We are tasked with finding $\frac{dy}{dx}$ by implicit differentiation for the equation:

$x e^y = x - y$

Step-by-step solution:

1. Differentiate both sides with respect to $x$:

   On the left-hand side, we have the product of $x$ and $e^y$, so we'll apply the product rule: $\frac{d}{dx}(x e^y) = \frac{d}{dx}(x) \cdot e^y + x \cdot \frac{d}{dx}(e^y)$

   • The derivative of $x$ with respect to $x$ is 1, so the first term becomes $e^y$.
   • The derivative of $e^y$ with respect to $x$ involves the chain rule: $\frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx}$ Thus, the derivative of the left-hand side is: $e^y + x e^y \frac{dy}{dx}$

2. Differentiate the right-hand side:

   The right-hand side is $x - y$. The derivative of this with respect to $x$ is: $\frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}$

3. Set up the equation:

   Now, equating the derivatives of both sides: $e^y + x e^y \frac{dy}{dx} = 1 - \frac{dy}{dx}$

4. Solve for $\frac{dy}{dx}$:

   Collect all terms involving $\frac{dy}{dx}$ on one side: $x e^y \frac{dy}{dx} + \frac{dy}{dx} = 1 - e^y$

   Factor out $\frac{dy}{dx}$ from the left-hand side: $\frac{dy}{dx}(x e^y + 1) = 1 - e^y$

   Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1 - e^y}{x e^y + 1}$

Final Answer:

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

Would you like more details, or do you have any questions about this solution?

Here are 5 related questions to deepen your understanding:

1. How does the product rule apply when differentiating two functions of $x$?
2. Why do we need to use the chain rule when differentiating $e^y$?
3. What is the geometric interpretation of implicit differentiation?
4. How would the solution change if the equation involved a different function like $\sin(y)$ instead of $e^y$?
5. What happens if $y$ is a constant? How does that simplify the differentiation?

Tip: Always remember to apply both the product and chain rules when differentiating composite and product terms in implicit differentiation.