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lgbasallote
 2 years ago
Best ResponseYou've already chosen the best response.1y = x^x ln both sides... lny = lnx^x via reverse power rule... lny = xlnx you can use implicit differentiation now...you do know that right/

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0so d/dx(lny) = 1/y * y' and d/dx(xlnx) = x*1/x .....????

The_geek
 2 years ago
Best ResponseYou've already chosen the best response.1it goes lyk tis logy= xlogx diff wrt to x on bth sides so on lhs its 1/y dy/dx=1+logx by product rule on rhs u get 1+logx thn dy/dx =(1+logx)y wer y is nthin bt =x^x

gordonj005
 2 years ago
Best ResponseYou've already chosen the best response.0one of my favourite derivatives. an alternative to the ones suggested above: y = e^log(x^x) = e^xlog(x) which you can differentiate using the chain rule

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0thx guys you just walked me through my first ever log diff problem
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