Mixing
How to compute $u_x(sqrt {(x^2+y^2)})=$?
$begingroup$
This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.
complex-analysis derivatives pde
$endgroup$
add a comment |
$begingroup$
This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.
complex-analysis derivatives pde
$endgroup$
add a comment |
$begingroup$
This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.
complex-analysis derivatives pde
$endgroup$
This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(sqrt {(x^2+y^2)})=$?
It is given in the question that linked above
that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false
What is the correct way to solve it?
Any help.
complex-analysis derivatives pde
complex-analysis derivatives pde
asked Jan 20 at 5:19
user596656
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1 Answer
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$begingroup$
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
Then
$$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
=left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
=2x+frac{x}{sqrt{x^2+y^2}}.$$
I don't see any problem....
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
Then
$$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
=left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
=2x+frac{x}{sqrt{x^2+y^2}}.$$
I don't see any problem....
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
add a comment |
$begingroup$
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
Then
$$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
=left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
=2x+frac{x}{sqrt{x^2+y^2}}.$$
I don't see any problem....
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
add a comment |
$begingroup$
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
Then
$$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
=left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
=2x+frac{x}{sqrt{x^2+y^2}}.$$
I don't see any problem....
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
Then
$$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
=left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
=2x+frac{x}{sqrt{x^2+y^2}}.$$
I don't see any problem....
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 20 at 6:02
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
add a comment |
add a comment |
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