If `x+y=t-(1)/(t),x^(2)+y^(2)=t^(2)+(1)/(t^(2))`, what is `(dy)/(dx)` equal to?

Options

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

Explanation

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

### Step 1: Use the first equation to express \( x + y \)
We have:
\[ x + y = t - \frac{1}{t} \]

### Step 2: Square both sides
Squaring both sides gives:
\[ (x + y)^2 = \left(t - \frac{1}{t}\right)^2 \]
Expanding both sides, we get:
\[ x^2 + 2xy + y^2 = t^2 - 2 + \frac{1}{t^2} \]

### Step 3: Substitute \( x^2 + y^2 \)
From the second equation, we know:
\[ x^2 + y^2 = t^2 + \frac{1}{t^2} \]
Now substituting this into our equation from Step 2:
\[ t^2 + \frac{1}{t^2} + 2xy = t^2 - 2 + \frac{1}{t^2} \]

### Step 4: Simplify to find \( xy \)
Cancelling \( t^2 + \frac{1}{t^2} \) from both sides:
\[ 2xy = -2 \]
Thus,
\[ xy = -1 \]

### Step 5: Use the product rule to differentiate
Now, we differentiate the product \( xy = -1 \) implicitly with respect to \( x \):
\[ x \frac{dy}{dx} + y \cdot 1 = 0 \]
This simplifies to:
\[ x \frac{dy}{dx} = -y \]
So,
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 6: Substitute the values of \( x \) and \( y \)
From our earlier calculations, we know:
- \( xy = -1 \)
- We can express \( y \) in terms of \( x \) using \( y = -\frac{1}{x} \).

Substituting \( y = -\frac{1}{x} \) into the derivative:
\[ \frac{dy}{dx} = -\frac{-\frac{1}{x}}{x} = \frac{1}{x^2} \]

### Final Answer
Thus, we conclude that:
\[ \frac{dy}{dx} = \frac{1}{x^2} \]

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