# Xy + x - 2y -1 = 0 Help

1. Oct 16, 2009

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 !

2. Oct 16, 2009

### 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

3. Oct 16, 2009

### 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)

4. Oct 16, 2009

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?

5. Oct 16, 2009

### 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