https://file.aiquickdraw.com/mathbot/file/1725967888438uzzljcie.jpg

The equation in the image is as follows:

\[
\text{If } 2^x + 2^y = 2^{x + y}, \text{ then } \frac{dy}{dx} \text{ at } x = y = 1 \text{ is }
\]

To solve this problem, we can follow these steps:

### Step 1: Differentiate the equation implicitly
The given equation is:
\[
2^x + 2^y = 2^{x + y}
\]

Now differentiate both sides with respect to \(x\). We'll use the chain rule, and recall that the derivative of \(2^u\) with respect to \(u\) is \(2^u \ln(2)\).

\[
\frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x + y})
\]

Breaking this down:

- The derivative of \(2^x\) is \(2^x \ln(2)\).
- The derivative of \(2^y\) with respect to \(x\) is \(2^y \ln(2) \cdot \frac{dy}{dx}\) (chain rule, since \(y\) depends on \(x\)).
- The derivative of \(2^{x + y}\) is \(2^{x + y} \ln(2) \cdot (1 + \frac{dy}{dx})\), applying the product rule and chain rule.

This gives us the equation:

\[
2^x \ln(2) + 2^y \ln(2) \frac{dy}{dx} = 2^{x + y} \ln(2) \left(1 + \frac{dy}{dx}\right)
\]

### Step 2: Simplify the equation
Canceling out \(\ln(2)\) from all terms:

\[
2^x + 2^y \frac{dy}{dx} = 2^{x + y} \left(1 + \frac{dy}{dx}\right)
\]

### Step 3: Substitute \(x = y = 1\)
Now we substitute \(x = y = 1\). This simplifies the exponents:

- \(2^x = 2^1 = 2\)
- \(2^y = 2^1 = 2\)
- \(2^{x + y} = 2^{1 + 1} = 2^2 = 4\)

The equation becomes:

\[
2 + 2 \frac{dy}{dx} = 4 \left(1 + \frac{dy}{dx}\right)
\]

### Step 4: Solve for \(\frac{dy}{dx}\)
Expanding the right-hand side:

\[
2 + 2 \frac{dy}{dx} = 4 + 4 \frac{dy}{dx}
\]

Now, collect like terms:

\[
2 - 4 = 4 \frac{dy}{dx} - 2 \frac{dy}{dx}
\]

\[
-2 = 2 \frac{dy}{dx}
\]

Finally, solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = -1
\]

### Answer:
\[
\boxed{-1}
\]

Do you need more details or have any questions? Here are five related questions:

1. What is implicit differentiation, and when is it used?
2. How do you differentiate exponential functions like \(2^x\)?
3. Why does the chain rule apply when differentiating with respect to \(x\) for terms involving \(y\)?
4. Can this problem be solved using logarithmic differentiation?
5. How do the initial conditions \(x = y = 1\) simplify the calculation?

**Tip:** When differentiating complex functions, breaking down the expression step by step and applying relevant rules such as the chain rule or product rule is essential for accuracy.

If 2^x + 2^y = 2^(x + y), then dy/dx at x = y = 1 is

Learn how to solve the calculus problem: If 2^x + 2^y = 2^(x + y), then dy/dx at x = y = 1. This problem involves implicit differentiation, chain rule, and exponential functions, and is suitable for advanced calculus students in Grades 11-12.

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