Mixing
taking two times the derivative is squaring
$begingroup$
I don't understand why the following is true :
We have using the chain rule :
$$frac{partial}{partial x}=frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}$$
But then why :
$$frac{partial}{partial x} frac{partial}{partial x} = frac{partial^2}{partial x^2} = (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x})^2$$ ?
Normally it should be :
$$frac{partial^2}{partial x^2} = frac{partial}{partial x} (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}) = frac{partial}{partial x}frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial x}frac{partial}{partial theta}frac{partial theta}{partial x}$$
?
I really don't understand what is going on...
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
I don't understand why the following is true :
We have using the chain rule :
$$frac{partial}{partial x}=frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}$$
But then why :
$$frac{partial}{partial x} frac{partial}{partial x} = frac{partial^2}{partial x^2} = (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x})^2$$ ?
Normally it should be :
$$frac{partial^2}{partial x^2} = frac{partial}{partial x} (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}) = frac{partial}{partial x}frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial x}frac{partial}{partial theta}frac{partial theta}{partial x}$$
?
I really don't understand what is going on...
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49
add a comment |
$begingroup$
I don't understand why the following is true :
We have using the chain rule :
$$frac{partial}{partial x}=frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}$$
But then why :
$$frac{partial}{partial x} frac{partial}{partial x} = frac{partial^2}{partial x^2} = (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x})^2$$ ?
Normally it should be :
$$frac{partial^2}{partial x^2} = frac{partial}{partial x} (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}) = frac{partial}{partial x}frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial x}frac{partial}{partial theta}frac{partial theta}{partial x}$$
?
I really don't understand what is going on...
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
I don't understand why the following is true :
We have using the chain rule :
$$frac{partial}{partial x}=frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}$$
But then why :
$$frac{partial}{partial x} frac{partial}{partial x} = frac{partial^2}{partial x^2} = (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x})^2$$ ?
Normally it should be :
$$frac{partial^2}{partial x^2} = frac{partial}{partial x} (frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial theta}frac{partial theta}{partial x}) = frac{partial}{partial x}frac{partial}{partial r}frac{partial r}{partial x}+frac{partial}{partial x}frac{partial}{partial theta}frac{partial theta}{partial x}$$
?
I really don't understand what is going on...
Thank you !
real-analysis calculus multivariable-calculus partial-derivative
real-analysis calculus multivariable-calculus partial-derivative
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
asked Jan 27 at 16:42
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49
add a comment |
$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49
$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49
add a comment |
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$begingroup$
Are you sure these two aren't equal?
$endgroup$
– Yanko
Jan 27 at 16:44
$begingroup$
So now if you plug in the first line for $fracpartial{partial x}$ in the last ...
$endgroup$
– Hagen von Eitzen
Jan 27 at 16:45
$begingroup$
@Yanko I must say that I really don't know... In the second one I am really just applying the operator $frac{partial}{partial x}$ on the two elements inside the parenthesis, just because In one dimension $(u+v)' = u' + v'$ but then I don't know what to do with this expression and get the square...
$endgroup$
– dghkgfzyukz
Jan 27 at 16:47
$begingroup$
@HagenvonEitzen Got it ! Thank you. This is really strange because in one dimension taking the second derivative is not squaring. But ok ty I got it.
$endgroup$
– dghkgfzyukz
Jan 27 at 16:49