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?

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**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 \(x\).

Learn how to find dy/dx for the implicit equation 12e^x + 5e^y = 0 using implicit differentiation. This calculus problem involves the chain rule and exponential functions, providing a step-by-step solution ideal for Grades 11-12.

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