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

Options

A. `(2t)/(t^(2)-1)`
B. `(2t)/(t^(2)+1)`
C. `(2t)/(1-t^(2))`
D. `-1`

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 differentiation. The steps are as follows:

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

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

Using the quotient rule:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(0 - 2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]

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

Calculating 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:
\[ \frac{dy}{dt} = \frac{-4t}{(1 + t^2)^2} \]

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

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

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

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

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

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

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

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

### Step 6: Simplify \(\frac{dy}{dx}\)

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

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

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