Implicit Differentiation - What am I doing wrong?

Question

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?

Answer 1

$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.

Answer 2

I do this like this:

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

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

Answer 3

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.