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OpenStudy (anonymous):
anyone please help!
OpenStudy (anonymous):
do you know logarithmic differentiation?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
y = x^ln(x), then what?
OpenStudy (anonymous):
(i found the first derivative which is (2lnx/x)x^lnx but then deriveing it again is trouble for me
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OpenStudy (anonymous):
product rule
OpenStudy (anonymous):
but i did it a few times and got differnt answers
OpenStudy (anonymous):
here is a formula
(fgh)' = f' gh + f g' h + fg h'
(2lnx/x)x^lnx = (2lnx) (1/x) x^ln(x)
derivative of that is:
(2/x) (1/x) x^ln(x) + (2lnx) (-1/x^2) x^ln(x) + (2lnx) (1/x) (2lnx) (1/x) x^ln(x)
now simplify
OpenStudy (anonymous):
ok let me see if one of my answers the same
OpenStudy (anonymous):
(2x^lnx) (1/x^2) (2ln^2(x) - ln(x) + 1)
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OpenStudy (anonymous):
i got (x^lnx) (2-2lnx+4ln^2(x))/x^2
OpenStudy (anonymous):
samething if you were to factor out 2
OpenStudy (anonymous):
oh ok sorry! so now i solve for x when i set this to zero right
OpenStudy (anonymous):
yes.
OpenStudy (anonymous):
ok im sure im going to need help with this too please
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OpenStudy (anonymous):
its just that i make minor mistakes that have big affects
OpenStudy (anonymous):
ok im stuck at where i divide by 2x^lnx
then have 1-lnx+2lnx^2=0
OpenStudy (anonymous):
i got next -lnx+2lnx^v=-1
then im stuck
OpenStudy (anonymous):
can you solve 2u^2 - u + 1 = 0?
OpenStudy (anonymous):
oh yes so we will get two answers?
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OpenStudy (anonymous):
i don't know. What does the quadratic formula say?