- #1

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is it 0 or 1?

i have this problem in the book

y = e^(x^x)

and the ANSWER is

y' = (x^x) e^(x^x) [1 + lnx]

my FIRST solution and i assumed (ln e=1) ; i treated it a constant so it must be "0"

where (d/dx) c = 0

b^x = r ---> x = logb r

(d/dx) ln x = (1/x) (du/dx)

ln y = ln e^(x^x) <--- i use ln both sides

(y'/y) = [ (x^x) . (ln e) ] <--- product rule

u v ------> du v + u dv

(y'/y) = [(x^x ln x) (ln e)] + [(x^x) (1/e) (e.0)]<---must be zero ;(d/dx) ln e = 0 because ln e = 1

(y'/y) = [ (x^x ln x) (ln e) + 0 ]

y' = [(x^x ln x) (1)] (y) ---> multiply by (y/y) to the to both side to get y'

my final answerr is;

y' = (x^x ln x) [ e^(x^x) ] <------ where [y = e^(x^x)]

= which is in contradiction to the ANSWER IN THE BOOK

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and the other solution i came up with, to get the right answer in the book is;

IF (d/dx) (ln e) is = to 1

ln y = ln e^(x^x)

(y'/y) = [ (x^x) . (ln e) ]

(y'/y) = [(x^x ln x) (ln e)] + [(x^x) (1)]

(y'/y) = [(x^x ln x) (ln e)] + [(x^x)]

y' = [(x^x ln x) (1)] + [(x^x)] (y) ;(ln e) = 1

anwers is

y' = (x^x) e^(x^x) [1 + lnx] ;factored out (x^x) and y = e^(x^x)

which conforms to the answer in the book

is my solution correct? pls enlighten me.