If y = t^ 4/3 - 3 t^ - 2/3 , then dy dt = - Vidyadip

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
${{2(2{t^2} + 3)} \over {{t^{5/3}}}}$
✓
${{2(2{t^2} + 3)} \over {3{t^{5/3}}}}$

✓

Answer

Correct option: D.

${{2(2{t^2} + 3)} \over {3{t^{5/3}}}}$

d
(d) $y = {t^{4/3}} - 3{t^{ - 2/3}}$

$\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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