If` x^(2) +y^(2) =t-(1)/(t) andx^(4) +y^(4) =t^(2) +(1)/( t^(2)),then (dy)/(dx) =`

Options

A. ` x^(3) y`
B. ` -x^(3)y`
C. ` (1)/( x^(3)y)`
D. ` (-1)/( x^(3)y)`

Answer

To solve the given problem, we will follow a systematic approach using differentiation and algebraic manipulation. Let's break it down step by step.

### Given:
1. \( x^2 + y^2 = t - \frac{1}{t} \) (Equation 1)
2. \( x^4 + y^4 = t^2 + \frac{1}{t^2} \) (Equation 2)

### Step 1: Express \( x^4 + y^4 \) in terms of \( x^2 + y^2 \)
We know that:
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]
Substituting Equation 1 into this expression:
\[ x^4 + y^4 = (t - \frac{1}{t})^2 - 2x^2y^2 \]

### Step 2: Expand the square
Expanding \( (t - \frac{1}{t})^2 \):
\[ (t - \frac{1}{t})^2 = t^2 - 2 + \frac{1}{t^2} \]
Thus,
\[ x^4 + y^4 = t^2 - 2 + \frac{1}{t^2} - 2x^2y^2 \]

### Step 3: Set the two expressions for \( x^4 + y^4 \) equal
From Equation 2, we have:
\[ x^4 + y^4 = t^2 + \frac{1}{t^2} \]
Setting the two expressions equal:
\[ t^2 - 2 + \frac{1}{t^2} - 2x^2y^2 = t^2 + \frac{1}{t^2} \]

### Step 4: Simplify the equation
Cancelling \( t^2 + \frac{1}{t^2} \) from both sides:
\[ -2 + 2x^2y^2 = 0 \]
This simplifies to:
\[ 2x^2y^2 = 2 \quad \Rightarrow \quad x^2y^2 = 1 \]

### Step 5: Differentiate the first equation
Now we differentiate Equation 1 with respect to \( x \):
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(t - \frac{1}{t}) \]
Using the chain rule:
\[ 2x + 2y \frac{dy}{dx} = \frac{dt}{dx} + \frac{1}{t^2} \frac{dt}{dx} \]
Factoring out \( \frac{dt}{dx} \):
\[ 2x + 2y \frac{dy}{dx} = \frac{dt}{dx} \left(1 + \frac{1}{t^2}\right) \]

### Step 6: Express \( \frac{dt}{dx} \)
From the relation \( x^2y^2 = 1 \), we can differentiate:
\[ 2xy^2 \frac{dy}{dx} + 2x^2y = 0 \]
This gives:
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 7: Substitute \( \frac{dy}{dx} \) into the differentiated equation
Substituting \( \frac{dy}{dx} \) back into the differentiated Equation 1:
\[ 2x - 2y \frac{y}{x} = \frac{dt}{dx} \left(1 + \frac{1}{t^2}\right) \]
This simplifies to:
\[ 2x - \frac{2y^2}{x} = \frac{dt}{dx} \left(1 + \frac{1}{t^2}\right) \]

### Step 8: Solve for \( \frac{dy}{dx} \)
Using \( x^2y^2 = 1 \), we can express \( y^2 \) in terms of \( x^2 \):
\[ y^2 = \frac{1}{x^2} \]
Substituting this into the equation gives us:
\[ \frac{dy}{dx} = \frac{y}{x^3} \]

### Final Result
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{y}{x^3} \]