Xy'=yln y/x
y'= (2xy)/ (x^2- x^2)
please
thank you so much!...
Mathematics, 26.07.2021 19:10 xxtonixwilsonxx
Xy'=yln y/x
y'= (2xy)/ (x^2- x^2)
please
thank you so much!
Answers: 1
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