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

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = \frac{2t}{1 + t^2}\) and \(y = \frac{1 - t^2}{1 + t^2}\), we will use the chain rule for derivatives.

### Step 1: Find \(\frac{dx}{dt}\)

Starting with the equation for \(x\):
\[ x = \frac{2t}{1 + t^2} \]
We will differentiate \(x\) with respect to \(t\):
Using the quotient rule:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(2) - (2t)(2t)}{(1 + t^2)^2} \]
Simplifying this:
\[ \frac{dx}{dt} = \frac{2(1 + t^2) - 4t^2}{(1 + t^2)^2} = \frac{2 - 2t^2}{(1 + t^2)^2} = \frac{2(1 - t^2)}{(1 + t^2)^2} \]

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

Now, we differentiate \(y\):
\[ y = \frac{1 - t^2}{1 + t^2} \]
Using the quotient rule again:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]
Simplifying this:
\[ \frac{dy}{dt} = \frac{-2t(1 + t^2) - 2t(1 - t^2)}{(1 + t^2)^2} \]
\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2} \]

### Step 3: 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:
\[ \frac{dy}{dx} = \frac{\frac{-4t}{(1 + t^2)^2}}{\frac{2(1 - t^2)}{(1 + t^2)^2}} \]
The \((1 + t^2)^2\) terms cancel out:
\[ \frac{dy}{dx} = \frac{-4t}{2(1 - t^2)} = \frac{-2t}{1 - t^2} \]
This can also be expressed as:
\[ \frac{dy}{dx} = \frac{2t}{t^2 - 1} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{2t}{t^2 - 1} \]