First, using the quotient rule, we get
g ′ ( t ) = ( ln ( 3 t − 4 ) ) ′ t − ln ( 3 t − 4 ) ⋅ 1 t 2 {\displaystyle g'(t)={\frac {(\ln(3t-4))'t-\ln(3t-4)\cdot 1}{t^{2}}}} .
Then, to compute ( ln ( 3 t − 4 ) ) ′ {\displaystyle (\ln(3t-4))'} , we use the chain rule with y = ln ( 3 t − 4 ) {\displaystyle y=\ln(3t-4)} and u = 3 t − 4 {\displaystyle u=3t-4} ;
( ln ( 3 t − 4 ) ) ′ = d y d t = d y d u ⋅ d u d t = d d u ln u ⋅ d d t ( 3 t − 4 ) = 1 u ⋅ 3 = 3 3 t − 4 {\displaystyle (\ln(3t-4))'={\frac {dy}{dt}}={\frac {dy}{du}}\cdot {\frac {du}{dt}}={\frac {d}{du}}\ln u\cdot {\frac {d}{dt}}(3t-4)={\frac {1}{u}}\cdot 3={\frac {3}{3t-4}}} .
Plugging this back to the first equation, we get
g ′ ( t ) = 3 t 3 t − 4 − ln ( 3 t − 4 ) t 2 {\displaystyle \color {blue}g'(t)={\frac {{\frac {3t}{3t-4}}-\ln(3t-4)}{t^{2}}}}
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, we use the chain rule with 
and 
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