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

Options

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

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 \(y\) with respect to \(t\)

Using the quotient rule for differentiation, where \(y = \frac{u}{v}\) with \(u = 2t\) and \(v = 1 + t^2\):

\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \(\frac{du}{dt}\) and \(\frac{dv}{dt}\):

- \(\frac{du}{dt} = 2\)
- \(\frac{dv}{dt} = 2t\)

Now substituting into the quotient rule:

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

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 2: Differentiate \(x\) with respect to \(t\)

Using the same quotient rule for \(x\):

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

Here, \(u = 1 - t^2\) 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:

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

Simplifying the numerator:

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

### Step 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 values we found:

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

Simplifying:

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

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

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