Question

Need help solving this differentiation problem please. Thank you.

Answer 1

first convert the terms to fractional exponents

u  =  t^2/3  - 3t^3/2

differentiating

u'  =  2/3 t^ (2/3 - 1) - 3* 3/2  t^(3/2 - 1)

     =  2/3 t ^(-1/3) - 9/2 t ^(1/2)

     =  2 / (3∛t) -  9 √ t / 2   in radical form

Answer 2

$\bf u=\sqrt[3]{t^2}-3\sqrt{t^3}\implies u=t^{\frac{2}{3}}-3t^{\frac{3}{2}}\implies \cfrac{du}{dt}=\stackrel{power~rule}{\cfrac{2}{3}t^{\frac{2}{3}-1}~~-~~3\cdot \cfrac{3}{2}t^{\frac{3}{2}-1}} \\\\\\ \cfrac{du}{dt}=\cfrac{2}{3}t^{-\frac{1}{3}}~~-~~\cfrac{9}{2}t^{\frac{1}{2}}\implies \cfrac{du}{dt}=\cfrac{2}{3t^{\frac{1}{3}}}~~-~~\cfrac{9t^{\frac{1}{2}}}{2} \\\\\\ \cfrac{du}{dt}=\cfrac{2}{3\sqrt[3]{t}}-\cfrac{9\sqrt{t}}{2}$