Changing variable in a second derivative

Question

I want to convert the differentiation variable in a second derivative, but it's a bit more complicated than the case of the first derivative. For context, the variable $\eta$ is a dimensionless density and $V$ a volume. I have the expression for $d\eta/dV$ and $\frac{d^2}{dV^2}$. The first derivative conversion is the following:

$\frac{da}{dV}=\frac{da}{d\eta}\cdot \frac{d\eta}{dV}$

Then what are the functions to convert the following second derivatives in terms of $\eta$ to volume $V$?

$$\frac{d^2a}{dV^2}=\frac{d^2a}{d\eta^2}$$ then $$\frac{d^2a}{dxdV}=\frac{d^2a}{dxd\eta}$$

Answer 1

We can apply the chain rule to get higher order derivatives:

\begin{eqnarray*} \frac{\mathrm{d}y}{\mathrm{d}u} &=& \frac{\mathrm{d}x}{\mathrm{d}u}\cdot \frac{\mathrm{d}y}{\mathrm{d}x} \\ \\ \frac{\mathrm{d}^2y}{\mathrm{d}u^2} &=& \frac{\mathrm{d}^2x}{\mathrm{d}u^2} \cdot \frac{\mathrm{d}y}{\mathrm{d}x} + \left(\frac{\mathrm{d}x}{\mathrm{d}u}\right)^{\! \!2} \cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \end{eqnarray*}

The main idea to understand is that, as differential operators:

$$\frac{\mathrm{d}}{\mathrm{d}u} = \frac{\mathrm{d}x}{\mathrm{d}u}\cdot \frac{\mathrm{d}}{\mathrm{d}x}$$

Answer 2

Your question is unclear so I'll give a general answer.

$y$ is a function of $x$. we change the variables such that $x=g(t)$.

this means $dx=g'(t)dt$.

use this representation:

$$y''=\frac{d}{dx}\left(\frac{dy}{dx}\right)$$

now substitute $dx=g'(t)dt$:

$$y''=\frac{1}{g'(t)}\frac{d}{dt}\left(\frac{1}{g'(t)}\frac{dy}{dt}\right)$$