If `x=(2t)/(1+t^2),y=(1-t^2)/(1+t^2),t h e nfin d(dy)/(dx)a tt=2.`

Explanation

To find \(\frac{dy}{dx}\) at \(t = 2\), we start with the parametric equations given:

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

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

Using the quotient rule, where if \(u = 2t\) and \(v = 1 + t^2\), then:

\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):

- \(\frac{du}{dt} = 2\)
- \(\frac{dv}{dt} = 2t\)

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 \(t\)

Using the quotient rule again, where \(u = 1 - t^2\) and \(v = 1 + t^2\):

\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):

- \(\frac{du}{dt} = -2t\)
- \(\frac{dv}{dt} = 2t\)

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, we have:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

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 \(t = 2\)

Now substituting \(t = 2\):

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

Thus, the value of \(\frac{dy}{dx}\) at \(t = 2\) is:

\[ \frac{4}{3} \]