Find `(d^(2)y)/(dx^(2))` in the following
`x=(2at^(2))/(1+t),y=(3at)/(1+t)`

Answer

To find \(\frac{d^2y}{dx^2}\) given the parametric equations \(x = \frac{2at^2}{1+t}\) and \(y = \frac{3at}{1+t}\), we will follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

First, we differentiate \(x\) and \(y\) with respect to \(t\).

1. **Differentiate \(x\)**:
\[ x = \frac{2at^2}{1+t} \]
Using the quotient rule:
\[ \frac{dx}{dt} = \frac{(1+t)(4at) - 2at^2(1)}{(1+t)^2} = \frac{4at + 4at^2 - 2at^2}{(1+t)^2} = \frac{4at + 2at^2}{(1+t)^2} = \frac{2at(2+t)}{(1+t)^2} \]

2. **Differentiate \(y\)**:
\[ y = \frac{3at}{1+t} \]
Using the quotient rule:
\[ \frac{dy}{dt} = \frac{(1+t)(3a) - 3at(1)}{(1+t)^2} = \frac{3a + 3at - 3at}{(1+t)^2} = \frac{3a}{(1+t)^2} \]

### Step 2: Find \(\frac{dy}{dx}\)

Now, we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{3a}{(1+t)^2}}{\frac{2at(2+t)}{(1+t)^2}} = \frac{3a}{2at(2+t)} = \frac{3}{2t(2+t)} \]

### Step 3: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\)

Now, we differentiate \(\frac{dy}{dx}\) with respect to \(t\) to find \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} \]

1. **Differentiate \(\frac{dy}{dx}\)**:
\[ \frac{dy}{dx} = \frac{3}{2t(2+t)} \]
Using the quotient rule:
\[ \frac{d}{dt}\left(\frac{3}{2t(2+t)}\right) = \frac{0 \cdot (2t(2+t)) - 3 \cdot (2(2+t) + 2t)}{(2t(2+t))^2} = \frac{-3(4 + 2t)}{(2t(2+t))^2} \]

2. **Substituting \(\frac{dx}{dt}\)**:
\[ \frac{dx}{dt} = \frac{2at(2+t)}{(1+t)^2} \]

Putting it all together:
\[ \frac{d^2y}{dx^2} = \frac{-3(4 + 2t)}{(2t(2+t))^2} \cdot \frac{(1+t)^2}{2at(2+t)} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{-3(4 + 2t)(1+t)^2}{2a(2t(2+t))^3} \]

### Final Result

Thus, the second derivative \(\frac{d^2y}{dx^2}\) is:
\[ \frac{d^2y}{dx^2} = \frac{-3(4 + 2t)(1+t)^2}{2a(2t(2+t))^3} \]