If `x^(2)+y^(2)=t-(1)/(t)andx^(4)+y^(4)=t^(2)+(1)/(t_(2)),` then `((dy)/(dx))_((1.1))` is…………

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) at the point (1, 1) 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: Write down the equations.**
We have:
\[ x^2 + y^2 = t - \frac{1}{t} \quad (1) \]
\[ x^4 + y^4 = t^2 + \frac{1}{t^2} \quad (2) \]

**Step 2: Square the first equation.**
Squaring equation (1):
\[ (x^2 + y^2)^2 = \left(t - \frac{1}{t}\right)^2 \]
Expanding both sides:
\[ x^4 + y^4 + 2x^2y^2 = t^2 - 2 + \frac{1}{t^2} \]

**Step 3: Substitute equation (2) into the squared equation.**
From equation (2), we know:
\[ x^4 + y^4 = t^2 + \frac{1}{t^2} \]
Substituting this into the expanded equation gives:
\[ t^2 + \frac{1}{t^2} + 2x^2y^2 = t^2 - 2 + \frac{1}{t^2} \]

**Step 4: Simplify the equation.**
Cancelling \(t^2 + \frac{1}{t^2}\) from both sides:
\[ 2x^2y^2 = -2 \]
Dividing by 2:
\[ x^2y^2 = -1 \]

**Step 5: Solve for \(y^2\).**
Rearranging gives:
\[ y^2 = -\frac{1}{x^2} \]

**Step 6: Differentiate both sides with respect to \(x\).**
Differentiating:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}\left(-\frac{1}{x^2}\right) \]
Using the chain rule on the left:
\[ 2y \frac{dy}{dx} = 2x^{-3} \]

**Step 7: Solve for \(\frac{dy}{dx}\).**
Rearranging gives:
\[ \frac{dy}{dx} = \frac{x^{-3}}{y} \]

**Step 8: Evaluate \(\frac{dy}{dx}\) at the point (1, 1).**
Substituting \(x = 1\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{1^{-3}}{1} = 1 \]

Thus, the final answer is:
\[ \frac{dy}{dx} \text{ at } (1, 1) = 1 \]