Answer
$$ -\frac{1}{16}$$
$$y=-\frac{1}{16} x+\frac{3}{4}$$
Work Step by Step
Let $f(x)=\frac{1}{\sqrt{x}} .$ Then
\begin{aligned}
f^{\prime}(4) &=\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{4+h}}-\frac{1}{2}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{2-\sqrt{4+h}}{2 \sqrt{4+h}} \cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{4-4-h}{4 \sqrt{4+h}+2(4+h)}}{h} \\
&=\lim _{h \rightarrow 0} \frac{-1}{4 \sqrt{4+h}+2(4+h)}=-\frac{1}{16}
\end{aligned}
At $a=4$, the tangent line is
\begin{aligned}
y&=f^{\prime}(4)(x-4)+f(4)\\
&=-\frac{1}{16}(x-4)+\frac{1}{2}\\
&=-\frac{1}{16} x+\frac{3}{4}
\end{aligned}
