Answer
$G'(4)=\dfrac{1}{2}$
Work Step by Step
$G(x)=1+2\sqrt{x},$ at $4$
The derivative of a function $G$ at $a$ is $G'(a)=\lim_{h\to0}\dfrac{G(a+h)-G(a)}{h}$
In this case, $a=4$
Find $G(a+h)$ by substituting $x$ by $4+h$ in $G(x)$ and simplifying:
$G(4+h)=1+2\sqrt{4+h}$
Find $G(a)$ by substituting $x$ by $4$ in $G(x)$ and evaluating:
$G(4)=1+2\sqrt{4}=1+(2)(2)=5$
Subsitute the known values into the formula that gives $G'(a)$ and evaluate:
$G'(4)=\lim_{h\to0}\dfrac{1+2\sqrt{4+h}-5}{h}=\lim_{h\to0}\dfrac{2\sqrt{4+h}-4}{h}=...$
$...=\lim_{h\to0}\dfrac{2\sqrt{4+h}-4}{h}\cdot\dfrac{2\sqrt{4+h}+4}{2\sqrt{4+h}+4}=...$
$...=\lim_{h\to0}\dfrac{4(4+h)-16}{h(2\sqrt{4+h}+4)}=\lim_{h\to0}\dfrac{16+4h-16}{h(2\sqrt{4+h}+4)}=...$
$...=\lim_{h\to0}\dfrac{4h}{h(2\sqrt{4+h}+4)}=\lim_{h\to0}\dfrac{4}{2\sqrt{4+h}+4}=...$
$...=\dfrac{4}{2\sqrt{4+0}+4}=\dfrac{4}{(2)(2)+4}=\dfrac{4}{8}=\dfrac{1}{2}$
