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Mathematics 47 Online
OpenStudy (anonymous):

derivate: y=cube square of t(t^2+t+t^-1)

OpenStudy (anonymous):

\[\sqrt[3]{t}\left( t^{2} +t+t ^{-1}\right)\]

OpenStudy (anonymous):

This is done by product rule. $$y' = t^{1/3}(2t + 1 - t^{-2}) + \frac{1}{3}t^{-2/3}(t^2 + t + t^{-1})$$

OpenStudy (anonymous):

cool

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