`"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 parametric 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 \(t\)

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

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

Calculating the numerator:
\[ = 2(1 + t^2) - 4t^2 = 2 + 2t^2 - 4t^2 = 2 - 2t^2 \]

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

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

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

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

Calculating the numerator:
\[ = -2t(1 + t^2) - 2t(1 - t^2) = -2t - 2t^3 - 2t + 2t^3 = -4t \]

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

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

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

This simplifies to:
\[ \frac{dy}{dx} = \frac{-4t}{2 - 2t^2} \]

### Step 4: Simplify \(\frac{dy}{dx}\)

Factoring out \(-2\) from the denominator:
\[ \frac{dy}{dx} = \frac{-4t}{-2(1 - t^2)} = \frac{2t}{1 - t^2} \]

### Step 5: Evaluate \(\frac{dy}{dx}\) at \(t = 2\)

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

Thus, the final answer is:
\[ \frac{dy}{dx} \text{ at } t = 2 \text{ is } -\frac{4}{3}. \]

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