Question

attempt 1: 10 attempts remaining. evaluate \\(\frac{d}{dt}\ln(6t^2 + 11t + 8) =\\) submit answer next item

Explanation:

Step1: Apply chain rule

The derivative of $\ln(u)$ with respect to $t$ is $\frac{1}{u} \cdot \frac{du}{dt}$, where $u = 6t^2 + 11t + 8$.
So, $\frac{d}{dt}\ln(6t^2 + 11t + 8) = \frac{1}{6t^2 + 11t + 8} \cdot \frac{d}{dt}(6t^2 + 11t + 8)$

Step2: Differentiate the polynomial

Differentiate $6t^2 + 11t + 8$ with respect to $t$. The derivative of $6t^2$ is $12t$, the derivative of $11t$ is $11$, and the derivative of $8$ is $0$. So, $\frac{d}{dt}(6t^2 + 11t + 8) = 12t + 11$.

Step3: Combine the results

Substitute the derivative of the polynomial back into the chain rule formula: $\frac{1}{6t^2 + 11t + 8} \cdot (12t + 11) = \frac{12t + 11}{6t^2 + 11t + 8}$

Answer:

$\frac{12t + 11}{6t^2 + 11t + 8}$