Why is $\frac{\partial }{\partial \phi}(\frac{\partial^2 \phi}{\partial x^2})$ = 0?

I'm happy to read an article, but I don't know what keywords to search on.

Background: Apologies if this is a math question. The setup is physics and not sure if it's correct.

A text lists an example as $L=\frac{1}{2}(\partial _{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}$ and then asks for $\frac{\partial L}{\partial \phi}$. It lists the answer as $-m^{2}\phi$. The left-hand side of the equation, the d'Alembertian I believe, must have a derivative of 0, giving us the answer as the derivative of the right-hand side ($\frac{\partial }{\partial \phi}(\frac{1}{2}m^{2}\phi^{2}=-m^{2}\phi)$).

Why is the derivative of the d'Alembertian = 0? Expanding, it seems $\frac{\partial }{\partial \phi}\frac{1}{2}(\frac{\partial^2 \phi}{\partial x_{0}^2}-\frac{\partial^2 \phi}{\partial x_{1}^2}-\frac{\partial^2 \phi}{\partial x_{2}^2}-\frac{\partial^2 \phi}{\partial x_{3}^2})=0$.

Of course, the partial of a constant, or a variable that is not being differentiated with respect to, is 0. However, it seems to me all the terms depend on $\phi$, which is what we are differentiating with respect to. It probably has something to do with the chain rule of a first derivative applied to a second derivative, but I'm not clear how the chain rule works in that case. Further, I don't know what to search on.

Any help much appreciated! Thanks in advance!

Why is $\frac{\partial }{\partial \phi}(\frac{\partial^2 \phi}{\partial x^2})$ = 0?

I'm happy to read an article, but I don't know what keywords to search on.

Background: Apologies if this is a math question. The setup is physics and not sure if it's correct.

A text lists an example as $L=\frac{1}{2}(\partial _{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}$ and then asks for $\frac{\partial L}{\partial \phi}$. It lists the answer as $-m^{2}\phi$. The left-hand side of the equation, the d'Alembertian I believe, must have a derivative of 0, giving us the answer as the derivative of the right-hand side ($\frac{\partial }{\partial \phi}(\frac{1}{2}m^{2}\phi^{2}=-m^{2}\phi)$).

Why is the derivative of the d'Alembertian = 0? Expanding, it seems $\frac{\partial }{\partial \phi}\frac{1}{2}(\frac{\partial^2 \phi}{\partial x_{0}^2}-\frac{\partial^2 \phi}{\partial x_{1}^2}-\frac{\partial^2 \phi}{\partial x_{2}^2}-\frac{\partial^2 \phi}{\partial x_{3}^2})=0$.

Of course, the partial of a constant, or a variable that is not being differentiated with respect to, is 0. However, it seems to me all the terms depend on $\phi$, which is what we are differentiating with respect to. It probably has something to do with the chain rule of a first derivative applied to a second derivative, but I'm not clear how the chain rule works in that case. Further, I don't know what to search on.

Any help much appreciated!

Why is $\frac{\partial }{\partial \phi}(\frac{\partial^2 \phi}{\partial x^2})$ = 0?

I'm happy to read an article, but I don't know what keywords to search on.

Background: Apologies if this is a math question. The setup is physics and not sure if it's correct.

A text lists an example as $L=\frac{1}{2}(\partial _{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}$ and then asks for $\frac{\partial L}{\partial \phi}$. It lists the answer as $-m^{2}\phi$. The left-hand side of the equation, the d'Alembertian I believe, must have a derivative of 0, giving us the answer as the derivative of the right-hand side ($\frac{\partial }{\partial \phi}(\frac{1}{2}m^{2}\phi^{2}=-m^{2}\phi)$).

Why is the derivative of the d'Alembertian = 0? Expanding, it seems $\frac{\partial }{\partial \phi}\frac{1}{2}(\frac{\partial^2 \phi}{\partial x_{0}^2}-\frac{\partial^2 \phi}{\partial x_{1}^2}-\frac{\partial^2 \phi}{\partial x_{2}^2}-\frac{\partial^2 \phi}{\partial x_{3}^2})=0$ Of course, the partial of a constant, or a variable that is not being differentiated with respect to, is 0. However, it seems to me all the terms rely on phi, which is what we are differentiating with respect to. It probably has something to do with the chain rule of a first derivative applied to a second derivative, but I'm not clear how the chain rule works in that case. Further, I don't know what to search on.

Any help much appreciated! Thanks in advance!

Why is $\frac{\partial }{\partial \phi}(\frac{\partial^2 \phi}{\partial x^2})$ = 0?

I'm happy to read an article, but I don't know what keywords to search on.

Background: Apologies if this is a math question. The setup is physics and not sure if it's correct.

A text lists an example as $L=\frac{1}{2}(\partial _{\mu}\phi)^{2}-\frac{1}{2}m^{2}\phi^{2}$ and then asks for $\frac{\partial L}{\partial \phi}$. It lists the answer as $-m^{2}\phi$. The left-hand side of the equation, the d'Alembertian I believe, must have a derivative of 0, giving us the answer as the derivative of the right-hand side ($\frac{\partial }{\partial \phi}(\frac{1}{2}m^{2}\phi^{2}=-m^{2}\phi)$).

Why is the derivative of the d'Alembertian = 0? Expanding, it seems $\frac{\partial }{\partial \phi}\frac{1}{2}(\frac{\partial^2 \phi}{\partial x_{0}^2}-\frac{\partial^2 \phi}{\partial x_{1}^2}-\frac{\partial^2 \phi}{\partial x_{2}^2}-\frac{\partial^2 \phi}{\partial x_{3}^2})=0$.

Of course, the partial of a constant, or a variable that is not being differentiated with respect to, is 0. However, it seems to me all the terms depend on $\phi$, which is what we are differentiating with respect to. It probably has something to do with the chain rule of a first derivative applied to a second derivative, but I'm not clear how the chain rule works in that case. Further, I don't know what to search on.

Any help much appreciated! Thanks in advance!