- #1
dimasalang
- 21
- 1
please enlighten me, what is the derivative of (ln e) ; natural log e
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
=============================================================================
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) = 1anwers 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.
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
=============================================================================
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) = 1anwers 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.