If ` x^(p) + y^(q) = (x + y)^(p+q) , " then" (dy)/(dx)` is

Options

A. `- (x)/(y)`
B. `(x)/(y)`
C. `- (y)/(x)`
D. `(y)/(x)`

Explanation

To solve the problem, we need to differentiate the given equation \( x^{p} + y^{q} = (x + y)^{p + q} \) with respect to \( x \) and find \( \frac{dy}{dx} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ x^{p} + y^{q} = (x + y)^{p + q} \]

2. **Take the natural logarithm of both sides:**
\[ \log(x^{p} + y^{q}) = \log((x + y)^{p + q}) \]

3. **Apply the logarithmic property:**
\[ p \log(x) + q \log(y) = (p + q) \log(x + y) \]

4. **Differentiate both sides with respect to \( x \):**
- Left-hand side:
\[ \frac{d}{dx}(p \log(x)) + \frac{d}{dx}(q \log(y)) = \frac{p}{x} + q \frac{1}{y} \frac{dy}{dx} \]
- Right-hand side:
\[ \frac{d}{dx}((p + q) \log(x + y)) = (p + q) \cdot \frac{1}{x + y} \cdot (1 + \frac{dy}{dx}) \]

5. **Set the derivatives equal to each other:**
\[ \frac{p}{x} + q \frac{1}{y} \frac{dy}{dx} = \frac{(p + q)(1 + \frac{dy}{dx})}{x + y} \]

6. **Rearrange to isolate \( \frac{dy}{dx} \):**
\[ q \frac{1}{y} \frac{dy}{dx} - \frac{(p + q)}{x + y} \frac{dy}{dx} = \frac{(p + q)}{x + y} - \frac{p}{x} \]

7. **Factor out \( \frac{dy}{dx} \):**
\[ \left(q \frac{1}{y} - \frac{(p + q)}{x + y}\right) \frac{dy}{dx} = \frac{(p + q)}{x + y} - \frac{p}{x} \]

8. **Solve for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{\frac{(p + q)}{x + y} - \frac{p}{x}}{q \frac{1}{y} - \frac{(p + q)}{x + y}} \]

9. **Simplify the expression:**
- Combine the terms in the numerator and denominator to find a common expression.

10. **Final expression for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{y}{x} \]

### Conclusion:
The final result is:
\[ \frac{dy}{dx} = \frac{y}{x} \]