Answer
$$ \frac{-1}{27} $$
$$y=\frac{-x}{27}+\frac{13}{27}$$
Work Step by Step
Since
$$ f(x)= \frac{1}{\sqrt{2x+1}} ,\ \ a= 4 $$
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{2 h+9}}-\frac{1}{3}}{h}\\
&=\lim _{h \rightarrow 0} \frac{3-\sqrt{2 h+9}}{3 h \sqrt{2 h+9}}\\
&=\lim _{h \rightarrow 0} \frac{3-\sqrt{2 h+9}}{3 h \sqrt{2 h+9}} \times \frac{3+\sqrt{2 h+9}}{3+\sqrt{2 h+9}}\\
&=\lim _{h \rightarrow 0} \frac{9-2 h-9}{3 h \sqrt{2 h+9}(3+\sqrt{2 h+9})}\\
&=\lim _{h \rightarrow 0} \frac{-2}{3 \sqrt{2 h+9}(3+\sqrt{2 h+9})}\\
&=\frac{-1}{27}
\end{aligned}
and the tangent line is
\begin{aligned}
y-\frac{1}{3}&=-\frac{1}{27}(x-4)\\
y&=\frac{1}{3}-\frac{1}{27} x+\frac{4}{27}\\
y&=-\frac{1}{27} x+\frac{13}{27}\\
y&=\frac{-x+13}{27}\\
y&=\frac{-x}{27}+\frac{13}{27}\\
\end{aligned}
