Solving a Sound Wave Equation in Physics 1: Halliday, Resnick, and Krane

[Afo]

Homework Statement:: This is from 5 ed, Physics 1Halliday, Resnick, and Krane. page 428 about sound waves

I have highlighted the equation that I don't understand. How did the author get it? I understand how they get from the middle side to the RHS of the equation, but I don't understand how they get from the first equation to the middle equation.

I don't have any problem with anything before.
Relevant Equations:: Density: Rho = Mass / Volume

 

[Orodruin]

Differentiation.

 

[Afo]

Differentiation.

Is it like this?

dm/dv^2 = - m /(V^2)
so
dp = dm/dv = dv * (-m/(V^2))

 

[Orodruin]

No. Use $\Delta y/\Delta x \simeq dy/dx$

 

[Afo]

No. Use $\Delta y/\Delta x \simeq dy/dx$

Like this?
$d\rho = \frac{\Delta \rho}{\Delta V}dV$
but $\frac{\Delta \rho}{\Delta V} = -\frac{m}{V^2}$

 

[Afo]

Is it like this?

dm/dv^2 = - m /(V^2)
so
dp = dm/dv = dv * (-m/(V^2))

Okay, I know why this is false.
It doesn't work since $\frac{a}{b} = \frac{c}{d}$ doesn't imply $\frac{a}{b^2} = \frac{c}{d^2}.$ If that was true, that would mean $|dV| = |V|$ which is obviously false.

 

[Orodruin]

Do you know how a derivative works?

 

[Afo]

Do you know how a derivative works?

I am still in high school so this is what I know:

1. We can view $\frac{d}{dx}$ as an operator so $\frac{d}{dx}f(x) = \frac{df}{dx}$ is the derivative of $f$ w.r.t. $x$.

Another way is to view it in terms of differentials:
2. $dx$ and $dy$ are called differentials. The differential $dx$ is an independent variable. The differential $dy$ is defined to be $dy = f'(x) dx$

Both ways are rigorous.

 

[Orodruin]

So, let $x = V$, $y = \rho(V)$ and use 2 to relate $d\rho$ to $dV$.

 

[Afo]

So, let $x = V$, $y = \rho(V)$ and use 2 to relate $d\rho$ to $dV$.

Ok, I got it!

$\rho(V) = \frac{m}{V}$ is a function of volume and $m$ is a constant.

Thus $\frac{d}{dV} \rho(V) = \frac{d}{dV} \frac{m}{V} = m\frac{d}{dV} \frac{1}{V} = -\frac{m}{V^2}$