TITLE: Derivative of $\frac{1}{\sqrt{x-5}}$ via definition of derivative
QUESTION [5 upvotes]: A high school student has asked me to help with a limit. The teacher wants them to calculate the derivative of
$$\frac{1}{\sqrt{x-5}}$$
at the point $x=9$ using the definition of the derivative. AND! They don't know $(1+x)^\alpha \approx 1 + \alpha x$.
I'm puzzled since don't know how to proceed without it.
$$\left.\left(\dfrac{1}{\sqrt{x-5}}\right)'\,\right|_{x=9}  =
\lim_{h\to 0} \dfrac{1}{h}\left(\dfrac{1}{\sqrt{4+h}}-\dfrac{1}{2}\right)=
\lim_{h\to 0} \dfrac{1}{2h}\left((1+h/4)^{-1/2}-1\right)\color{red}{{\bf=}}-1/16
$$
Is there really a way to walk around?..
BTW, what is the easiest way to derive $(1+x)^\alpha \approx 1 + \alpha x$ for $\alpha \in \mathbb{R}$? I forgot how we did this in school.

REPLY [3 votes]: You have 
$$
\frac1h\,\left(\frac1{\sqrt{4+h}}-\frac12\right)
=\frac{2-\sqrt{4+h}}{2h\sqrt{4+h}}
=-\frac1{2\sqrt{4+h}(2+\sqrt{4+h})},
$$
after multiplying and dividing by $2+\sqrt{4+h}$.