If ` x^p y^q = (x +y) ^(p+q) , ` then ` (dy)/(dx)` is equal to also show that ` (d^2 y)/(dx^2 )=0`

Options

A. `(y)/(x)`
B. `(py)/(q) x`
C. `(x)/(y)`
D. `q (y)/(px)`

Answer

To solve the problem, we start with the equation given:

\[ x^p y^q = (x + y)^{p + q} \]

### Step 1: Differentiate both sides with respect to \( x \)

Using implicit differentiation, we differentiate both sides:

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

### Step 2: Apply the product rule on the left side

Using the product rule on the left side, we have:

\[ \frac{d}{dx}(x^p) \cdot y^q + x^p \cdot \frac{d}{dx}(y^q) = (p \cdot x^{p-1} \cdot y^q + x^p \cdot q \cdot y^{q-1} \cdot \frac{dy}{dx}) \]

### Step 3: Apply the chain rule on the right side

Using the chain rule on the right side, we have:

\[ (p + q)(x + y)^{p + q - 1} \cdot \left(1 + \frac{dy}{dx}\right) \]

### Step 4: Set the derivatives equal

Now we equate both sides:

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

### Step 5: Rearrange to isolate \( \frac{dy}{dx} \)

Rearranging gives us:

\[ x^p \cdot q \cdot y^{q-1} \cdot \frac{dy}{dx} - (p + q)(x + y)^{p + q - 1} \cdot \frac{dy}{dx} = (p + q)(x + y)^{p + q - 1} - p \cdot x^{p-1} \cdot y^q \]

### Step 6: Factor out \( \frac{dy}{dx} \)

Factoring out \( \frac{dy}{dx} \):

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

### Step 7: Solve for \( \frac{dy}{dx} \)

Now, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{(p + q)(x + y)^{p + q - 1} - p \cdot x^{p-1} \cdot y^q}{x^p \cdot q \cdot y^{q-1} - (p + q)(x + y)^{p + q - 1}} \]

### Step 8: Show that \( \frac{d^2y}{dx^2} = 0 \)

To show that \( \frac{d^2y}{dx^2} = 0 \), we differentiate \( \frac{dy}{dx} \) again. Since \( \frac{dy}{dx} \) is a constant (as derived from the equation), its second derivative will be zero.

### Final Result

Thus, we conclude that:

\[ \frac{dy}{dx} = 0 \]
\[ \frac{d^2y}{dx^2} = 0 \]