differentiate the function. f(t) = ln((3t + 1)^4 / (2t - 1)^5) f(t) =

Explanation:

Step1: Use logarithm properties

First, use the property $\ln(\frac{a}{b})=\ln(a)-\ln(b)$ and $\ln(a^n)=n\ln(a)$. So $F(t) = 4\ln(3t + 1)-5\ln(2t - 1)$.

Step2: Differentiate term - by - term

The derivative of $\ln(u)$ with respect to $t$ is $\frac{u'}{u}$ by the chain - rule. For $y = 4\ln(3t + 1)$, let $u = 3t+1$, then $y'=\frac{4\times3}{3t + 1}=\frac{12}{3t + 1}$. For $y=-5\ln(2t - 1)$, let $u = 2t - 1$, then $y'=\frac{-5\times2}{2t - 1}=\frac{-10}{2t - 1}$.

Step3: Combine the derivatives

$F'(t)=\frac{12}{3t + 1}-\frac{10}{2t - 1}$.

Answer:

$\frac{12}{3t + 1}-\frac{10}{2t - 1}$