Respuesta :

biancamilburn3

hmm - i think thr answer is 2 ln t + (ln t)^2
[tex]y=t(\ln{t})^2 \\y'=[t(\ln{t})^2]'=t'(\ln{t})^2+t[(\ln{t})^2]'=(\ln{t})^2+2t\ln{t}(\ln{t})'= \\ \\=(\ln{t})^2+2t\frac{1}{t}\ln{t} =(\ln{t})^2+2\ln{t} =\ln{t}(\ln{t}+2) [/tex]

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