Question

1. find the derivative of $f(x)=sqrt{3x + 1}$ using the definition of the derivative. then, find the equation of the tangent line at $x = 8$.

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ is $f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\sqrt{3x + 1}$, then $f(x + h)=\sqrt{3(x + h)+1}=\sqrt{3x+3h + 1}$.

Step2: Substitute into derivative - definition formula

\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{\sqrt{3x + 3h+1}-\sqrt{3x + 1}}{h}\\ &=\lim_{h ightarrow0}\frac{(\sqrt{3x + 3h+1}-\sqrt{3x + 1})(\sqrt{3x + 3h+1}+\sqrt{3x + 1})}{h(\sqrt{3x + 3h+1}+\sqrt{3x + 1})}\\ &=\lim_{h ightarrow0}\frac{(3x + 3h+1)-(3x + 1)}{h(\sqrt{3x + 3h+1}+\sqrt{3x + 1})}\\ &=\lim_{h ightarrow0}\frac{3h}{h(\sqrt{3x + 3h+1}+\sqrt{3x + 1})}\\ &=\lim_{h ightarrow0}\frac{3}{\sqrt{3x + 3h+1}+\sqrt{3x + 1}} \end{align*}$$

\]
Evaluating the limit as $h
ightarrow0$, we get $f'(x)=\frac{3}{2\sqrt{3x + 1}}$.

Step3: Find the slope of the tangent - line at $x = 8$

Substitute $x = 8$ into $f'(x)$. $f'(8)=\frac{3}{2\sqrt{3\times8 + 1}}=\frac{3}{2\sqrt{25}}=\frac{3}{10}$.

Step4: Find the point on the function at $x = 8$

When $x = 8$, $f(8)=\sqrt{3\times8 + 1}=\sqrt{25}=5$. So the point on the function is $(8,5)$.

Step5: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(8,5)$ and $m=\frac{3}{10}$.
\[

$$\begin{align*} y-5&=\frac{3}{10}(x - 8)\\ y-5&=\frac{3}{10}x-\frac{12}{5}\\ y&=\frac{3}{10}x-\frac{12}{5}+5\\ y&=\frac{3}{10}x+\frac{13}{5} \end{align*}$$

\]

Answer:

The derivative of $f(x)$ is $f'(x)=\frac{3}{2\sqrt{3x + 1}}$, and the equation of the tangent line at $x = 8$ is $y=\frac{3}{10}x+\frac{13}{5}$.