If $\frac{x}{y^\frac{n-1}{n}}$ is constant, how do I prove $\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$? From: 
$\frac{x}{y^\frac{n-1}{n}}=constant$
To: 
$\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$
It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.
 A: Let  $x=c\cdot y^\frac{n-1}{n}$. Then 
\begin{align}
\frac{dx}{dy}&=\frac{n-1}{n}c\cdot y^\frac{-1}{n}\\
&=\frac{n-1}{n}c\cdot y^\frac{-1}{n} \cdot \frac{y}{y}\\
&=\frac{n-1}{n}c\cdot y^\frac{n-1}{n} \cdot \frac{1}{y}\\
&=\frac{n-1}{n} \frac{x}{y}\\
\end{align}
A: We want
$\dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{y}{x}; \tag 0$
from
$\dfrac{x}{y^{\frac{n - 1}{n}}} = \alpha = \text{constant}, \tag 1$
assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:
$x = \alpha y^{\frac{n - 1}{n}}; \tag 2$
$x^n = \alpha^n y^{n - 1}; \tag 3$
$nx^{n - 1} = \alpha^n (n - 1)y^{n - 2} \dfrac{dy}{dx}; \tag 4$
$\alpha^n \dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{x^{n - 1}}{y^{n - 2}}; \tag 5$
from (3),
$\alpha^n = \dfrac{x^n}{y^{n - 1}}; \tag 6$
thus, dividing (5) by (6),
$\dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{x^{n - 1}}{y^{n - 2}} \dfrac{y^{n - 1}}{x^n} =  \dfrac{n}{n - 1} \dfrac{y}{x}, \tag 7$
which we may write in differential form to obtain
$\dfrac{dx}{x} = \dfrac{n - 1}{n} \dfrac{dy}{y}; \tag 8$
as per request.  $OE\Delta$
