Question

Need help solving this differentiation problem please. Thank you.

Answer 1

`\bf u=\sqrt[3]{t^2}-3√(t^3)\implies u=t^{(2)/(3)}-3t^{(3)/(2)}\implies \cfrac{du}{dt}=\stackrel{power~rule}{\cfrac{2}{3}t^{(2)/(3)-1}~~-~~3\cdot \cfrac{3}{2}t^{(3)/(2)-1}} \\\\\\ \cfrac{du}{dt}=\cfrac{2}{3}t^{-(1)/(3)}~~-~~\cfrac{9}{2}t^{(1)/(2)}\implies \cfrac{du}{dt}=\cfrac{2}{3t^{(1)/(3)}}~~-~~\cfrac{9t^{(1)/(2)}}{2} \\\\\\ \cfrac{du}{dt}=\cfrac{2}{3\sqrt[3]{t}}-\cfrac{9√(t)}{2}`

Answer 2

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