9 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Idk how that's differentiation but $log|x| /in (-1,1)$ so x$/in$[-e,-1/e]U[1/e,e]
SirLancelotDuLac
why put diffentiation as heading then
Yeah mb...wrong heading :/
x E [-e, -1/e] U [1/e, e]
+solved @serenity. 力 @SirLancelotDuLac
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