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how to differentiate f(x)=lnx/1+ln(2x) using natural log functions/properties?
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Well, ln(2x) = ln 2 + ln x. Does that help you?
\[\ln x+ \ln*(2x)=\ln x + \ln2 +\ln x\]
Log (ab) = log (a) + log (b)
apply formula for (u/v)' = (u' v + v' u) / v^2 . u= ln x; v = 1 + ln2 + lnx. just using for v' in numerator. in denominator, you better let the whole thing (1+ ln2x)^2. It's easier than break it down to take exponent. the rest is for you to practice. hope this help
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