If `x=phi(t),y=Psi(t)," then "(d^(2)y)/(dx^(2))` is equal to

Options

A. `(phi'psi''-psi'phi'')/((phi')^(2))`
B. `(phi'psi''-psi'phi'')/((phi')^(3))`
C. `(psi'')/(phi') -(psi'phi'')/((phi')^(2))`
D. `(psi'')/((phi')^(2))-(psi'phi'')/((phi')^(3))`

Answer

To find \(\frac{d^2y}{dx^2}\) when \(x = \phi(t)\) and \(y = \psi(t)\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dx}\)

Using the chain rule, we can express the derivative of \(y\) with respect to \(x\) as:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\psi'(t)}{\phi'(t)} \]

### Step 2: Find \(\frac{d^2y}{dx^2}\)

To find the second derivative, 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 quotient rule for differentiation, where \(u = \psi'(t)\) and \(v = \phi'(t)\):

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Substituting \(u\) and \(v\):

\[ \frac{d^2y}{dx^2} = \frac{\phi'(t) \psi''(t) - \psi'(t) \phi''(t)}{(\phi'(t))^2} \]

### Step 3: Change of Variables

Since \(\frac{dx}{dt} = \phi'(t)\), we can express \(\frac{d^2y}{dx^2}\) in terms of \(t\):

\[ \frac{d^2y}{dx^2} = \frac{\phi'(t) \psi''(t) - \psi'(t) \phi''(t)}{(\phi'(t))^2} \]

### Step 4: Simplify the Expression

Now, we can simplify the expression further:

\[ \frac{d^2y}{dx^2} = \frac{\phi'(t) \psi''(t)}{(\phi'(t))^2} - \frac{\psi'(t) \phi''(t)}{(\phi'(t))^2} \]

This can be rewritten as:

\[ \frac{d^2y}{dx^2} = \frac{\psi''(t)}{\phi'(t)} - \frac{\psi'(t) \phi''(t)}{(\phi'(t))^2} \]

### Final Result

Thus, the final expression for \(\frac{d^2y}{dx^2}\) is:

\[ \frac{d^2y}{dx^2} = \frac{\psi''(t)}{\phi'(t)} - \frac{\psi'(t) \phi''(t)}{(\phi'(t))^2} \]

This matches with the options provided in the question.