For an $O(N)$ symmetric function, is the 4-derivative simply equal to the derivative w.r.t the $N$-dimensional norm?

Question

I am working on an $O(4)$-symmetric instanton which has the Lagrangian: $$L = \frac{M^6}{4E^2T^2} \left[ \frac{1}{2}(\partial_\mu\Phi)^2 - \frac{1}{2} \Phi^2 + \frac{1}{2}\Phi^3 - \frac{\alpha}{8} \Phi^4\right]$$ The prefactor here can largely be ignored; it represents the different values of the terms in the effective potential. Crucially here the 4-derivative has been reparameterised and $\partial_\mu = \frac{\partial}{\partial X^\mu}$ via $X =Mx$ s.t. $$ \frac{\partial}{\partial X^\mu} = \frac{1}{M} \frac{\partial}{\partial x^\mu}$$ The prefactor doesn't affect the field equation which is: $$\frac{d^2 \Phi}{d \rho^2} + \frac{3}{\rho} \frac{d \Phi}{d \rho} = \Phi - \frac{3}{2}\Phi^2 - \frac{\alpha}{8} \Phi^3$$ where $\rho = \sqrt{x_\mu x^\mu }$ By numerically solving this 2nd equation, subject to boundary I have found a high order polynomial that approximates $\Phi$ well for a given value of $\alpha$ and I'd like to know how I can go from this to the action. Normally this would be a simple integral of the Lagrangian but with the sign of the potential reversed (as this process is in imaginary time). Doing this the prefactor pops out the front and I have the integral: $$S =2 \pi^2 \frac{M^6}{4E^2T^2}\int \rho^3 d\rho \left[ \frac{1}{2}(\partial_\mu\Phi)^2 - \frac{1}{2} \Phi^2 + \frac{1}{2}\Phi^3 - \frac{\alpha}{8} \Phi^4\right]$$ How can I turn $\partial_\mu\Phi$ into a derivative involving $\frac{d \Phi}{d \rho}$. If some factor of $M$ pops out is there a way to paramaterise this such that the prefactor exists only outside the integral?