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When trying to differentiate $\frac{x+y}{xy}=x$, I get different results. If I use the quotient rule: $$ \begin{array}{rcl} \dfrac{xy(1+y´)-(x+y)(xy´+y)}{(xy)^2} &=& 1 \\ \dfrac{xy+xyy´-x^2y´-xy-xyy´-y^2}{(xy)^2} &=& 1 \\ -x^2y´ -y^2 &=& x^2y^2 \\ y´ &=& \dfrac{-(y^2+x^2y^2)}{x^2} \end{array} $$ On the other hand if I do the following $$ \begin{array}{rcl} \dfrac{x+y}{xy} &=& x \\ x+y &=& x^2y \\ 1+y´ &=& x^2y´ +2xy \\ y´ -x^2y´ &=& 2xy-1 \\ y´ &=& \dfrac{2xy-1}{1-x^2} \end{array} $$ I don't understand what's happening.

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  • $\begingroup$ You have two versions of the same result, only in the former you would now seek to substitute $y$ in terms of $x$. Re-write the initial equation as $y = \ldots$ and insert to your first result. $\endgroup$ – Kevin Nov 18 '16 at 8:59
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$\frac{x+y}{xy}=x$ so that $y=\frac{x}{x^2-1}$ If we use this we can show that two terms for $y'$ are equal

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If you can get y in terms of x it's better to avoid getting different versions of the same answer.

One may think of this $$ln(x+y)-ln(xy)=lnx $$ y' in terms of x,y you will get different version of the same answer than yours .

So it's better to do this : $$ \frac{x+y}{xy}=x \Rightarrow y=\frac{x}{x^2-1}\Rightarrow y'=\frac{-(x^2-1)}{(x^2-1)^2}$$

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