To find $\frac{dy}{dx}$ for the equation:

$12e^x + 5e^y = 0$

we'll differentiate both sides of the equation with respect to $x$.

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

Since $y$ is a function of $x$, we apply implicit differentiation.

1. Differentiate $12e^x$ with respect to $x$: $\frac{d}{dx}(12e^x) = 12e^x$

2. Differentiate $5e^y$ with respect to $x$: $\frac{d}{dx}(5e^y) = 5e^y \cdot \frac{dy}{dx}$

3. Differentiating the right side $0$ with respect to $x$ gives $0$.

So, our differentiated equation becomes: $12e^x + 5e^y \cdot \frac{dy}{dx} = 0$

Step 2: Solve for $\frac{dy}{dx}$

Rearrange the equation to isolate $\frac{dy}{dx}$: $5e^y \cdot \frac{dy}{dx} = -12e^x$

Now, divide by $5e^y$: $\frac{dy}{dx} = -\frac{12e^x}{5e^y}$

Final Answer

$\frac{dy}{dx} = -\frac{12e^x}{5e^y}$

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

---

Related Questions:

1. How would the derivative change if we had $12e^x - 5e^y = 0$ instead?
2. How do you perform implicit differentiation with trigonometric functions?
3. What happens to $\frac{dy}{dx}$ if we substitute a value for $x$ and $y$ here?
4. How does the Chain Rule apply to implicit differentiation problems?
5. Can implicit differentiation be used for equations with logarithmic functions?

Tip: In implicit differentiation, always apply the chain rule when differentiating terms involving $y$ with respect to $x$.