Calculus Help: Solving xy + x - 2y - 1 = 0 with Schaum's Calculus 5th Edition

[sponsoredwalk]

Bought Schaum's Calculus 5th edition to clean up my calculus worries but I haven't cleared this one up yet;

xy + x - 2y -1 = 0

Solving for (y);
xy - 2y = - x + 1
y(x - 2) = - x + 1
y = \frac {-x+1}{x-2}

Taking the derivative of y = \frac {-x+1}{x-2} using the quotient rule

\frac {dy}{dx} = \frac {- 1 (x - 2) - 1 (- x + 1)} {(x - 2)^2}

\frac {dy}{dx} = \frac {- x + 2 + x - 1)} {(x - 2)^2}

\frac {dy}{dx} = \frac {1} {(x - 2)^2}

Now, my problem is that when I do this calculation via Implicit Differentiation, this is what I get;

xy + x - 2y -1 = 0

y + x\frac {dy}{dx} + 1 - 2\frac {dy}{dx} = 0
x\frac {dy}{dx}- 2\frac {dy}{dx} = - y - 1
\frac {dy}{dx}( x - 2) = - y - 1
\frac {dy}{dx} = \frac {- y - 1} {x - 2}

So;
1.My two answers don't agree, aren't they supposed to? If not, why don't they?

2.My book gives a different answer which I cannot get,
\frac{1+ y} {2 - x}
Am I careless with signs or is my book wrong?

Gratias tibi ago !

 

[lanedance]

i haven't checked all your working yet but the main differnce is the y in the implictly derived equation if you sub in for y and have done everything correctly they will be the same

 

[lanedance]

xy + x - 2y -1 = 0
differntiating
y + xy' + 1 - 2y'= 0
re-arranging
y'(x-2)= -(y+1)
so
y'= \frac{(y+1)}{2-x}

your answer is in fact the same as the book and above, try multiplying through by (-1)/(-1)

 

[sponsoredwalk]

That's great, I got very worried I was still getting wrong answers (but also excited at the thought I found another wrong answer in a book lol) But is the quotient rule derivation wrong? It seems very plausible the way I worked it out yet there's a different answer. How so?

 

[Chairman Lmao]

They're both the same thing. You've got y=(1-x)/(x-2) So,

y+1=(1-x)/(x-2)+1=(1-x+x-2)/(x-2)=1/(2-x) So

(y+1)/(2-x)=1/(2-x).^2