Partial Derivative of $G(t,x(t)) = e^{-pt} frac{u_x(x(t))}{u_x(x(0))}$
$begingroup$
I have given the following function for some constant $p$ and function $u(x(t))$
$$G(t,x(t)) = e^{-pt} frac{u_x(x(t))}{u_x(x(0))}$$
where $$ u_x(x(t)) = frac{partial u(x(t))}{partial x} $$
Now I want to calculate
$$frac{partial G(t,x(t)) }{partial x} $$
But I just cannot wrap my head around it. Better said I don't know how to start to write it down - any help would be appreciated!
derivatives partial-derivative
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
$endgroup$
add a comment |
$begingroup$
I have given the following function for some constant $p$ and function $u(x(t))$
$$G(t,x(t)) = e^{-pt} frac{u_x(x(t))}{u_x(x(0))}$$
where $$ u_x(x(t)) = frac{partial u(x(t))}{partial x} $$
Now I want to calculate
$$frac{partial G(t,x(t)) }{partial x} $$
But I just cannot wrap my head around it. Better said I don't know how to start to write it down - any help would be appreciated!
derivatives partial-derivative
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
$endgroup$
add a comment |
$begingroup$
I have given the following function for some constant $p$ and function $u(x(t))$
$$G(t,x(t)) = e^{-pt} frac{u_x(x(t))}{u_x(x(0))}$$
where $$ u_x(x(t)) = frac{partial u(x(t))}{partial x} $$
Now I want to calculate
$$frac{partial G(t,x(t)) }{partial x} $$
But I just cannot wrap my head around it. Better said I don't know how to start to write it down - any help would be appreciated!
derivatives partial-derivative
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
$endgroup$
I have given the following function for some constant $p$ and function $u(x(t))$
$$G(t,x(t)) = e^{-pt} frac{u_x(x(t))}{u_x(x(0))}$$
where $$ u_x(x(t)) = frac{partial u(x(t))}{partial x} $$
Now I want to calculate
$$frac{partial G(t,x(t)) }{partial x} $$
But I just cannot wrap my head around it. Better said I don't know how to start to write it down - any help would be appreciated!
derivatives partial-derivative
derivatives partial-derivative
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
asked Dec 16 '18 at 17:01
MethamortixMethamortix
306210
306210
add a comment |
add a comment |
2 Answers
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$begingroup$
So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} dfrac{u_{xx}(x(t))}{u_x(x(0))}.$
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
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$begingroup$
Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus
$$frac{partial G(t,x(t)) }{partial x} = e^{-pt} frac{u_{xx}(x(t))}{u_x(x(0))} $$
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} dfrac{u_{xx}(x(t))}{u_x(x(0))}.$
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
$endgroup$
add a comment |
$begingroup$
So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} dfrac{u_{xx}(x(t))}{u_x(x(0))}.$
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
$endgroup$
add a comment |
$begingroup$
So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} dfrac{u_{xx}(x(t))}{u_x(x(0))}.$
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
$endgroup$
So you are considering a function $G$ of the vector (presumably): $(t, x).$ So wherever $x(t)$ appears, change it to $x$ and remark that $u_x(x(0))$ is simply a constant, say $c,$ so your function $G$ takes the form $G(t, x) = c^{-1} e^{-pt} u'(x),$ where $u$ is a function of $x$ alone and hence, $u'$ makes sense. The partial derivative of $G$ would be simply $G_x(t, x) = c^{-1} e^{-pt} u''(x) = c^{-1} e^{-pt} u_{xx}(x).$ Hence, $G(t, x(t)) = e^{-pt} dfrac{u_{xx}(x(t))}{u_x(x(0))}.$
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
answered Dec 16 '18 at 17:41
Will M.Will M.
2,855315
2,855315
add a comment |
add a comment |
$begingroup$
Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus
$$frac{partial G(t,x(t)) }{partial x} = e^{-pt} frac{u_{xx}(x(t))}{u_x(x(0))} $$
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
$endgroup$
add a comment |
$begingroup$
Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus
$$frac{partial G(t,x(t)) }{partial x} = e^{-pt} frac{u_{xx}(x(t))}{u_x(x(0))} $$
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
$endgroup$
add a comment |
$begingroup$
Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus
$$frac{partial G(t,x(t)) }{partial x} = e^{-pt} frac{u_{xx}(x(t))}{u_x(x(0))} $$
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
$endgroup$
Ok my solution would be to treat $u_x(x(0))$ as a constant because $x(0)=1$ and thus
$$frac{partial G(t,x(t)) }{partial x} = e^{-pt} frac{u_{xx}(x(t))}{u_x(x(0))} $$
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
edited Dec 16 '18 at 18:12
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
answered Dec 16 '18 at 17:30
MethamortixMethamortix
306210
306210
add a comment |
add a comment |
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