
multivariable-calculus derivatives partial-derivative
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
edited Jan 5 at 15:15
Martin Sleziak
44.9k10121273
44.9k10121273
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
asked Feb 1 '16 at 19:30
whatwhatwhatwhatwhatwhat
782821
782821
1
$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
1
$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13
1
1
$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
764821
$endgroup$
add a comment |
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1 Answer
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$begingroup$
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
764821
$endgroup$
add a comment |
$begingroup$
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
764821
$endgroup$
add a comment |
$begingroup$
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
764821
$endgroup$
When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.
Now, you can use the product rule if you choose; you just have to rewrite:
$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$
Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$
which you can simplify.
answered Feb 1 '16 at 20:05
IcemanIceman
764821
edited Feb 1 '16 at 20:38
answered Feb 1 '16 at 20:05
IcemanIceman
764821
answered Feb 1 '16 at 20:05
IcemanIceman
764821
answered Feb 1 '16 at 20:05
IcemanIceman
764821
764821
add a comment |
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$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11
$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13