https://file.aiquickdraw.com/mathbot/file/1727814832164l67tuing.jpg

The problem in the image asks to find \(\frac{dy}{dx}\) using implicit differentiation for the equation:

\[
xe^y = x - y
\]

Let’s solve it step-by-step:

### Step 1: Differentiate both sides with respect to \(x\).

Differentiate the left-hand side:  
\[
\frac{d}{dx}(xe^y) = \frac{d}{dx}(x) \cdot e^y + x \cdot \frac{d}{dx}(e^y)
\]
Since \(y\) is a function of \(x\), we apply the chain rule:
\[
= e^y + x \cdot e^y \cdot \frac{dy}{dx}
\]

Differentiate the right-hand side:
\[
\frac{d}{dx}(x - y) = 1 - \frac{dy}{dx}
\]

### Step 2: Set up the differentiated equation.
Now, combining the derivatives from both sides:
\[
e^y + x \cdot e^y \cdot \frac{dy}{dx} = 1 - \frac{dy}{dx}
\]

### Step 3: Solve for \(\frac{dy}{dx}\).
First, collect the \(\frac{dy}{dx}\) terms on one side:
\[
x \cdot e^y \cdot \frac{dy}{dx} + \frac{dy}{dx} = 1 - e^y
\]

Factor out \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} (x \cdot e^y + 1) = 1 - e^y
\]

Finally, solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{1 - e^y}{x \cdot e^y + 1}
\]

### Final Answer:
\[
\frac{dy}{dx} = \frac{1 - e^y}{x \cdot e^y + 1}
\]

Let me know if you'd like more details or have any questions!

### Further questions:
1. How is implicit differentiation different from explicit differentiation?
2. What is the significance of the chain rule in this problem?
3. Can you apply implicit differentiation to higher-order derivatives?
4. How would the result change if the equation were more complex, e.g., \(xe^{y^2} = x - y\)?
5. How can implicit differentiation be applied in real-world scenarios?

**Tip:** Implicit differentiation is crucial when variables cannot be easily isolated.

Find dy/dx by implicit differentiation for the equation xe^y = x - y.

Find the derivative dy/dx using implicit differentiation for the equation xe^y = x - y. This solution demonstrates the application of the chain rule and implicit differentiation techniques, suitable for calculus students.

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