## Calculus: Early Transcendentals (2nd Edition)

$a.$ $f'(x)=\frac{a}{2\sqrt{ax+b}}$; $b.$ $f'(x)=\frac{5}{2\sqrt{5x+9}}$; $c.$ $f'(-1)=\frac{5}{4}.$
$a.$ By the definition of the derivative we have $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{a(x+h)+b}-\sqrt{ax+b}}{h}=\lim_{h\to0}\frac{\sqrt{ax+ah+b}-\sqrt{ax+b}}{h}=\lim_{h\to0}\frac{\sqrt{ax+ah+b}-\sqrt{ax+b}}{h}\cdot\frac{\sqrt{ax+ah+b}+\sqrt{ax+b}}{\sqrt{ax+ah+b}+\sqrt{ax+b}}=\lim_{h\to0}\frac{\sqrt{ax+ah+b}^2-\sqrt{ax+b}^2}{h(\sqrt{ax+ah+b}+\sqrt{ax+b})}=\lim_{h\to0}\frac{ax+ah+b-(ax+b)}{h(\sqrt{ax+ah+b}+\sqrt{ax+b})}=\lim_{h\to0}\frac{ah}{h(\sqrt{ax+ah+b}+\sqrt{ax+b})}=\lim_{h\to0}\frac{a}{\sqrt{ax+ah+b}+\sqrt{ax+b}} = \frac{a}{\sqrt{ax+a\cdot0+b}+\sqrt{ax+b}}=\frac{a}{2\sqrt{ax+b}}.$$ $b.$ We have shown that the derivative of a function of the form of $\sqrt{ax+b}$ is $\frac{a}{2\sqrt{ax+b}}.$ Applying this to the given function we get $$f'(x)=\frac{5}{2\sqrt{5x+9}}.$$ $c.$ We just have to put $x=-1$ into the calculated derivative from the previous part $$f'(-1) =\frac{5}{2\sqrt{5\cdot(-1)+9}}=\frac{5}{2\sqrt{-5+9}}=\frac{5}{8}.$$