Finding the second derivative; What am I doing wrong? Original Question:
$xy+y-x=1$
Find the second derivative;
$d^2y\over{dx^2}$$(xy+y-x=1)$
We are allowed to use either notation as far as I know: ${dy\over{dx}}$ or ${y'}$. Because ${dy\over{dx}}=y'$ according to my Math 1000 prof.
$(xy)'+y'-x'-1'=0'$
$(y+xy')+y'-1=0$
$y'(x+1)-1+y=0$
$y'(x+1)=1-y$
$y'={1-y\over{x+1}}$
Now we're supposed to find $y''$ but, that's where I mess things up. Quotient rule:
$y''={{(1-y)'(x+1)-(1-y)(x+1)'}\over[(x+1)]^2}$
$y''={{(-y')(x+1)-(1-y)(1)}\over[(x+1)]^2}$
$y''={{-({{1-y}\over{x+1}}){\cdot}(x+1)-(1-y)}\over{(x+1)^2}}$
$y''={{-(1-y)-(1-y)}\over{(x+1)^2}}={-2(1-y)\over{(x+1)^2}}={2(y-1)\over{(x+1)^2}}$
But, The answer is supposed to be ${y''={{y-1}\over{(x+1)^2}}}$
Why am I getting a 2? That's all I want to know.
Also, I took out the unnecessary equal signs, my prof wants us to have those cause he's basically insane but, it's confusing everyone so I'm taking them out.
The improper method! According to my prof:
$(xy)'+y'-x'-1'=0'$
$(y+xy')+y'-1=0$
$y'+(xy')'+y''=0$
$y'+(y'+xy'')+y''=0$
$y''(x+1)=-2y'$
${y''={{-2y'}\over{x+1}}}$
Plugin $y'$ and it's the same answer i'm getting. Supposedly the wrong one everywhere I've looked.

Edit: After your equation changes, your work is correct and the "supposed" answer is wrong.
