Question

given the function $f(x)=3ln(x + 2)$, determine the instantaneous rate of change of $f$ at $x = 5$ using the limit shown below. you do not have to simplify your answer.

Explanation:

Step1: Recall the derivative definition

The instantaneous rate of change of a function $y = f(x)$ at $x=a$ is given by $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here $a = 5$ and $f(x)=3\ln(x + 2)$. So $f(5)=3\ln(5 + 2)=3\ln(7)$ and $f(5 + h)=3\ln((5 + h)+ 2)=3\ln(7 + h)$.

Step2: Substitute into the limit formula

$f^{\prime}(5)=\lim_{h
ightarrow0}\frac{3\ln(7 + h)-3\ln(7)}{h}=3\lim_{h
ightarrow0}\frac{\ln(7 + h)-\ln(7)}{h}$.

Step3: Use the logarithm property $\ln m-\ln n=\ln\frac{m}{n}$

$3\lim_{h
ightarrow0}\frac{\ln\frac{7 + h}{7}}{h}=3\lim_{h
ightarrow0}\frac{\ln(1+\frac{h}{7})}{h}$. Let $t=\frac{h}{7}$, then $h = 7t$ and as $h
ightarrow0$, $t
ightarrow0$.

Step4: Rewrite the limit

$3\lim_{t
ightarrow0}\frac{\ln(1 + t)}{7t}=\frac{3}{7}\lim_{t
ightarrow0}\frac{\ln(1 + t)}{t}$.

Step5: Recall the important limit $\lim_{t

ightarrow0}\frac{\ln(1 + t)}{t}=1$
$\frac{3}{7}\times1=\frac{3}{7}$.

Answer:

$\frac{3}{7}$