Thread: how to know here on what variable its a derivative of..(diff)

$\displaystyle (2x^2ylny-x)y'=y$
$\displaystyle (2x^2ylny-x)dy=ydx$
then i divide both sides by dy
$\displaystyle (2x^2ylny-x)=yx'$
then i divide both sides by y
$\displaystyle (2x^2lny-\frac{x}{y})=x'$
$\displaystyle x'+\frac{x}{y}=2x^2lny$
so i have here a bernuly foruma
i divide both sides by $\displaystyle x^2$
$\displaystyle \frac{x'}{x^2}+\frac{1}{xy}=2lny$
$\displaystyle z=x^{-1}$
$\displaystyle z'=-1x^{-2}x'$
$\displaystyle -z'+\frac{z}{y}=2ln y$

z is defined to be a function of x
so $\displaystyle z'=\frac{dz}{dx}$
why the book interprets $\displaystyle z'=\frac{dz}{dy}$
??

$\displaystyle z=\frac{1}{x}$ and x is a function of y, therefore z is also a function of y
This is already what you have considered when you wrote $\displaystyle z' = -x^{-2}x'$

z is linked to y not in a direct way .
but z linked to x in a direct way
z and x are more close to each other.

i cant see a mathematical way of figuring it out
its all intuition.and i my intuition is very bad

When you write $\displaystyle z=\frac{1}{x}$ with x function of y (in other words x = x(y)), you write a relation between two functions

It is the same as writing $\displaystyle z(y)=\frac{1}{x(y)}$

ok thanks