How would you differentiate 2exlnx?

Question

The question is: use calculus to find the minimum value of 2exlogex. I know how to find the minimum value, but i keep getting the answer wrong, so i must be differentiate it wrong. which is what i always do :/

Any help is appreciated (in just differentiating, not finding minimum value), please show steps :) thanks

Answer 1

y=2e^xlnx

y'=2e^xlnx+(2e^x)/x

PRODUCT RULE

Answer 2

Are you asking how to differentiate 2e^x*ln(x)?  or  2e^x*log(e^x) or  2e^(xlog(e^x))

?

Answer 3

f(x)== 2 e x Ln(x)

f'(x)== 2e (ln(x)+1)== 2e Ln(x)+2e

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mmm

f'(x) ==0 ===> 2e Ln(x)+2e==0 then:

Ln(x)== -1

x== e^(-1)

how do you get your solution? i will attend your answer

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Maybe the solution in the book is not correct. i think this is the method to get the solution and x== e^(-1) is minimum value.

sorry for english, i'm italian

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WAIT the solution is right:

let x== e^(-1) then

f(e^(-1))== 2e (e^(-1))Ln(e^(-1))== 2*(-1) == -2

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fiuuuuu, have a good time :)