steel11
differentiate
ln(7e^x+xe^x)



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steel11
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my answer was 1/f(x) * f'(x). but it keeps saying wrong

hartnn
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first take e^(x) common from (7e^x+xe^x)
then use the property of log.
log ab=log a + log b

steel11
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so, ln e^x(7+x)?

cinar
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answer=(7e^x+e^xxe^x)/(7e^x+xe^x)

hartnn
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ln [e^x(7+x)]
now use log ab = log a +log b

steel11
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so, lne^x+ln(7+x), and go from there?

hartnn
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u can simplify ln(e^(x))
do u know how ?

steel11
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no, im afraid not goku. :l

hartnn
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\(\huge ln e^{(a)}=aln e=a\)
now ?

steel11
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oh ye, makes sense.

hartnn
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so lne^x+ln(7+x) = ?

steel11
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so, x+ln(7+x)? is that right so far?

hartnn
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yes, that is correct :)
now its differentiation is simple....

steel11
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1+...ermm 1/(7+x)*1?

hartnn
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that is absolutely correct (except for ...ermm part)
simplify it further

steel11
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lol, wow. thank you goku san.

hartnn
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welcome ^_^