Getting different answers when using product rule and limit substitution than I do with quotient rule I'm trying to differentiate $$y = \frac{x+1}{x-1}.$$
Using quotient rule:
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Using product:
http://prntscr.com/mt91dd
Using limit definition:
http://prntscr.com/mt91ih
I get $-2$ for product and limit, but I get $\frac{-2}{x^{2}-2x+1}$ using quotient.
 A: It means that you're doing something wrong.
$$
\left(\frac{x+1}{x-1}\right)'=
\frac{(x+1)'(x-1)-(x+1)(x-1)'}{(x-1)^2}=\\
\frac{x-1-(x+1)}{(x-1)^2}=
\frac{x-1-x-1}{(x-1)^2}=-\frac{2}{(x-1)^2}
$$

$$
[(x+1)(x-1)^{-1}]'=\\
(x+1)'(x-1)^{-1}+(x+1)[(x-1)^{-1}]'=\\
(x-1)^{-1}+(x+1)(-1)(x-1)^{-2}=\\
\frac{1}{x-1}-\frac{x+1}{(x-1)^2}=
\frac{x-1}{(x-1)^2}-\frac{x+1}{(x-1)^2}=\\
\frac{x-1-x-1}{(x-1)^2}=-\frac{2}{(x-1)^2}
$$

$$
\lim\limits_{\Delta x \rightarrow 0}\frac{\frac{x+\Delta x+1}{x+\Delta x-1}-\frac{x+1}{x-1}}{\Delta x}=
\lim\limits_{\Delta x \rightarrow 0}\frac{\frac{(x+\Delta x+1)(x-1)-(x+1)(x+\Delta x-1)}{(x+\Delta x-1)(x-1)}}{\Delta x}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2-x+x\Delta x-\Delta x+x-1-(x^2+x\Delta x-x+x+\Delta x-1)}{\Delta x(x+\Delta x-1)(x-1)}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2+x\Delta x-\Delta x-1-(x^2+x\Delta x+\Delta x-1)}{\Delta x(x+\Delta x-1)(x-1)}=
\lim\limits_{\Delta x \rightarrow 0}\frac{x^2+x\Delta x-\Delta x-1-x^2-x\Delta x-\Delta x+1}{\Delta x(x+\Delta x-1)(x-1)}=\\
\lim\limits_{\Delta x \rightarrow 0}\frac{-2\Delta x}{\Delta x(x+\Delta x-1)(x-1)}=
\lim\limits_{\Delta x \rightarrow 0}\frac{-2}{(x+\Delta x-1)(x-1)}=\\
\frac{-2}{(x+0-1)(x-1)}=-\frac{2}{(x-1)^2}
$$
A: By the way, the simplest way to solve this problem is probably to observe
$$y= \frac{x+1}{x-1} = 1+\frac{2}{x-1} \Rightarrow \frac{dy}{dx} = -\frac{2}{(x-1)^2}.$$
I'm guessing the point of the problem was to convince yourself that the various techniques result in the same answer, but I didn't want you to overlook the value of simple division.
