If `3^(x)+3^(y)=3^(x+y)`, then what is `(dy)/(dx)` equal to ?

Options

A. `(3^(x+y)-3^(x))/(3^(y))`
B. `(3^(x-y)(3^(y)-1))/(1-3^(x))`
C. `(3^(x)+3^(y))/(3^(x)-3^(y))`
D. `(3^(x)+3^(y))/(1+3^(x+y))`

Answer

To solve the equation \(3^x + 3^y = 3^{x+y}\) and find \(\frac{dy}{dx}\), we will follow these steps:

### Step 1: Differentiate both sides with respect to \(x\)
We start with the equation:
\[ 3^x + 3^y = 3^{x+y} \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(3^x) + \frac{d}{dx}(3^y) = \frac{d}{dx}(3^{x+y}) \]

### Step 2: Apply the chain rule
Using the chain rule, we know that the derivative of \(3^u\) is \(3^u \ln(3) \cdot \frac{du}{dx}\). Thus, we have:
\[ 3^x \ln(3) + 3^y \ln(3) \frac{dy}{dx} = 3^{x+y} \ln(3) \left(1 + \frac{dy}{dx}\right) \]

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ 3^x \ln(3) + 3^y \ln(3) \frac{dy}{dx} = 3^{x+y} \ln(3) + 3^{x+y} \ln(3) \frac{dy}{dx} \]

### Step 4: Rearrange the terms
Rearranging the terms to isolate \(\frac{dy}{dx}\):
\[ 3^y \ln(3) \frac{dy}{dx} - 3^{x+y} \ln(3) \frac{dy}{dx} = 3^{x+y} \ln(3) - 3^x \ln(3) \]

### Step 5: Factor out \(\frac{dy}{dx}\)
Factoring out \(\frac{dy}{dx}\) from the left side:
\[ \left(3^y \ln(3) - 3^{x+y} \ln(3)\right) \frac{dy}{dx} = 3^{x+y} \ln(3) - 3^x \ln(3) \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{3^{x+y} \ln(3) - 3^x \ln(3)}{3^y \ln(3) - 3^{x+y} \ln(3)} \]

### Step 7: Simplify the expression
We can simplify this further:
\[ \frac{dy}{dx} = \frac{\ln(3) (3^{x+y} - 3^x)}{\ln(3) (3^y - 3^{x+y})} \]
Cancelling \(\ln(3)\) from the numerator and denominator:
\[ \frac{dy}{dx} = \frac{3^{x+y} - 3^x}{3^y - 3^{x+y}} \]

### Step 8: Final expression
This can be expressed as:
\[ \frac{dy}{dx} = \frac{3^x(3^y - 1)}{3^y(1 - 3^x)} \]

### Final Result
Thus, we have:
\[ \frac{dy}{dx} = \frac{3^x(3^y - 1)}{3^y(1 - 3^x)} \]