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

Options

A. `(-y)/(x)`
B. `(y)/(x)`
C. `(-x)/(y)`
D. `(x)/(y)`

Explanation

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

### Step 1: Differentiate \(x\) with respect to \(t\)
We start by finding \(\frac{dx}{dt}\).

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

Using the quotient rule for differentiation, where \(f(t) = 1 - t^2\) and \(g(t) = 1 + t^2\):

\[ \frac{dx}{dt} = \frac{f' \cdot g - g' \cdot f}{g^2} \]

Calculating \(f'\) and \(g'\):

- \(f' = -2t\)
- \(g' = 2t\)

Now substituting into the quotient rule:

\[ \frac{dx}{dt} = \frac{(-2t)(1 + t^2) - (2t)(1 - t^2)}{(1 + t^2)^2} \]

Simplifying the numerator:

\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2} \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Next, we find \(\frac{dy}{dt}\).

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

Using the quotient rule again, where \(f(t) = 2t\) and \(g(t) = 1 + t^2\):

\[ \frac{dy}{dt} = \frac{f' \cdot g - g' \cdot f}{g^2} \]

Calculating \(f'\) and \(g'\):

- \(f' = 2\)
- \(g' = 2t\)

Now substituting into the quotient rule:

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

Simplifying the numerator:

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

### Step 3: Find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the chain rule:

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

Substituting the derivatives we found:

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

The \((1 + t^2)^2\) terms cancel out:

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

### Step 4: Express in terms of \(x\) and \(y\)
From the original equations, we have:

- \(x = \frac{1 - t^2}{1 + t^2}\)
- \(y = \frac{2t}{1 + t^2}\)

Thus, we can express \(1 - t^2\) in terms of \(x\):

\[ 1 - t^2 = x(1 + t^2) \implies 1 - t^2 = x + xt^2 \]

Rearranging gives:

\[ t^2(1 + x) = 1 - x \implies t^2 = \frac{1 - x}{1 + x} \]

Substituting \(t^2\) back into our expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{-(1 - t^2)}{2t} = \frac{-(1 - \frac{1 - x}{1 + x})}{2\sqrt{\frac{1 - x}{1 + x}}} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{-\frac{2x}{1 + x}}{2\sqrt{\frac{1 - x}{1 + x}}} = -\frac{x}{\sqrt{(1 - x)(1 + x)}} \]

### Final Answer
Thus, the final answer is:

\[ \frac{dy}{dx} = -\frac{x}{\sqrt{(1 - x)(1 + x)}} \]