If \(x = 2t + 1\) and \(y = \displaystyle\frac{4}{t}\), where \(t \ne 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{2}{t^{2}}\)
Differentiating each with respect to t:
\( \displaystyle \frac{dx}{dt} = 2 \)
\( \displaystyle \frac{dy}{dt} = -\frac{4}{t^{2}} \)
By the parametric derivative:
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{4}{t^{2}} \cdot \frac{1}{2} = -\frac{2}{t^{2}} \)