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steel11 Group TitleBest ResponseYou've already chosen the best response.0
my answer was 1/f(x) * f'(x). but it keeps saying wrong
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
first take e^(x) common from (7e^x+xe^x) then use the property of log. log ab=log a + log b
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
so, ln e^x(7+x)?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
answer=(7e^x+e^xxe^x)/(7e^x+xe^x)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
ln [e^x(7+x)] now use log ab = log a +log b
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
so, lne^x+ln(7+x), and go from there?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
u can simplify ln(e^(x)) do u know how ?
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
no, im afraid not goku. :l
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
\(\huge ln e^{(a)}=aln e=a\) now ?
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
oh ye, makes sense.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
so lne^x+ln(7+x) = ?
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
so, x+ln(7+x)? is that right so far?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
yes, that is correct :) now its differentiation is simple....
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
1+...ermm 1/(7+x)*1?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
that is absolutely correct (except for ...ermm part) simplify it further
 2 years ago

steel11 Group TitleBest ResponseYou've already chosen the best response.0
lol, wow. thank you goku san.
 2 years ago
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