The second derivative of a single valued function parametrically represented by `x=phi(t) and y=psi(t)` (where `phi(t) and psi(t)` are different function and `phi'(t)ne0)` is given by

Options

A. `(d^(2)y)/(dx^(2))=(((dx)/(dt))((d^(2)y)/(dt^(2)))-((d^(2)x)/(dt^(2)))((dy)/(dt)))/(((dx)/(dt))^(3))`
B. `(d^(2)y)/(dx^(2))=(((dx)/(dt))((d^(2)y)/(dt^(2)))-((d^(2)x)/(dt^(2)))((dy)/(dt)))/(((dx)/(dt))^(2))`
C. `(d^(2)y)/(dx^(2))=(((d^(2)x)/(dt))((dy)/(dt))-(dx)/(dt)((d^(2)y)/(dt^(2))))/(((dx)/(dt))^(3))`
D. `(d^(2)y)/(dx^(2))=(((d^(2)x)/(dt))((dy)/(dt))-((d^(2)y)/(dt^(2)))((dy)/(dt)))/(((dy)/(dt))^(3))`

Explanation

To find the second derivative of a single-valued function parametrically represented by \(x = \phi(t)\) and \(y = \psi(t)\), we follow these steps:

### Step 1: Find the first derivatives
We start by finding the first derivatives of \(x\) and \(y\) with respect to \(t\):
\[ \frac{dx}{dt} = \phi'(t) \]
\[ \frac{dy}{dt} = \psi'(t) \]

### Step 2: Find the first derivative of \(y\) with respect to \(x\)
Using the chain rule, we can express the first derivative of \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi'(t)}{\phi'(t)} \]

### Step 3: Find the second derivative of \(y\) with respect to \(x\)
To find the second derivative \(\frac{d^2y}{dx^2}\), we need to differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \]
Using the chain rule again, we can rewrite this as:
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

### Step 4: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)
Now, we differentiate \(\frac{dy}{dx}\) with respect to \(t\):
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{\psi'(t)}{\phi'(t)}\right) \]
Using the quotient rule:
\[ \frac{d}{dt}\left(\frac{\psi'(t)}{\phi'(t)}\right) = \frac{\phi'(t) \cdot \psi''(t) - \psi'(t) \cdot \phi''(t)}{(\phi'(t))^2} \]

### Step 5: Substitute back into the expression for \(\frac{d^2y}{dx^2}\)
Now we substitute this back into our expression for \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \left(\frac{\phi'(t) \cdot \psi''(t) - \psi'(t) \cdot \phi''(t)}{(\phi'(t))^2}\right) \cdot \frac{1}{\phi'(t)} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{\phi'(t) \cdot \psi''(t) - \psi'(t) \cdot \phi''(t)}{(\phi'(t))^3} \]

### Final Result
Thus, the second derivative of \(y\) with respect to \(x\) is given by:
\[ \frac{d^2y}{dx^2} = \frac{\psi''(t) \cdot \phi'(t) - \psi'(t) \cdot \phi''(t)}{(\phi'(t))^3} \]