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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)?

answer=(-7e^-x+e^-x-xe^-x)/(7e^-x+xe^-x)

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 ^_^