#### Answer

The slope of the tangent line to the curve at $x=4$ is $-1/16$.

#### Work Step by Step

$$y=f(x)=\frac{1}{\sqrt x}$$
The slope of the tangent line to the curve at $x=4$ is $$y'=\lim_{h\to0}\frac{f(4+h)-f(4)}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+4}}-\frac{1}{\sqrt 4}}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+4}}-\frac{1}{2}}{h}$$
$$y'=\lim_{h\to0}\frac{2-\sqrt{h+4}}{2h\sqrt{h+4}}$$
Multiply both numerator and denominator by $2+\sqrt{h+4}$:
$$y'=\lim_{h\to0}\frac{(2-\sqrt{h+4})(2+\sqrt{h+4})}{2h\sqrt{h+4}(2+\sqrt{h+4})}=\lim_{h\to0}\frac{4-(h+4)}{2h\sqrt{h+4}(2+\sqrt{h+4})}$$
$$y'=\lim_{h\to0}\frac{-h}{2h\sqrt{h+4}(2+\sqrt{h+4})}=\lim_{h\to0}\frac{-1}{2\sqrt{h+4}(2+\sqrt{h+4})}$$
$$y'=\frac{-1}{2\sqrt{0+4}(2+\sqrt{0+4})}=\frac{-1}{2\sqrt4(2+\sqrt4)}$$
$$y'=\frac{-1}{2\times2\times4}=-\frac{1}{16}$$