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

Options

A. `(x)/(y)`
B. `(y)/(x)`
C. `(x^(p-1))/(y^(q -1))`
D. none of these

Explanation

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

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides of the equation:
\[ \log(x^p y^q) = \log((x + y)^{p + q}) \]

### Step 2: Apply logarithmic properties
Using the properties of logarithms, we can simplify both sides:
\[ p \log x + q \log y = (p + q) \log(x + y) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(p \log x + q \log y) = \frac{d}{dx}((p + q) \log(x + y)) \]

Using the chain rule, we get:
\[ \frac{p}{x} + q \frac{1}{y} \frac{dy}{dx} = (p + q) \frac{1}{x + y} \left(1 + \frac{dy}{dx}\right) \]

### Step 4: Rearranging the equation
Now, we will rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{p}{x} + q \frac{1}{y} \frac{dy}{dx} = \frac{(p + q)}{x + y} + \frac{(p + q)}{x + y} \frac{dy}{dx} \]

### Step 5: Collect terms involving \(\frac{dy}{dx}\)
Rearranging gives:
\[ q \frac{1}{y} \frac{dy}{dx} - \frac{(p + q)}{x + y} \frac{dy}{dx} = \frac{(p + q)}{x + y} - \frac{p}{x} \]

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

### Step 6: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{(p + q)}{x + y} - \frac{p}{x}}{q \frac{1}{y} - \frac{(p + q)}{x + y}} \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is given by:
\[ \frac{dy}{dx} = \frac{(p + q) \frac{1}{x + y} - \frac{p}{x}}{q \frac{1}{y} - \frac{(p + q)}{x + y}} \]