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

Options

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

Answer

To find \(\frac{dy}{dx}\) 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 \(t\)

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

Using the quotient rule, where if \(u = 1 - t^2\) and \(v = 1 + t^2\), then:
\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating the derivatives:
- \(\frac{du}{dt} = -2t\)
- \(\frac{dv}{dt} = 2t\)

Now substituting:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]
\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} \]
\[ = \frac{-4t}{(1 + t^2)^2} \]

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

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

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

### Step 3: 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 \((1 + t^2)^2\) cancels out:
\[ = \frac{2(1 - t^2)}{-4t} \]
\[ = \frac{-(1 - t^2)}{2t} \]

### Step 4: Express in terms of \(x\) and \(y\)

Recall:
\[ x = \frac{1 - t^2}{1 + t^2} \quad \Rightarrow \quad 1 - t^2 = x(1 + t^2) \]
Thus:
\[ 1 - t^2 = x + xt^2 \]
Rearranging gives:
\[ t^2(1 + x) = 1 - x \quad \Rightarrow \quad t^2 = \frac{1 - x}{1 + x} \]

Substituting \(t^2\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-(1 - t^2)}{2t} = \frac{-(x + xt^2)}{2t} \]
Using \(y = \frac{2t}{1 + t^2}\):
\[ t = \frac{y(1 + t^2)}{2} \]
Thus:
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{x}{y} \]