If \(x = \displaystyle\frac{2t}{1 + t^{2}}\) and \(y = \displaystyle\frac{1 - t^{2}}{1 + t^{2}}\), where \(0 \lt t \lt 1\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{x}{y}\)
Differentiating each with respect to t by the quotient rule:
\( \displaystyle \frac{dx}{dt} = \frac{2(1 + t^{2}) - 2t \cdot 2t}{(1 + t^{2})^{2}} = \frac{2(1 - t^{2})}{(1 + t^{2})^{2}} \)
\( \displaystyle \frac{dy}{dt} = \frac{-2t(1 + t^{2}) - (1 - t^{2}) \cdot 2t}{(1 + t^{2})^{2}} = \frac{-4t}{(1 + t^{2})^{2}} \)
By the parametric derivative:
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-4t}{2(1 - t^{2})} = -\frac{2t}{1 - t^{2}} \)
Since \(\frac{x}{y} = \frac{2t}{1 - t^{2}}\):
\( \displaystyle \frac{dy}{dx} = -\frac{x}{y} \)
A quick check: \(x^{2} + y^{2} = 1\), and differentiating gives \(2x + 2yy' = 0\), the same result.