## Calculus: Early Transcendentals (2nd Edition)

$a.$ The derivative is $f'(x)=\frac{3}{10}.$ $b.$ The equation is $y=\frac{3}{10}x+\frac{13}{5}$.
$a.$ Using the definition of the derivative at $x$ we have $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}=\lim_{h\to0}\frac{\sqrt{3x+3h+1}-\sqrt{3x+1}}{h}\cdot\frac{\sqrt{3x+3h+1}+\sqrt{3x+1}}{\sqrt{3x+3h+1}+\sqrt{3x+1}}=\lim_{h\to0}\frac{\sqrt{3x+3h+1}^2-\sqrt{3x+1}^2}{h(\sqrt{3x+3h+1}+\sqrt{3x+1})}=\lim_{h\to0}\frac{3x+3h+1-(3x+1)}{h(\sqrt{3x+3h+1}+\sqrt{3x+1})}=\lim_{h\to0}\frac{3h}{{h(\sqrt{3x+3h+1}+\sqrt{3x+1})}}=\lim_{h\to0}\frac{3}{\sqrt{3x+3h+1}+\sqrt{3x+1}}=\frac{3}{\sqrt{3x+3\cdot0+1}+\sqrt{3x+1}}=\frac{3}{2\sqrt{3x+1}}.$$ $b.$ The equation of the tanget at the point $P(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using this with $a=8$, $f(a)=f(8)=\sqrt{3\cdot 8+1}=5$ and $f'(a)=f'(8)=\frac{3}{2\cdot\sqrt{3\cdot8+1}}=\frac{3}{10}$ we have $$y-5=\frac{3}{10}(x-8)\Rightarrow y-5=\frac{3}{10}x-\frac{12}{5}$$ which gives $$y=\frac{3}{10}x-\frac{12}{5}+5=\frac{3}{10}x+\frac{13}{5}.$$