Let x=f(t) and y=g(t), where x and y are twice differentiable function. If f'(0)= g'(0) =f''(0) = 2. g''(0) = 6, then the value of `((d^(2)y)/(dx^(2)))_(t=0)` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the value of \(\left(\frac{d^2y}{dx^2}\right)_{t=0}\), we start with the given functions \(x = f(t)\) and \(y = g(t)\), where both functions are twice differentiable.

### Step 1: Find \(\frac{dy}{dx}\)
Using the chain rule, we can express \(\frac{dy}{dx}\) in terms of \(t\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \]

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