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I have the function $f(x,y,y')= \sqrt{1+y'(x)^2}$ from which I want to calculate $\frac{\partial f}{\partial y}$ as well as $\frac{\partial f}{\partial y'}$ and $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$.

I tried using the SymbolicRing of Sagemath with

 x = SR.var('x')
 y = function('y')
 f = sqrt(1+y(x).diff(x,1)^2)

which only allows me to calculate $\frac{df}{dx}$. So I tried with the following sample

 x,y = SR.var('x,y')
 y = function('y')(x)
 y1 = y.diff(x,1)
 f(x,y,y1) = sqrt(1+y1^2)
 fy1 = f.diff(y1)
 fy = f.diff(y)
 fxy1 = fy1.diff(x)

which works for calculating $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial y'}$, but not for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$. Because fxy1 returns 0 when it should return $\frac{y''}{\sqrt{1+y'(x)^2}^{3/2}}$.

I'd really appreciate any idea. Thanks in advance!

I have the function $f(x,y,y')= \sqrt{1+y'(x)^2}$ from which I want to calculate $\frac{\partial f}{\partial y}$ as well as $\frac{\partial f}{\partial y'}$ and $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$.

I tried using the SymbolicRing of Sagemath with

 x = SR.var('x')
 y = function('y')
 f = sqrt(1+y(x).diff(x,1)^2)

which only allows me to calculate $\frac{df}{dx}$. So I tried with the following sample

 x,y = SR.var('x,y')
 y = function('y')(x)
 y1 = y.diff(x,1)
 f(x,y,y1) = sqrt(1+y1^2)
 fy1 = f.diff(y1)
 fy = f.diff(y)
 fxy1 dxfy1 = fy1.diff(x)

which works for calculating $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial y'}$, but not for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$. Because fxy1 dxfy1 which stands for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$ returns 0 when it should return $\frac{y''}{\sqrt{1+y'(x)^2}^{3/2}}$.

I'd really appreciate any idea. Thanks in advance!

I have the function $f(x,y,y')= \sqrt{1+y'(x)^2}$ from which I want to calculate $\frac{\partial f}{\partial y}$ as well as $\frac{\partial f}{\partial y'}$ and $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$.

I tried using the SymbolicRing of Sagemath with

 x = SR.var('x')
 y = function('y')
 f = sqrt(1+y(x).diff(x,1)^2)

which only allows me to calculate $\frac{df}{dx}$. So I tried with the following sample

 x,y = SR.var('x,y')
 y = function('y')(x)
 y1 = y.diff(x,1)
 f(x,y,y1) = sqrt(1+y1^2)
 fy1 = f.diff(y1)
 fy = f.diff(y)
 dxfy1 = fy1.diff(x)

which works for calculating $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial y'}$, but not for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$. Because dxfy1 which stands for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$ returns 0 when it should return $\frac{y''}{\sqrt{1+y'(x)^2}^{3/2}}$.

Another unseccesful try would be

x,y = SR.var('x,y')
y = function('y')(x)
y1 = y.diff(x,1)
f(x,y,y1) = sqrt(1+y.diff(x)^2)
fy1(x,y,y1) = f.diff(y1); fy1
fy(x,y,y1) = f.diff(y); fy
fy1.diff(x)

I'd really appreciate any idea. Thanks in advance!

I have the function $f(x,y,y')= \sqrt{1+y'(x)^2}$ from which I want to calculate $\frac{\partial f}{\partial y}$ as well as $\frac{\partial f}{\partial y'}$ and $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$.

I tried using the SymbolicRing of Sagemath with

 x = SR.var('x')
 y = function('y')
 f = sqrt(1+y(x).diff(x,1)^2)

which only allows me to calculate $\frac{df}{dx}$. So I tried with the following sample

 x,y = SR.var('x,y')
 y = function('y')(x)
 y1 = y.diff(x,1)
 f(x,y,y1) = sqrt(1+y1^2)
 fy1 = f.diff(y1)
 fy = f.diff(y)
 dxfy1 = fy1.diff(x)

which works for calculating $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial y'}$, but not for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$. Because dxfy1 which stands for $\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)$ returns 0 when it should return $\frac{y''}{\sqrt{1+y'(x)^2}^{3/2}}$.

Another unseccesful try would be

x,y = SR.var('x,y')
y = function('y')(x)
y1 = y.diff(x,1)
y1=function('y1')(x)
f(x,y,y1) = sqrt(1+y.diff(x)^2)
fy1(x,y,y1) = f.diff(y1); fy1
fy(x,y,y1) = f.diff(y); fy
fy1.diff(x)

I'd really appreciate any idea. Thanks in advance!