Question

ln(sqrt(x)/(x^2+e^x))

Answer 1

\[\ln(\frac{ \sqrt{x} }{ x^2+e^x }\]

Answer 2

i know that it's suppose to be u'/u but not sure how to do it

Answer 3

We want to apply this rule of logarithms before we differentiate.
\[\huge \log(\frac{ a }{ b })=\log(a)-\log(b)\]

Answer 4

ok im stuck

Answer 5

Using the rule above, we get the following:
\[\huge \ln(\frac{ \sqrt x }{ x^2+e^x })=\ln(\sqrt x)-\ln(x^2+e^x)\]
Understand that part? :D

Answer 6

yes

Answer 7

\[\large \frac{ d }{ dx }\ln(\sqrt x)=\frac{ 1 }{ \sqrt x }(\sqrt x)'\]

Answer 8

If you prefer to write it as u'/u that's fine also, I'm writing it as (1/u)u' because we're going to get some ugly fractions. (Unless you're converting to a fractional exponent, then ignore the words coming out of my mouth :333)