Is this the best route to go in this implicit derivation problem?

Question

I have this equation:

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

I seem to be going down the wrong path. Is this right so far?

$$\frac{dy}{dx} = e^{\frac{x}{y}} \cdot \frac{dy}{dx} (x \cdot y^{-1}) = 1 - y'$$

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

WolframAlpha has this as

$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

Answer 1

Your result $$-e^{\frac{x}{y}}xy^{-2}\color{blue}{y'}+e^{\frac{x}{y}}y^{-1}=1-\color{blue}{y'}$$ is correct.

You can continue and simplify this by collecting all the terms with $y'$ as a factor:

$$\begin{align}\color{blue}{y'}\left (1-e^{\frac{x}{y}}xy^{-2}\right )=1-e^{\frac{x}{y}}y^{-1}\end{align}\\\color{blue}{y'}=\frac{1-e^{\frac{x}{y}}y^{-1}}{1-e^{\frac{x}{y}}xy^{-2}}\frac{y^2}{y^2}=\frac{y(y-e^{\frac{x}{y}})}{y^2-e^{\frac{x}{y}}x}$$

This is the same answer that WolframAlpha gave you, only that they have multiplied numerator and denominator by $-1$.

Answer 2

Please check your line 2, it should be something like this:

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

Proceed using this to get the WolframAlpha output

Answer 3

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

differentiate wrt $y$ and use the chain rule and Quotient rule on $e^{\frac xy}$ in that order ;

$ e^{\frac xy}\cdot \frac{y-\frac{dy}{dx}x}{y^2}=1-\frac{dy}{dx} $

$e^{\frac xy} y -\frac{dy}{dx}x e^{\frac xy} = y^2-y^2\frac{dy}{dx}$

$\frac{dy}{dx}\big(y^2-x\cdot e^{\frac xy}\big) = y^2-e^{\frac xy}y$

$\frac{dy}{dx} = \frac{y(y-e^{\frac xy})}{y^2-x\cdot e^{\frac xy}}$

Answer 4

I don't know why your subtraction sign became an equals sign, but your goal, after implicitly differentiating, is to isolate the $y'$ terms so you can factor and divide:

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

Then factor:

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

And divide. After some simplification, you should reach:

$$y' = {{y - e^{x/y} \over y} \cdot -{y^2 \over e^{x/y} x - y^2}} = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$

Answer 5

Why is nobody suggesting logarithmic differentiation ?

First take log base $e$ of both sides to get

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

Now differentiate implicitly:

$$\begin{align} 1&=y\cdot \frac{1}{x-y}\cdot \left(1-\frac{dy}{dx}\right)+\ln(x-y)\cdot \frac{dy}{dx} \\ \Rightarrow 1&=\frac{y}{x-y}-\frac{y}{x-y}\cdot \frac{dy}{dx}+\ln(x-y)\cdot \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\ln(x-y)-\frac{y}{x-y}\right)&=\frac{x-2y}{x-y} \\ \Rightarrow \frac{dy}{dx}&=\frac{\frac{x-2y}{x-y}}{\ln(x-y)-\frac{y}{x-y}} \\ \Rightarrow \frac{dy}{dx}&=\frac{x-2y}{(x-y)\ln(x-y)-y}. \end{align}$$

OK, maybe now I know why...

Answer 6

We have

$$ f(x,y) = e^{\frac{x}{y}}-x+y=0 $$

so we have

$$ df = f_x dx + f_y dy = 0\Rightarrow \frac{dy}{dx}=-\frac{f_x}{f_y} = \frac{y (x-2 y)}{x^2-x y-y^2} $$