Resposta:

[tex]g"(x)=-\frac{78}{(3x+2)^3}[/tex]

Explicação passo a passo:

[tex]g(x)=\frac{2x-3}{3x+2} \\\\g'(x)=\frac{(3x+2).2-(2x-3).3}{(3x+2)^2} \\\\g'(x)=\frac{6x+4-6x+9}{(3x+2)^2} \\\\g'(x)=\frac{13}{(3x+2)^2} \\\\g"(x) = \frac{(3x+2)^2*0-13.2(3x+2)*3}{(3x+2)^4} \\\\g"(x)=-\frac{78(3x+2)}{(3x+2)^4} \\\\g"(x)=-\frac{78}{(3x+2)^3}[/tex]