If ` x^(2)+y^(2) = t +1//t and x^(4) +y^(4) = t^(2) + 1//t^(2) ` then ` (dy)/(dx)` =______________

Options

A. ` 1/x`
B. ` -y/x`
C. ` x/y^(2)`
D. ` -1/y^(2)`

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the equations:

1. \(x^2 + y^2 = t + \frac{1}{t}\)
2. \(x^4 + y^4 = t^2 + \frac{1}{t^2}\)

### Step-by-Step Solution:

**Step 1: Express \(x^4 + y^4\) in terms of \(x^2 + y^2\)**

We can use the identity:
\[ x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \]
Substituting \(x^2 + y^2 = t + \frac{1}{t}\):
\[ x^4 + y^4 = \left(t + \frac{1}{t}\right)^2 - 2x^2y^2 \]
Expanding the square:
\[ x^4 + y^4 = t^2 + 2 + \frac{1}{t^2} - 2x^2y^2 \]

**Step 2: Set the two expressions for \(x^4 + y^4\) equal to each other**

From the problem, we have:
\[ t^2 + \frac{1}{t^2} = t^2 + 2 + \frac{1}{t^2} - 2x^2y^2 \]
Cancelling \(t^2 + \frac{1}{t^2}\) from both sides gives:
\[ 0 = 2 + 2x^2y^2 \]
Thus:
\[ 2x^2y^2 = 2 \implies x^2y^2 = 1 \]

**Step 3: Differentiate the equation \(x^2y^2 = 1\) with respect to \(x\)**

Using the product rule:
\[ \frac{d}{dx}(x^2y^2) = 0 \]
This gives:
\[ 2xy^2 + 2x^2y\frac{dy}{dx} = 0 \]

**Step 4: Solve for \(\frac{dy}{dx}\)**

Rearranging the equation:
\[ 2x^2y\frac{dy}{dx} = -2xy^2 \]
Dividing both sides by \(2x^2y\) (assuming \(x \neq 0\) and \(y \neq 0\)):
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{y}{x} \]