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I need to find $y'$for the following equation:

$$ e^{\frac{x}{y}} = x-y $$

Before differentiating I decided to perform a quick rewrite: $$ \begin{align*} e^{\frac{x}{y}} &= x-y \newline \ln(e^{\frac{x}{y}}) &= \ln(x-y) \newline \frac{x}{y} &= \ln(x-y) \newline x &= \ln(x-y) \cdot y \newline \end{align*} $$

However, differentiating the above re-write doesn't yield the correct answer.

What am I doing wrong?

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    $\begingroup$ What answer did you get when you differentiated the above? $\endgroup$
    – ervx
    Apr 25 '16 at 22:45
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$e^{\frac{x}{y}} = x-y $

$e^{\frac{x}{y}} \frac d{dx}\left( \frac xy \right) = 1-y' $

$e^{\frac{x}{y}}\left[ x\frac d{dx}\left( \frac 1y \right) + \frac 1{y} \frac d{dx}\left( x \right) \right] = 1-y' $

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

$\frac 1{y} e^{\frac{x}{y}} - \frac x{y^2} e^{\frac{x}{y}} y' = 1-y' $

$y' - \frac x{y^2} e^{\frac{x}{y}} y' = 1- \frac 1{y} e^{\frac{x}{y}}$

$y' \left(1 - \frac x{y^2} e^{\frac{x}{y}} \right) = 1- \frac 1{y} e^{\frac{x}{y}}$

$y' =\frac { 1- \frac 1{y} e^{\frac{x}{y}}}{1 - \frac x{y^2} e^{\frac{x}{y}}}= \frac { y^2- y e^{\frac{x}{y}}}{y^2 - x e^{\frac{x}{y}}}$

$e^{\frac{x}{y}}= x-y $

$\ln(e^{\frac{x}{y}}) = \ln(x-y)$

$\frac{x}{y} = \ln(x-y)$

$\frac d{dx}\left( \frac xy \right) = \frac d{dx}\left(\ln(x-y) \right) $

$x \left(-\frac 1{y^2} y' \right) + \frac 1{y} = \frac 1{x-y} \frac d{dx}\left( x-y \right) $

$x \left(-\frac 1{y^2} y' \right) + \frac 1{y} = \frac 1{x-y} \left( 1-y' \right) $

Recall that $e^{\frac{x}{y}}= x-y $

$x \left(-\frac 1{y^2} y' \right) + \frac 1{y} = \frac 1{e^{\frac{x}{y}}} \left( 1-y' \right) $

$-\frac x{y^2} e^{\frac{x}{y}} y' + \frac 1{y} e^{\frac{x}{y}}= 1-y'$

$\frac 1{y} e^{\frac{x}{y}} - \frac x{y^2} e^{\frac{x}{y}} y' = 1-y' $ as above.

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I do this like this:

$$e^{x/y}=x-y\implies \frac1ye^{x/y}-\frac x{y^2}e^{x/y}\,y'=1-y'\implies$$

$$\left(\frac x{y^2}e^{x/y}-1\right)y'=\frac1ye^{x/y}-1\implies y'=\frac{\frac1ye^{x/y}-1}{\frac x{y^2}e^{x/y}-1}=\frac{ye^{x/y}-y^2}{xe^{x/y}-y^2}$$

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  • $\begingroup$ This is essentially how my textbook solves it. However, I'm confused as to why the rewrite does not yield an equivalent result. $\endgroup$
    – kylemart
    Apr 25 '16 at 22:57
  • $\begingroup$ I think perhaps because re-write is not original function, it is the logarithm of original function. $\endgroup$
    – user312943
    Apr 25 '16 at 22:59
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    $\begingroup$ The re-write should yield the same result, but perhaps in a form where you don't recognize it is the same result. Just like it may not seem, at first sight, that your original equation had anything to do with the final form you derived (before you differentiated). $\endgroup$
    – mathguy
    Apr 25 '16 at 23:03
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Consider the implicit function $$F=e^{\frac{x}{y}} - x+y=0$$ and compute its derivatives $$F'_x=\frac{e^{x/y}}{y}-1$$ $$F'_y=1-\frac{x e^{x/y}}{y^2}$$ Now, use the implicit function theorem $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{\frac{e^{x/y}}{y}-1} {1-\frac{x e^{x/y}}{y^2}}$$ which can rewrite in different ways.

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