Answer
a) $f(x)=\sqrt x$, $a=9$
b) $f'(a)=\frac{1}{6}$
Work Step by Step
a) The derivative $f'(a)$ is defined as
$f'(a)=\lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$
Comparing $\lim\limits_{h \to 0}\frac{\sqrt {9+h}-\sqrt {9}}{h}$ with the above definition, we get
$a=9$ and $f(x)=\sqrt x$
b) $f'(9)=\lim\limits_{h \to 0}\frac{\sqrt {9+h}-\sqrt {9}}{h}=\lim\limits_{h \to 0}\frac{(\sqrt {9+h}-\sqrt {9})(\sqrt {9+h}+\sqrt {9})}{h(\sqrt {9+h}+\sqrt {9})}$
$=\lim\limits_{h \to 0}\frac{9+h-9}{h(\sqrt {9+h}+\sqrt {9})}=\lim\limits_{h \to 0}\frac{1}{\sqrt {9+h}+\sqrt {9}}=\frac{1}{2\sqrt {9}}=\frac{1}{6}$
