Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=(1-t^(2))/(1+t^(2)),y=(2t)/(1+t^(2))`

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 will follow these steps:

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

Given:
\[ x = \frac{1 - t^2}{1 + t^2} \]

We will use the quotient rule for differentiation, which states that if \(u\) and \(v\) are functions of \(t\), then:
\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^2} \]

Here, \(u = 1 - t^2\) and \(v = 1 + t^2\).

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):
\[ \frac{du}{dt} = -2t, \quad \frac{dv}{dt} = 2t \]

Now applying the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]

### Step 2: Simplify \(\frac{dx}{dt}\)

Expanding the numerator:
\[ = \frac{-2t(1 + t^2) - 2t(1 - t^2)}{(1 + t^2)^2} \]
\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} \]
\[ = \frac{-4t}{(1 + t^2)^2} \]

Thus, we have:
\[ \frac{dx}{dt} = \frac{-4t}{(1 + t^2)^2} \]

### Step 3: Differentiate \(y\) with respect to \(t\)

Given:
\[ y = \frac{2t}{1 + t^2} \]

Using the quotient rule again:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(2) - (2t)(2t)}{(1 + t^2)^2} \]
\[ = \frac{2(1 + t^2) - 4t^2}{(1 + t^2)^2} \]
\[ = \frac{2 - 2t^2}{(1 + t^2)^2} \]

Thus, we have:
\[ \frac{dy}{dt} = \frac{2(1 - t^2)}{(1 + t^2)^2} \]

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

### Step 5: Simplify \(\frac{dy}{dx}\)

The \((1 + t^2)^2\) terms cancel out:
\[ \frac{dy}{dx} = \frac{2(1 - t^2)}{-4t} \]
\[ = \frac{-(1 - t^2)}{2t} \]

### Final Result

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