the implicit derivative of $x+2y+1=\frac{y^2}{x-1}$ at the point $(2,-1)$

Question

So as the title states, I have to find the points $(2,-1)$ for the equation $x+2y+1=\frac{y^2}{x-1}$.

first I multiply both side with $(x-1)$ since when taking the derivative I try to avoid fractions:

I get $(x-1)(x+2y+1)=y^2$.

distributing the equation I get:

$x^2-x+2yx-2y+x-1=y^2$.

So after taking the implicit derivative(not sure of the nomenclature of this kind)

I get:

$2x+2(y'x+y)-2y'-0=2yy'$.

after substituting x and y with (2,-1)

I get $y'=-\frac{3}{2}$

It's supposed to be $y'=-\frac{1}{2}$

I know there are different approaches, like using the quotient rule instead. However, this should also be a valid approach.

So I would appriciate if somebody could employ my approach, because it discourages me a little when this does't work as I'd wished it to work:)

Thank you in advance!

Answer 1

The derivative with respect to $x$ of $2y$ is $2y'$ not $2yy'$.

Answer 2

Given $x+2y+1=\frac{y^2}{x-1}$, we implicitly assume (thus the name) that $y$ is a differentiable function of $x$. Then since the left and right hand sides are two ways of writing the same function, we can differentiate each of them and get another equality (the derivative of the same function). We can differentiate $y$ as simply $y'$ - since we assumed $y$ is differentiable - and for any function $f$, we can differentiate $f(y)$ as $f'(y)y'$, using the chain rule.

So,

\begin{eqnarray} x+2y+1 &=& \frac{y^2}{x-1} \newline 1 + 2y' &=&\frac{(2y\cdot y')(x-1) - y^2}{(x-1)^2} \newline 2y' - \frac{2y\cdot y'}{x-1}&=& \frac{-y^2}{(x-1)^2} - 1 \newline \left(2 - \frac{2y}{x-1}\right)y' &=& \frac{-y^2}{(x-1)^2} - 1 \newline y' &=& \frac{\left(\frac{-y^2}{(x-1)^2} - 1\right)}{\left(2 - \frac{2y}{x-1}\right)} \end{eqnarray}

where the first step is the differentiation and the rest is algebra. You can check that this indeed gives the answer $y' = -1/2$ when $x = 2, y = -1$.

Answer 3

On the left side, the derivative of $x^2-x$ is $2x-1$, not $2x$.