If x = 3t2 + 5t + 6 and y = –4t3 &

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

If x = 3t² + 5t + 6 and y = –4t³ – 2t² + 5t + 7, t ≠ $\frac{-5}{6}$then the value of $\frac{dy}{dx}$is

Options:

–2t + 1

$\frac{-12 t^2-4 t+5}{6 t+5} $

$\frac{-4 t^3-2 t^2+5 t+7}{3 t^2+5 t+6} $

$ \frac{-4 t^3-2 t^2+5 t+7}{6 t+5} $

Correct Answer:

$\frac{-12 t^2-4 t+5}{6 t+5} $

Explanation:

$\frac{dx}{dt}=6t+5$, $\frac{dy}{dt}=-12t^2-4t+5$

$∴\frac{dy}{dx}= \frac{dy/dt}{dx/dt} = \frac{-12t^2-4t+5}{6t+5}$

So, option 2 is correct.