Answer
$ g^{'}(x)=3e^{3x}-x^{-1} $
$g^{''}(x)=9e^{3x}+x^{-29}$
Work Step by Step
$g(x)=e^{3x}-\ln 3x$
Taking derivative with respect to x
$ g^{'}(x)=\frac{d(g(x))}{dx}= \frac{d( e^{3x}-\ln 3x )}{dx}$
$ g^{'}(x)=\frac{d(e^{3x})}{dx}-\frac{d(\ln 3x )}{dx}$
$ g^{'}(x)=3e^{3x}-\frac{1}{3x} \frac{d(3x)}{dx}$
$ g^{'}(x)=3e^{3x}-3\times\frac{1}{3x} \frac{d(x)}{dx}$
$ g^{'}(x)=3e^{3x}-\frac{1}{x} $
$ g^{'}(x)=3e^{3x}-x^{-1} $
Differentiating again with respect to x
$g^{''}(x)=\frac{ d g{'}(x)}{dx}= \frac{d( 3e^{3x}-x^{-1} )}{dx}$
$g^{''}(x)=\frac{d(3e^{3x})}{dx}- \frac{d(x^{-1})}{dx}$
$g^{''}(x)=3\frac{d(e^{3x})}{dx}- (-1)x^{-2}$
$g^{''}(x)=3 e^{3x}\frac{d(3x)}{dx}+\frac{1}{x^2}$
$g^{''}(x)=3 \times 3e^{3x}\frac{d(x)}{dx}+\frac{1}{x^2}$
$g^{''}(x)=9e^{3x}+\frac{1}{x^2}$
