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