I was doing a physics exercise, and all of a sudden, in the solution, it says that: $$ \Delta(V^y) = y\Delta(V)V^{y-1} $$ Why is this equation correct? How am I able to conclude that $\Delta(V^y)$ is indeed what it's saying that it is?!
UPDATE:
$V$ is volume
$y$ is a constant
$\Delta$ is change in volume, hence $\Delta(V) = V_f - V_i$
Thanks!
Note that we can write
$$\Delta(v^y)=v^y\left(\left(1+\frac{\Delta v}{v}\right)^y-1\right)$$
Using the Generalized Binomial Theorem yields
$$\left(1+\frac{\Delta v}{v}\right)^y-1=y\left(\frac{\Delta v}{v}\right)+\underbrace{O\left(\left(\frac{\Delta v}{v}\right)^2\right)}_{\text{Higher Order Terms}}$$
where $f(x)=O(x)$ means that there exists a number $C>0$ such that $|f(x)|\le C|x|$ as $x\to 0$. Here, we are taking $\Delta v$ to be "small."
Then, we have
$$\Delta(v^y)=yv^{y-1}\Delta v+v^yO\left(\left(\frac{\Delta v}{v}\right)^2\right)$$
Neglecting the Higher Order Terms gives
$$\Delta(v^y)=yv^{y-1}\Delta v$$
which is the sought expression.