I am working through this example from Peter Olver's Application of Lie Groups to Differential Equations on page252 and I am stuck on the highlighted part.
everything else in the example I was able to do and I know what I am confused about is rather simple. Now, this is a question about how the Lagrangian operator and Euler Operator $E$ change under a change of variables. Now if we make a change of variable $x = \tilde{u} $ and $\tilde{x} = u$ then from what I know from a previous ODE's course I have that $u_x = (\tilde{u}_{\tilde{x}})^{-1}$ since $\frac{du}{dx} = \frac{d \tilde{x}}{d\tilde{u}} = (\frac{d\tilde{u}}{d\tilde{x}})^{-1}$ which checks out with the example and the change in the Lagrangian. However, it is the $u_{xx}$ that I am not getting. I computed the new Euler operator using the new Lagrangian and I got the highlighted part. However, I am assuming that you can find $u_{xx}$ using just the change of variables formula except I haven't been able to do it. Could somebody help me out? I know it's rather basic but I have been stuck on this for some time and would really like to move on.
If more background info is needed I can provide but I think for my question the top half of the example is not necessary since my question is really about a change of variable.
We know that $\frac d{dx}=\left(\frac{dx}{du}\right)^{-1}\frac d{du}$.
Let's apply this operator twice:
$$ \left(\frac{dx}{du}\right)^{-1}\frac d{du} \left[\left(\frac{dx}{du}\right)^{-1}\underbrace{\frac d{du}u}_{=1}\right] = \left(\frac{dx}{du}\right)^{-1}\color{red}{\frac d{du} \left(\frac{dx}{du}\right)^{-1}} = \left(\frac{dx}{du}\right)^{-1}\left(\color{red}{-\left(\frac{dx}{du}\right)^{-2}\frac d{du}\frac{dx}{du}}\right) \\= - \left(\frac{dx}{du}\right)^{-1} \left(\frac{dx}{du}\right)^{-2} \frac{d^2x}{du^2} = -x_u^{-3}x_{uu} $$
In other words, if you consider $dx/du = f(u)$, then $\frac d{du}f(u)^{-1} = -f(u)^{-2}f'(u)$