How do we compute laplacian of implicit function? For example, if given a function $f(x,t)=g(ax,a^{2}t)$, how do we compute the laplacian of $f$? and how to differentiate with respect to $a$?
The answer for the latter is $f_{a}=xDg(ax,a^{2}t)+2atg_{t}(ax,a^{2}t)$. I really don't understand when to write a capital $D$ to represent a first-order derivative and when to write a subscript like $g_{t}$. And one thing I don't understand here either is that why it is $g_{t}$ instead of "$g_{a^{2}t}$" as we do in a chain rule.
 A: Note that the Laplacian of $f(x,t)$ is defined as:
$$ \Delta f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial t^2} $$
Since all that we know about $f(x,t) = g(ax,a^2 t)$ depends on the otherwise unknown function $g(u,v)$, we should assume that $g$ has second partial derivatives with respect to its (formal) arguments $u,v$.
The chain rule for partial derivatives can be appealed to for computation of the first partial of $f$ with respect to $x$:
$$ \frac{\partial f}{\partial x} = D_u g |_{u=ax,v=a^2 t} \cdot a $$
In other words, evaluate the partial derivative of $g$ with respect to its first argument $u$ at $u=ax$, and multiply by the derivative of $ax$ with respect to $x$ (giving the final factor $a$), and that gives the first partial derivative of $f$ with respect to $x$.  Here we assume that $t$ (and $a$) are independent of $x$.
Continuing in this way we get the second partial of $f$ with respect to $x$:
$$ \frac{\partial^2 f}{\partial x^2} = D_{uu} g |_{u=ax,v=a^2 t} \cdot a^2 $$
In this expression we have gotten a second factor of $a$ because again that is the derivative of $ax$ with respect to $x$.
Computing the partials of $f$ with respect to $t$ is a similar task, but now $f$ depends on $t$ only through the assignment $v=a^2 t$ in the second argument of $g$:
$$ \frac{\partial f}{\partial t} = D_v g |_{u=ax,v=a^2 t} \cdot a^2 $$
$$ \frac{\partial^2 f}{\partial t^2} = D_{vv} g |_{u=ax,v=a^2 t} \cdot a^4 $$
Putting these expressions for the second partials of $f$ together, we get the Laplacian of $f$:
$$ \Delta f = D_{uu} g |_{u=ax,v=a^2 t} \cdot a^2 + D_{vv} g |_{u=ax, v=a^2 t} \cdot a^4 $$
If I understand what you want, the next step is to differentiate $\Delta f$ with respect to $a$.  
This is a little tricky because the expression not only has $a$ appearing in the coefficients of the second partials of $g$, $a$ also affects the expression through its appearance in the arguments of the second partials of $g$.  
Why not give this (differentiating $\Delta f$ with respect to $a$) a try yourself, and I will fill in any gaps in your result?
