If \(x = 2t^{2} + 3\) and \(y = t^{3} - 4t\), then the value of \(\displaystyle\frac{dy}{dx}\) at \(t = 2\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
For a curve in parametric form, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Differentiating each with respect to t:
\( \displaystyle \frac{dx}{dt} = 4t \)
\( \displaystyle \frac{dy}{dt} = 3t^{2} - 4 \)
So
\( \displaystyle \frac{dy}{dx} = \frac{3t^{2} - 4}{4t} \)
At t = 2:
\( \displaystyle \frac{dy}{dx} = \frac{12 - 4}{8} \)
\( = 1 \)