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I was learning implicit differentiation and came across a discrepancy in finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ of $y=x+\frac{1}{y}$: when differentiating without simplifying there is no $x$ in $\frac{\mathrm{d}y}{\mathrm{d}x}$, but after simplifying there is.

When I simplified the equation to $y^2=xy+1$ I got the right answer. Why did this happen?

I could only conclude it has to do something with how implicit functions are defined. I could not find anything on the internet.

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  • $\begingroup$ What's the discrepancy you are facing? $\endgroup$
    – Anurag A
    Aug 6 '21 at 7:43
  • $\begingroup$ @Anurag A when differentiating without simplifying you see there will be no x in dy/dx but after simplifying there is $\endgroup$ Aug 6 '21 at 7:46
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By differentiating $y=x+\frac{1}{y}$ we find $$y'=1-\frac{y'}{y^2} \tag{1}$$ On the other hand, $y^2=xy+1$ gives $$2yy'=y+xy'\tag{2}$$ (1) and (2) are both valid equations.

Infact, by using $x=y-\frac{1}{y}$ in (2), we remove $x$ from (2) and we get (1): $$2yy'=y+xy'=y+\left(y-\frac{1}{y}\right)y'$$ that is $$yy'=y-\frac{y'}{y}$$ and after dividing by $y$ (recall that here $y\not=0$), $$y'=1-\frac{y'}{y^2}.$$ By reversing the process we may obtain (2) from (1).

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