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Mathematics 19 Online
OpenStudy (anonymous):

whats the derivative of (lnt)^2

OpenStudy (dumbcow):

use chain rule =2*ln(t)*1/t

OpenStudy (anonymous):

srry my question was actually the integral of this problem

OpenStudy (anonymous):

lol srry i was asking for the derivative

OpenStudy (anonymous):

thank you for your help

OpenStudy (dumbcow):

oh ok use integration by parts then u=ln(t) dv=ln(t) du= 1/t v = tlnt - t

OpenStudy (anonymous):

can i just say that my u is (lnt)^2 and that my dv is 1

OpenStudy (dumbcow):

yeah i guess that will work too

OpenStudy (anonymous):

k ty

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