We study the one dimensional heat equation
$$u_{t}(x, t) = u_{xx}(x, t), \quad x \in \mathbb{R}, t> 0.$$
I our textbook, in order to prove that the solution is a function of one variable, they define a function
$$v(x, t) = u(cx, c^2t)$$
for a $c > 0$ and claims
$$v_t = c^2 u_t \quad \text{and} \quad v_{xx} = c^2 u_{xx}.$$
This is probably completely trivial, but I fail to see how they have applied the chain rule here. Here is my reasoning. If we let $\phi_1 = cx$ and $\phi_2 = c^2 t$, we can apply the chain rule to find $v_t$ as follows:
$$v_t = \frac{\partial u}{\partial \phi_1}\frac{\partial \phi_1}{\partial t} + \frac{\partial u}{\partial \phi_2}\frac{\partial \phi_2}{\partial t} = c^2\frac{\partial u}{\partial \phi_2}.$$
What I don't quite understand is how $\frac{\partial u}{\partial \phi_2}$, the partial with respect to $\phi_2 = c^2 t$, is equal to $u_t = \frac{\partial u}{\partial t}$. What am I missing?
Let us use the following notation. Suppose $u(X, T)$ solves the heat equation, i.e. \begin{align} u_{T}(a, b)= u_{XX}(a, b). \end{align} for any $(a, b) \in \mathbb{R}\times \mathbb{R}$.
Now, consider \begin{align} v(x, t) = u(cx, c^2t) \end{align} i.e. set $X = cx$ and $T = c^2t$. Then we see that \begin{align} \frac{\partial v}{\partial t} = \frac{\partial u}{\partial T}\frac{\partial T}{\partial t} = c^2u_T(cx, c^2t) \end{align} and \begin{align} \frac{\partial v}{\partial x} = \frac{\partial u}{\partial X}\frac{\partial X}{\partial x} = cu_X(cx, c^2t) \end{align} which means \begin{align} \frac{\partial}{\partial x}(cu_X(cx, c^2t)) = c^2u_{XX}(cx, c^2t). \end{align} Hence we have that \begin{align} v_t-v_{xx} = c^2\left(u_T(cx, c^2t)-u_{XX}(cx, c^2t)\right) = 0. \end{align}