Find the stationary points - algebra

Question

I've been struggling with this for days and still haven't found the answer given in the book. Hope someone can help.

Deduce the coordinates of the stationary point of the following curve:

$$y = \frac{2}{x-1}\ - \frac{2}{(x-1)^2}$$

The answer in the book is $\left(3,\frac{1}{2}\right)$ when $x - 1 = 2$, but I can't get there.

Here's my working:

$$\begin{align} \frac{2}{x-1}\ - \frac{2}{(x-1)^2} &= \frac{2}{x-1}\ - \frac{2}{x^2 - 2x + 1} \\ &= 2(x-1)^{-1} - 2(x^2-2x+1)^{-1} \\ &= 2x^{-1} - 2^{-1} -2x^{-2}+4x^{-1}+2^{-1} \\ &= 6x^{-1} - 2x^{-2} \end{align}$$

$$\frac{\mathrm dy}{\mathrm dx} = 4x^{-3} - 6x^{-2} = 2x^{-2}(2x^{-1} - 3)$$

Then $2x^{-1} - 3 = 0$ and so $\frac{2}{x} = 3$ giving $x = \frac{2}{3}$

Answer 1

Hint: You've fallen victim to Freshman's Dream by assuming:

$$(a + b)^{-1} = a^{-1} + b^{-1}$$

You would need to, instead, find a common denominator and combine, or apply the power rule when differentiating, for example:

$${2\over x-1} = 2(x-1)^{-1} \Rightarrow {\mathrm d\over \mathrm dx}2(x-1)^{-1} = -1\cdot 2(x-1)^{-2}\cdot 1$$

Answer 2

$\frac {dy}{dx}=\frac{-2}{(x-1)^2}+\frac4{(x-1)^3}=0 \implies \frac {-2(x-1)+4}{(x-1)^3}=0 \implies -2(x-1)+4=0 \implies x-1=2 \implies x=3$

*remember, $(x+y)^n\not =x^n+y^n$ in general; this is known as the freshman's dream...