Derivative of $\frac{1}{\sqrt{x-5}}$ via definition of derivative

Question

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.

Answer 1

Hint

Multiply numerator and denominator of $$ \frac{2 - \sqrt{4+h}}{2\sqrt{4+h}} $$ by the conjugate $$ 2 + \sqrt{4+h}\ . $$

This trick will work for square roots but not for other exponents.

Answer 2

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 $2+\sqrt{4+h}$.

Answer 3

Hint. One may use, as $h \to 0$, $$ \dfrac{1}{h}\left(\dfrac{1}{\sqrt{4+h}}-\dfrac{1}{2}\right)=\dfrac{1}{h}\left(\dfrac{2-\sqrt{4+h}}{2\sqrt{4+h}}\right)=\dfrac{1}{h}\cdot\dfrac{4-(4+h)}{2\sqrt{4+h}\cdot(2+\sqrt{4+h})}. $$