## Calculus: Early Transcendentals (2nd Edition)

$a.$ The derivative is $f'(x)=\frac{1}{2\sqrt{x+2}}$. $b.$ The equation is $y=\frac{1}{6}x+\frac{11}{6}$.
$a.$ Using the definition of the derivative at $x$ we have $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{(x+h)+2}-\sqrt{x+2}}{h}=\lim_{h\to0}\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}\cdot\frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}=\lim_{h\to0}\frac{\sqrt{x+h+2}^2-\sqrt{x+2}^2}{h(\sqrt{x+h+2}+\sqrt{x+2})}=\lim_{h\to0}\frac{x+h+2-(x+2)}{h(\sqrt{x+h+2}+\sqrt{x+2})}=\lim_{h\to0}\frac{h}{{h(\sqrt{x+h+2}+\sqrt{x+2})}}=\lim_{h\to0}\frac{1}{\sqrt{x+h+2}+\sqrt{x+2}}=\frac{1}{\sqrt{x+0+2}+\sqrt{x+2}}=\frac{1}{2\sqrt{x+2}}.$$ $b.$ 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 $a=7$, $f(a)=f(7)=\sqrt{7+2}=3$ and $f'(a)=f'(7)=\frac{1}{2\cdot\sqrt{7+2}}=\frac{1}{6}.$ we have $$y-3=\frac{1}{6}(x-7)\Rightarrow y-3=\frac{1}{6}x-\frac{7}{6}$$ which gives $$y=\frac{1}{6}x-\frac{7}{6}+3=\frac{1}{6}x+\frac{11}{6}.$$