MCQ
If $x = {t^2}$, $y = {t^3}$, then ${{{d^2}y} \over {d{x^2}}} =$
  • A
    ${3 \over 2}$
  • ${3 \over {(4t)}}$
  • C
    ${3 \over {2(t)}}$
  • D
    ${{3t} \over 2}$

Answer

Correct option: B.
${3 \over {(4t)}}$
b
(b) $\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{3{t^2}}}{{2t}} = \frac{3}{2}t = \frac{3}{2}\sqrt x $

$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{3}{{4\sqrt x }} = \frac{3}{{4t}}$.

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