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

Options

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

Explanation

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 will follow these steps:

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

We will first differentiate \(x\) and \(y\) with respect to \(t\).

1. **Differentiate \(x\)**:
\[ x = \frac{1 - t^2}{1 + t^2} \]
Using the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(0 - 2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]
Simplifying:
\[ \frac{dx}{dt} = \frac{-2t(1 + t^2) - 2t(1 - t^2)}{(1 + t^2)^2} = \frac{-2t(1 + t^2 + 1 - t^2)}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2} \]

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

### 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:
\[ \frac{dy}{dx} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} = \frac{2(1 - t^2)}{-4t} = \frac{-(1 - t^2)}{2t} \]

### Final Result

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{-(1 - t^2)}{2t} \]