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differentiate ln(7e^-x+xe^-x)

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my answer was 1/f(x) * f'(x). but it keeps saying wrong
first take e^(-x) common from (7e^-x+xe^-x) then use the property of log. log ab=log a + log b
so, ln e^-x(7+x)?

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Other answers:

ln [e^-x(7+x)] now use log ab = log a +log b
so, lne^-x+ln(7+x), and go from there?
u can simplify ln(e^(-x)) do u know how ?
no, im afraid not goku. :l
\(\huge ln e^{(a)}=aln e=a\) now ?
oh ye, makes sense.
so lne^-x+ln(7+x) = ?
so, -x+ln(7+x)? is that right so far?
yes, that is correct :) now its differentiation is simple....
-1+...ermm 1/(7+x)*1?
that is absolutely correct (except for ...ermm part) simplify it further
lol, wow. thank you goku san.
welcome ^_^

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