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

**transgalactic** · August 16th 2009, 01:49 AM

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

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

**running-gag** · August 16th 2009, 02:24 AM

$z=\frac{1}{x}$ 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 $z' = -x^{-2}x'$

**transgalactic** · August 16th 2009, 02:30 AM

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

**running-gag** · August 16th 2009, 04:21 AM

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

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

**transgalactic** · August 16th 2009, 04:26 AM

ok thanks

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