MCQ
If $y = {t^{4/3}} - 3{t^{ - 2/3}}$, then $\frac{{dy}}{{dt}}=$
- A${{2{t^2} + 3} \over {3{t^{5/3}}}}$
- B${{2{t^2} + 3} \over {{t^{5/3}}}}$
- C${{2(2{t^2} + 3)} \over {{t^{5/3}}}}$
- ✓${{2(2{t^2} + 3)} \over {3{t^{5/3}}}}$
$\therefore \frac{{dy}}{{dt}} = \frac{4}{3}{t^{1/3}} + 3 \times \frac{2}{3}{t^{ - 5/3}} $
$= \frac{{4{t^2} + 6}}{{3{t^{5/3}}}} = \frac{{2(2{t^2} + 3)}}{{3{t^{5/3}}}}$.
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