If I have a functional $I[y] = \int{({y^\prime}^2- 1)^2}dx$ , since $f(x) = x^2$ is an increasing function for $ x > 0$, can I make the conclusion that $I[y]$ is extremized for at the same functions that extremise the functional
$J[y] = \int{({y^\prime}^2 - 1)}dx$ ?
No, it is altogether a different function, If the action is defined as $$I=\int L(y,y',x) \mathrm dx$$
this extremizes when $$\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}=0$$
So for the case where another action is defined as
$$J=\int [L(y,y',x)]^n \mathrm dx$$
gives $$\frac{\partial L^n}{\partial y}-\frac{d}{dx}\frac{\partial L^n}{\partial y'}=0$$
So this of-course doesnt give the same equations as before