Differentiate.

Jason Administrator Staff member Administrator Moderator Jan 25, 2018 2,521 547 113 jasonyankee.info Oct 11, 2018 #2 \(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\) Without product rule: \(\displaystyle y = x^{2/3}(2x - x^{2})\) \(\displaystyle y = 2x^{5/3} - x^{8/3}\) \(\displaystyle y' = \dfrac{5}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\) Last edited: Oct 11, 2018

\(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\) Without product rule: \(\displaystyle y = x^{2/3}(2x - x^{2})\) \(\displaystyle y = 2x^{5/3} - x^{8/3}\) \(\displaystyle y' = \dfrac{5}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\)

MarkFL La Villa Strangiato Staff member Administrator Moderator Math Helper Jan 25, 2018 3,573 4,322 113 St. Augustine Oct 11, 2018 #3 Jason, the first term on the RHS of your final result has the original coefficient of 2 left off.

Jason Administrator Staff member Administrator Moderator Jan 25, 2018 2,521 547 113 jasonyankee.info Oct 11, 2018 #4 MarkFL said: Jason, the first term on the RHS of your final result has the original coefficient of 2 left off. Click to expand... \(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\) Without product rule: \(\displaystyle y = x^{2/3}(2x - x^{2})\) \(\displaystyle y = 2x^{5/3} - x^{8/3}\) \(\displaystyle y' = 2(\dfrac{5}{3})x^{2/3} - \dfrac{8}{3}x^{5/3}\) \(\displaystyle y' = \dfrac{10}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\) Reactions: harpazo and MarkFL

MarkFL said: Jason, the first term on the RHS of your final result has the original coefficient of 2 left off. Click to expand... \(\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})\) Without product rule: \(\displaystyle y = x^{2/3}(2x - x^{2})\) \(\displaystyle y = 2x^{5/3} - x^{8/3}\) \(\displaystyle y' = 2(\dfrac{5}{3})x^{2/3} - \dfrac{8}{3}x^{5/3}\) \(\displaystyle y' = \dfrac{10}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}\)