Question

students in a zoology class took a final exam. they took equivalent forms of the exam at monthly intervals thereafter. after t months, the average score s(t), as a percentage, was found to be given by the following equation, where t≥0. complete parts (a) through (e) below. s(t)=75 - 12 ln(t + 1), 0≤t≤80

a) what was the average score when they initially took the test?
the average score was 75 %. (round to one decimal place as needed.)
b) what was the average score after 4 months?
the average score after 4 months was 55.7 %. (round to one decimal place as needed.)
c) what was the average score after 24 months?
the average score after 24 months was 36.4 %. (round to one decimal place as needed.)
d) find s(t).
s(t)=□

Explanation:

Step1: Recall the derivative formula for $\ln(u)$

The derivative of $\ln(u)$ with respect to $t$ is $\frac{u'}{u}$, where $u = t + 1$ and the derivative of a constant is 0.

Step2: Differentiate $S(t)$

Given $S(t)=75 - 12\ln(t + 1)$. The derivative of the constant 75 is 0. For the $- 12\ln(t + 1)$ part, using the chain - rule, if $u=t + 1$, then $u'=1$. The derivative of $\ln(u)$ is $\frac{1}{u}$, so the derivative of $-12\ln(t + 1)$ is $-12\times\frac{1}{t + 1}$.
So $S'(t)=-\frac{12}{t + 1}$.

Answer:

$S'(t)=-\frac{12}{t + 1}$