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

Options

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

Answer

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

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

We start by differentiating both \(x\) and \(y\) with respect to \(t\).

1. **Differentiate \(x\)**:
\[ x = \frac{1 - t^2}{1 + t^2} \]
Using the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]
Simplifying the numerator:
\[ = \frac{-2t(1 + t^2) - 2t(1 - t^2)}{(1 + t^2)^2} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2} \]

2. **Differentiate \(y\)**:
\[ y = \frac{2t}{1 + t^2} \]
Again using the quotient rule:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(2) - (2t)(2t)}{(1 + t^2)^2} = \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2 - 2t^2}{(1 + t^2)^2} = \frac{2(1 - t^2)}{(1 + t^2)^2} \]

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

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} = \frac{2(1 - t^2)}{-4t} = \frac{-(1 - t^2)}{2t} \]

### Step 3: Simplify the expression

Thus, we have:
\[ \frac{dy}{dx} = -\frac{1 - t^2}{2t} \]

### Step 4: Final expression in terms of \(x\) and \(y\)

To express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\):
From the original equations, we know:
- \(x = \frac{1 - t^2}{1 + t^2}\)
- \(y = \frac{2t}{1 + t^2}\)

We can express \(1 - t^2\) and \(2t\) in terms of \(x\) and \(y\):
- \(1 - t^2 = x(1 + t^2)\)
- \(2t = y(1 + t^2)\)

From \(y = \frac{2t}{1 + t^2}\), we can express \(t\) in terms of \(y\) and \(x\):
\[ t = \frac{y(1 + t^2)}{2} \]

Substituting back into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{x(1 + t^2)}{2t} \]

### Final Result

Thus, the final answer is:
\[ \frac{dy}{dx} = -\frac{x}{y} \]