If `x=(2t)/(1+t^(2))` and `y=(1-t^(2))/(1+t^(2))`, then `(dy)/(dx)` is equal to

Options

A. `(2t)/(t^(2)+1)`
B. `(2t)/(t^(2)-1)`
C. `(2t)/(1-t^(2))`
D. None

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 \(x\) with respect to \(t\)
We start with the expression for \(x\):
\[ x = \frac{2t}{1 + t^2} \]
Using the quotient rule, where if \(u = 2t\) and \(v = 1 + t^2\), we have:
\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
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 \(t\)
Next, we differentiate \(y\):
\[ y = \frac{1 - t^2}{1 + t^2} \]
Using the same quotient rule:
- Let \(u = 1 - t^2\) and \(v = 1 + t^2\)

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 \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4t/(1 + t^2)^2}{2(1 - t^2)/(1 + t^2)^2} \]
The \((1 + t^2)^2\) cancels out:
\[ \frac{dy}{dx} = \frac{-4t}{2(1 - t^2)} = \frac{-2t}{1 - t^2} \]

### Final Answer
Thus, we find that:
\[ \frac{dy}{dx} = \frac{-2t}{1 - t^2} \]