Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$ Derivative at 4, when $f(x)=\frac{1}{\sqrt{2x+1}}$
I choose to use the formula $\displaystyle f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$
Which after some work I found to be 
$\frac{3-\sqrt{2x+1}}{3\sqrt{2x+1}(x-4)}$
Which is basically back to where I started. Is there any way to get the $x-4$ out of the denominator? 
 A: Hint: Try multiplying by the conjugate:
$$
\frac{3 + \sqrt{2x+1}}{3 + \sqrt{2x+1}}
$$
Then the numerator becomes:
$$
(3)^2 - (\sqrt{2x + 1})^2 = 9 - (2x + 1) = -2x + 8 = -2(x - 4)
$$
so that you can cancel the $(x - 4)$ term.

At the request of the comments, here's all the work:
\begin{align*}
\left. \frac{d}{dx}\right|_{x=4} \frac{1}{\sqrt{2x+1}}
&= \lim_{x\to 4} \frac{\frac{1}{\sqrt{2x+1}} - \frac{1}{\sqrt{2(4)+1}}}{x - 4} \\
&= \lim_{x\to 4} \frac{\frac{1}{\sqrt{2x+1}} - \frac{1}{3}}{x - 4} \\
&= \lim_{x\to 4} \frac{3-\sqrt{2x+1}}{3\sqrt{2x+1}(x-4)} \\
&= \lim_{x\to 4} \frac{-2(x - 4)}{3\sqrt{2x+1}(x-4)(3 + \sqrt{2x+1})} &\text{using my hint}\\
&= \lim_{x\to 4} \frac{-2}{3\sqrt{2x+1}(3 + \sqrt{2x+1})} \\
&= \frac{-2}{3\sqrt{2(4)+1}(3 + \sqrt{2(4)+1})} \\
&= \frac{-1}{27}
\end{align*}
A: $\displaystyle f'(x)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$
$\displaystyle=\frac{\dfrac1{\sqrt{2x+1}}-\dfrac1{\sqrt{2a+1}}}{x-a}$
$\displaystyle=\frac{\sqrt{2a+1}-\sqrt{2x+1}}{(x-a)\sqrt{2x+1}\sqrt{2a+1}}$
$\displaystyle=\frac{(2a+1)-(2x+1)}{(\sqrt{2a+1}+\sqrt{2x+1})(x-a)\sqrt{2x+1}\sqrt{2a+1}}$
Now cancel out $x-a$ as $x\to a,x\ne a\iff x-a\ne0$
Then set $x=a$ and finally $a=4$
