Using Implicit Differientation to find a formula for dy/dx

Question

I already have all the steps down, one thing I couldn't understand was how to get the final answer (which I have also):

The equation given is $y=xe^y$

Here is the steps I've taken so far:

$\frac{d}{dx}y=xe^{y}$

$\frac{dy}{dx}=\frac{d}{dx}e^y * x + \frac{d}{dx} x * e^y$

$\frac{dy}{dx}=e^y * \frac{d}{dx}y*x+1*e^y$

$\frac{dy}{dx}=e^y*\frac{dy}{dx}x+e^y$

$\frac{dy}{dx}=e^y(\frac{dy}{dx}x+1)$

The answer is: $\frac{dy}{dx}=-\frac{e^y}{xe^y-1}$

I'm stuck as to how I'm getting from the last step to the answer.

If I divide $e^y$ and then ($x+1$) I'd still have $\frac{dy}{dx}$ on both sides. So how is it that the answer is correct? What am I doing wrong here?

Answer 1

"Treat" $\dfrac{dy}{dx}$ as a variable and we want to isolate it.

So, $\dfrac{dy}{dx}-e^y\dfrac{dy}{dx}x=e^y$

$\dfrac{dy}{dx}(1-e^yx)=e^y \Rightarrow \dfrac{dy}{dx}=\dfrac{e^y}{1-e^yx}=-\dfrac{e^y}{e^yx-1}$

Answer 2

An easy way to do implicit differentiation is to write the function as $F(x,y) = 0,$ meaning move everything to one side of the equation, and take the partial derivatives $\frac{\partial F}{\partial x} = F_x$ and $F_y.$ Then $dy/dx = -F_x / F_y.$

So $F_x = e^y,$ $F_y = xe^y - 1,$ and $dy/dx = -e^y / (xe^y - 1).$

Answer 3

$$y'=e^y(y'x+1) = y'e^yx + e^y \Longleftrightarrow y'(e^yx-1) = -e^y ...$$

We are just rearranging terms, and then factoring out the derivative $y'$.