If `y=((alphax+beta)/(gammax+delta)),` then `2(dy)/(dx).(d^(3)y)/(dx^(3))` is

Options

A. `7((d^(2)y)/(dx^(2)))^(2)`
B. `5((d^(2)y)/(dx^(2)))^(2)`
C. `3((d^(2)y)/(dx^(2)))^(2)`
D. `((d^(2)y)/(dx^(2)))^(2)`

Explanation

To solve the problem, we need to find the expression for \( 2 \frac{dy}{dx} \cdot \frac{d^3y}{dx^3} \) given that \( y = \frac{\alpha x + \beta}{\gamma x + \delta} \).

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
Let \( u = \alpha x + \beta \) and \( v = \gamma x + \delta \).

2. **First Derivative \( \frac{dy}{dx} \)**:
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where \( \frac{du}{dx} = \alpha \) and \( \frac{dv}{dx} = \gamma \).
Thus,
\[ \frac{dy}{dx} = \frac{(\gamma x + \delta)(\alpha) - (\alpha x + \beta)(\gamma)}{(\gamma x + \delta)^2} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{\alpha \gamma x + \alpha \delta - \alpha \gamma x - \beta \gamma}{(\gamma x + \delta)^2} = \frac{\alpha \delta - \beta \gamma}{(\gamma x + \delta)^2} \]

3. **Second Derivative \( \frac{d^2y}{dx^2} \)**:
We differentiate \( \frac{dy}{dx} \) again using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(\gamma x + \delta)^2 \cdot 0 - (\alpha \delta - \beta \gamma) \cdot 2(\gamma x + \delta)(\gamma)}{(\gamma x + \delta)^4} \]
Simplifying this gives:
\[ \frac{d^2y}{dx^2} = -\frac{2\gamma(\alpha \delta - \beta \gamma)}{(\gamma x + \delta)^3} \]

4. **Third Derivative \( \frac{d^3y}{dx^3} \)**:
Differentiate \( \frac{d^2y}{dx^2} \):
\[ \frac{d^3y}{dx^3} = -\frac{d}{dx} \left( \frac{2\gamma(\alpha \delta - \beta \gamma)}{(\gamma x + \delta)^3} \right) \]
Using the quotient rule again:
\[ \frac{d^3y}{dx^3} = -\frac{0 - 2\gamma(\alpha \delta - \beta \gamma)(-3\gamma)}{(\gamma x + \delta)^4} \]
Simplifying gives:
\[ \frac{d^3y}{dx^3} = \frac{6\gamma^2(\alpha \delta - \beta \gamma)}{(\gamma x + \delta)^4} \]

5. **Calculate \( 2 \frac{dy}{dx} \cdot \frac{d^3y}{dx^3} \)**:
Now we can calculate:
\[ 2 \frac{dy}{dx} \cdot \frac{d^3y}{dx^3} = 2 \left( \frac{\alpha \delta - \beta \gamma}{(\gamma x + \delta)^2} \right) \left( \frac{6\gamma^2(\alpha \delta - \beta \gamma)}{(\gamma x + \delta)^4} \right) \]
This simplifies to:
\[ = \frac{12\gamma^2(\alpha \delta - \beta \gamma)^2}{(\gamma x + \delta)^6} \]

### Final Result:
Thus, the expression for \( 2 \frac{dy}{dx} \cdot \frac{d^3y}{dx^3} \) is:
\[ \frac{12\gamma^2(\alpha \delta - \beta \gamma)^2}{(\gamma x + \delta)^6} \]