If `x=t+(1)/(t),y=t-(1)/(t)," where "tne0," then: "(d^(2)y)/(dx^(2))=`

Options

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

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) given the equations:

\[ x = t + \frac{1}{t}, \quad y = t - \frac{1}{t} \]

where \( t \neq 0 \).

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

First, we differentiate \( x \) with respect to \( t \):

\[ \frac{dx}{dt} = \frac{d}{dt}\left(t + \frac{1}{t}\right) = 1 - \frac{1}{t^2} \]

Next, we differentiate \( y \) with respect to \( t \):

\[ \frac{dy}{dt} = \frac{d}{dt}\left(t - \frac{1}{t}\right) = 1 + \frac{1}{t^2} \]

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

Using the chain rule, we find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \]

### Step 3: Simplify \( \frac{dy}{dx} \)

To simplify \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{(1 + \frac{1}{t^2})}{(1 - \frac{1}{t^2})} = \frac{t^2 + 1}{t^2 - 1} \]

### Step 4: Differentiate \( \frac{dy}{dx} \) with respect to \( t \)

Now we need to differentiate \( \frac{dy}{dx} \) with respect to \( t \):

Let \( u = t^2 + 1 \) and \( v = t^2 - 1 \). Then:

\[ \frac{dy}{dx} = \frac{u}{v} \]

Using 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 \( \frac{dv}{dt} \):

\[ \frac{du}{dt} = 2t, \quad \frac{dv}{dt} = 2t \]

So we have:

\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{(t^2 - 1)(2t) - (t^2 + 1)(2t)}{(t^2 - 1)^2} \]

### Step 5: Simplify \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \)

Simplifying the numerator:

\[ = \frac{2t(t^2 - 1 - t^2 - 1)}{(t^2 - 1)^2} = \frac{2t(-2)}{(t^2 - 1)^2} = \frac{-4t}{(t^2 - 1)^2} \]

### Step 6: Find \( \frac{dx}{dt} \)

We already calculated \( \frac{dx}{dt} \):

\[ \frac{dx}{dt} = 1 - \frac{1}{t^2} \]

### Step 7: Find \( \frac{d^2y}{dx^2} \)

Now we can find \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{-4t}{(t^2 - 1)^2}}{1 - \frac{1}{t^2}} \]

### Step 8: Final simplification

Substituting \( \frac{dx}{dt} \):

\[ \frac{d^2y}{dx^2} = \frac{-4t}{(t^2 - 1)^2} \cdot \frac{t^2}{t^2 - 1} = \frac{-4t^3}{(t^2 - 1)^3} \]

Thus, the final answer is:

\[ \frac{d^2y}{dx^2} = -\frac{4t^3}{(t^2 - 1)^3} \]