if ` x=t +(1)/(t) , y=t-(1)/(t) ,` then `(d^2 y)/( dx^2)` is equal to

Options

A. `-4t(t^2 -1)^(-2)`
B. `-4t^3 (t^2 -1)^(-3)`
C. `(t^2 +1)(t^2-1)^(-1)`
D. `-4t^2 (t^2 -1)^(-2)`

Answer

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

### Step 1: Differentiate \(y\) with respect to \(t\)
We start by differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}\left(t - \frac{1}{t}\right) = 1 + \frac{1}{t^2} \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Next, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}\left(t + \frac{1}{t}\right) = 1 - \frac{1}{t^2} \]

### Step 3: Find \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the formula:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{(1 + \frac{1}{t^2})}{(1 - \frac{1}{t^2})} = \frac{t^2 + 1}{t^2 - 1} \]

### Step 4: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)
Next, we differentiate \(\frac{dy}{dx}\) with respect to \(t\):
Let \(u = t^2 + 1\) and \(v = t^2 - 1\). We will use the quotient rule:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
\[ \frac{du}{dt} = 2t, \quad \frac{dv}{dt} = 2t \]
Now substituting back:
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(t^2 - 1)(2t) - (t^2 + 1)(2t)}{(t^2 - 1)^2} \]
This simplifies to:
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2t(t^2 - 1 - t^2 - 1)}{(t^2 - 1)^2} = \frac{-4t}{(t^2 - 1)^2} \]

### Step 5: Find \(\frac{d^2y}{dx^2}\)
Now we can find \(\frac{d^2y}{dx^2}\) using the chain rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{\frac{dx}{dt}} = \frac{-4t}{(t^2 - 1)^2} \cdot \frac{1}{1 - \frac{1}{t^2}} \]
Substituting \(\frac{dx}{dt}\):
\[ \frac{d^2y}{dx^2} = \frac{-4t}{(t^2 - 1)^2} \cdot \frac{t^2}{t^2 - 1} = \frac{-4t^3}{(t^2 - 1)^3} \]

### Final Result
Thus, the final result is:
\[ \frac{d^2y}{dx^2} = \frac{-4t^3}{(t^2 - 1)^3} \]