Partial derivative of function of inverse function
I have got a probelm with the following task:
$frac{partial }{partial x}f(f^{-1}(x,t),tau)$, where $finmathscr{C}^{infty}(mathbb{R^2})$. My attemp is
$frac{partial }{partial x}f(f^{-1}(x,t),tau)=frac{partial }{partial x}f(f^{-1}(x,t),tau)frac{partial }{partial x}f^{-1}(x,t)$, but I am not sure, if it holds. Many thanks for any hints.
functions partial-derivative inverse-function
asked Nov 21 '18 at 20:52
lojdmoj
847
add a comment |
I have got a probelm with the following task:
$frac{partial }{partial x}f(f^{-1}(x,t),tau)$, where $finmathscr{C}^{infty}(mathbb{R^2})$. My attemp is
$frac{partial }{partial x}f(f^{-1}(x,t),tau)=frac{partial }{partial x}f(f^{-1}(x,t),tau)frac{partial }{partial x}f^{-1}(x,t)$, but I am not sure, if it holds. Many thanks for any hints.
functions partial-derivative inverse-function
asked Nov 21 '18 at 20:52
lojdmoj
847
add a comment |
I have got a probelm with the following task:
$frac{partial }{partial x}f(f^{-1}(x,t),tau)$, where $finmathscr{C}^{infty}(mathbb{R^2})$. My attemp is
$frac{partial }{partial x}f(f^{-1}(x,t),tau)=frac{partial }{partial x}f(f^{-1}(x,t),tau)frac{partial }{partial x}f^{-1}(x,t)$, but I am not sure, if it holds. Many thanks for any hints.
functions partial-derivative inverse-function
asked Nov 21 '18 at 20:52
lojdmoj
847
I have got a probelm with the following task:
$frac{partial }{partial x}f(f^{-1}(x,t),tau)$, where $finmathscr{C}^{infty}(mathbb{R^2})$. My attemp is
$frac{partial }{partial x}f(f^{-1}(x,t),tau)=frac{partial }{partial x}f(f^{-1}(x,t),tau)frac{partial }{partial x}f^{-1}(x,t)$, but I am not sure, if it holds. Many thanks for any hints.
functions partial-derivative inverse-function
functions partial-derivative inverse-function
asked Nov 21 '18 at 20:52
lojdmoj
847
asked Nov 21 '18 at 20:52
lojdmoj
847
asked Nov 21 '18 at 20:52
lojdmoj
847
asked Nov 21 '18 at 20:52
lojdmoj
847
asked Nov 21 '18 at 20:52
lojdmoj
847
847
add a comment |
add a comment |
1 Answer
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Unless I have misinterpreted your notation, the problem does not make sense. If $f$ is a function with domain $mathbb{R}^2$, and $f^{-1}$ is supposed to be its inverse, then the codomain of $f^{-1}$ must be $mathbb{R}^2$. But then it does not make sense to write
$$f(f^{-1}(x,t),tau).$$
To elaborate: $f^{-1}(x,t)inmathbb{R}^2$ and so ($f^{-1}(x,t),tau)inmathbb{R}^2timesmathbb{R}$. This cannot be an argument of $f$ because $f$ only takes elements of $mathbb{R}^2$ as arguments.
answered Nov 21 '18 at 21:43
smcc
4,297517
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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oldest
votes
Unless I have misinterpreted your notation, the problem does not make sense. If $f$ is a function with domain $mathbb{R}^2$, and $f^{-1}$ is supposed to be its inverse, then the codomain of $f^{-1}$ must be $mathbb{R}^2$. But then it does not make sense to write
$$f(f^{-1}(x,t),tau).$$
To elaborate: $f^{-1}(x,t)inmathbb{R}^2$ and so ($f^{-1}(x,t),tau)inmathbb{R}^2timesmathbb{R}$. This cannot be an argument of $f$ because $f$ only takes elements of $mathbb{R}^2$ as arguments.
answered Nov 21 '18 at 21:43
smcc
4,297517
add a comment |
Unless I have misinterpreted your notation, the problem does not make sense. If $f$ is a function with domain $mathbb{R}^2$, and $f^{-1}$ is supposed to be its inverse, then the codomain of $f^{-1}$ must be $mathbb{R}^2$. But then it does not make sense to write
$$f(f^{-1}(x,t),tau).$$
To elaborate: $f^{-1}(x,t)inmathbb{R}^2$ and so ($f^{-1}(x,t),tau)inmathbb{R}^2timesmathbb{R}$. This cannot be an argument of $f$ because $f$ only takes elements of $mathbb{R}^2$ as arguments.
answered Nov 21 '18 at 21:43
smcc
4,297517
add a comment |
Unless I have misinterpreted your notation, the problem does not make sense. If $f$ is a function with domain $mathbb{R}^2$, and $f^{-1}$ is supposed to be its inverse, then the codomain of $f^{-1}$ must be $mathbb{R}^2$. But then it does not make sense to write
$$f(f^{-1}(x,t),tau).$$
To elaborate: $f^{-1}(x,t)inmathbb{R}^2$ and so ($f^{-1}(x,t),tau)inmathbb{R}^2timesmathbb{R}$. This cannot be an argument of $f$ because $f$ only takes elements of $mathbb{R}^2$ as arguments.
answered Nov 21 '18 at 21:43
smcc
4,297517
Unless I have misinterpreted your notation, the problem does not make sense. If $f$ is a function with domain $mathbb{R}^2$, and $f^{-1}$ is supposed to be its inverse, then the codomain of $f^{-1}$ must be $mathbb{R}^2$. But then it does not make sense to write
$$f(f^{-1}(x,t),tau).$$
To elaborate: $f^{-1}(x,t)inmathbb{R}^2$ and so ($f^{-1}(x,t),tau)inmathbb{R}^2timesmathbb{R}$. This cannot be an argument of $f$ because $f$ only takes elements of $mathbb{R}^2$ as arguments.
answered Nov 21 '18 at 21:43
smcc
4,297517
answered Nov 21 '18 at 21:43
smcc
4,297517
answered Nov 21 '18 at 21:43
smcc
4,297517
answered Nov 21 '18 at 21:43
smcc
4,297517
4,297517
add a comment |
add a comment |
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