QuestionWhat is the first derivative of $t(x)=\ln \left(4 x^{3}+4 x^{2}+6 x+4\right)$ ?
Select the correct answer below: $\frac{1}{4 x^{3}+4 x^{2}+6 x+4}$ $\frac{12 x^{2}+8 x+6}{4 x^{3}+4 x^{2}+6 x+4}$ $\frac{12 x^{2}+8 x+6}{\ln \left(4 x^{3}+4 x^{2}+6 x+4\right)}$ $\ln \left(12 x^{2}+8 x+6\right)$ $\frac{4 x^{3}+4 x^{2}+6 x+4}{12 x^{2}+8 x+6}$