Question If $g(x)=\frac{x^{6}lnx}{(x-1)e^x},$ find $g^{\text{'}}(x)$ using logarithmic differentiation.

Select the correct answer below:

$g^{\text{'}}(x)=\frac{6}{x}\frac{1}{xlnx}-\frac{1}{x-1}-1$

$g^{\text{'}}(x)=\frac{x^{6}lnx}{(x-1)e^x}(\frac{6}{x}\frac{1}{xlnx}-\frac{1}{x-1}-1)$

$g^{\text{'}}(x)=\frac{x^{6}lnx}{(x-1)e^x}(\frac{6}{x}-\frac{1}{xlnx}-\frac{1}{x-1}-1)$

$g^{\text{'}}(x)=\frac{x^{6}lnx}{(x-1)e^x}(\frac{6}{x}\frac{1}{xlnx}-\frac{1}{x-1}1)$