Answer
$-\dfrac{1}{16}$
Work Step by Step
Consider $y=f(x)=\dfrac{1}{\sqrt x}$
Now, the slope of the tangent line to the curve at $x=4$ is given by:
$y'=\lim\limits_{h\to 0}\dfrac{f(4+h)-f(4)}{h}=\lim\limits_{h\to 0}\dfrac{\dfrac{1}{\sqrt{h+4}}-\dfrac{1}{\sqrt 4}}{h}=\lim\limits_{h\to 0}\dfrac{2-\sqrt{h+4}}{2h\sqrt{(h+4)}}$
$\implies \lim\limits_{h\to 0}\dfrac{(2-\sqrt{(h+4)})(2+\sqrt{(h+4)})}{2h\sqrt{h+4}(2+\sqrt{h+4})}=\lim\limits_{h\to 0} \dfrac{-h}{2h\sqrt{h+4}(2+\sqrt{(h+4)})}$
and $\dfrac{-1}{2\sqrt{(0)+4}(2+\sqrt{(0)+4})}=\dfrac{-1}{2\sqrt 4(2+\sqrt4)}$
Hence, $y'=\dfrac{-1}{(2)(2)(4)}=-\dfrac{1}{16}$
