Respuesta :

[tex]\bf y=ln(3x+2)^k\implies y=k\cdot ln(3x+2)\implies \cfrac{dy}{dx}=k\cdot \stackrel{chain~rule}{\cfrac{1}{3x+2}\cdot 3} \\\\\\ \cfrac{dy}{dx}=\cfrac{3k}{3x+2}\impliedby \textit{we know that }y'(2)=3\quad or\quad \begin{cases} x=2\\ y'(x)=3 \end{cases} \\\\\\ \textit{therefore}\qquad 3=\cfrac{3k}{3(2)+2}\implies 3=\cfrac{3k}{8}\implies \cfrac{3\cdot 8}{3}=k\implies 8=k[/tex]

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