`"If "x=(2t)/(1+t^(2)),y=(1-t^(2))/(1+t^(2))," then find "(dy)/(dx)" at "t=2.`

Explanation

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

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

The formula for \(x\) is:
\[ x = \frac{2t}{1+t^2} \]

Using the quotient rule, where \(u = 2t\) and \(v = 1+t^2\):
\[ \frac{dx}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \]

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

Now substituting into the quotient rule:
\[ \frac{dx}{dt} = \frac{(1+t^2)(2) - (2t)(2t)}{(1+t^2)^2} \]
\[ = \frac{2 + 2t^2 - 4t^2}{(1+t^2)^2} = \frac{2 - 2t^2}{(1+t^2)^2} = \frac{2(1-t^2)}{(1+t^2)^2} \]

### Step 2: Differentiate \(y\) with respect to \(t\)

The formula for \(y\) is:
\[ y = \frac{1-t^2}{1+t^2} \]

Using the quotient rule again, where \(u = 1-t^2\) and \(v = 1+t^2\):
\[ \frac{dy}{dt} = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \]

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

Now substituting into the quotient rule:
\[ \frac{dy}{dt} = \frac{(1+t^2)(-2t) - (1-t^2)(2t)}{(1+t^2)^2} \]
\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1+t^2)^2} = \frac{-4t}{(1+t^2)^2} \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{-4t/(1+t^2)^2}{2(1-t^2)/(1+t^2)^2} \]
\[ = \frac{-4t}{2(1-t^2)} = \frac{-2t}{1-t^2} \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(t=2\)

Substituting \(t=2\):
\[ \frac{dy}{dx} = \frac{-2(2)}{1-(2^2)} = \frac{-4}{1-4} = \frac{-4}{-3} = \frac{4}{3} \]

### Final Answer
\[ \frac{dy}{dx} \text{ at } t=2 \text{ is } \frac{4}{3} \]