Question

If $2^x + 2^y = 2^{x+y}$, then the value of $\frac{dy}{dx}$ at $(1, 1)$ is equal to:

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$
• E. $2$

Hint:

For equations of the form $a^x + a^y = a^{x+y}$, the derivative is always $\frac{dy}{dx} = -\frac{a^x(a^y - 1)}{a^y(a^x - 1)}$. At $(1,1)$, if $x=y$, this always simplifies to $-1$.

Answer 1

The Correct Option is B

Concept: For an implicit function $f(x, y) = 0$, we use implicit differentiation with respect to $x$. Recall that the derivative of $a^u$ is $a^u \ln(a) \frac{du}{dx}$.

Step 1: Differentiate both sides with respect to $x$.
Given: $2^x + 2^y = 2^{x+y}$ \[ \frac{d}{dx}(2^x) + \frac{d}{dx}(2^y) = \frac{d}{dx}(2^{x+y}) \] \[ 2^x \ln 2 + 2^y \ln 2 \frac{dy}{dx} = 2^{x+y} \ln 2 \left( 1 + \frac{dy}{dx} \right) \]

Step 2: Simplify the equation.
Divide the entire equation by $\ln 2$: \[ 2^x + 2^y \frac{dy}{dx} = 2^{x+y} + 2^{x+y} \frac{dy}{dx} \]

Step 3: Substitute the point $(1, 1)$.
Substitute $x = 1$ and $y = 1$: \[ 2^1 + 2^1 \frac{dy}{dx} = 2^{1+1} + 2^{1+1} \frac{dy}{dx} \] \[ 2 + 2 \frac{dy}{dx} = 4 + 4 \frac{dy}{dx} \]

Step 4: Solve for $\frac{dy}{dx}$.
\[ 2 - 4 = 4 \frac{dy}{dx} - 2 \frac{dy}{dx} \] \[ -2 = 2 \frac{dy}{dx} \quad \Rightarrow \quad \frac{dy}{dx} = -1 \]