Derivative of $\frac{1}{\sqrt{x-1}}$

Question

I want to check whether if my way is correct or not, and why?

\begin{align*} f'(x)&=\left(\dfrac{1}{\sqrt{x-1}}\right)'\\ &=\dfrac{-\left(\sqrt{x-1}\right)'}{\left(\sqrt{x-1}\right)^{2}}=\dfrac{-\left(\dfrac{(x-1)'}{2\sqrt{x-1}}\right)}{\left(\sqrt{x-1}\right)^{2}}\\ &=\dfrac{-\left(\dfrac{(x)'-(1)'}{2\sqrt{x-1}}\right)}{\left(\sqrt{x-1}\right)^{2}}=\dfrac{-\left(\dfrac{1-0}{2\sqrt{x-1}}\right)}{\left(\sqrt{x-1}\right)^{2}}=\dfrac{-\left(\dfrac{1}{2\sqrt{x-1}}\right)}{\left(\sqrt{x-1}\right)^{2}}\\ &=-\dfrac{1}{2\sqrt{(x-1)}\left(\sqrt{x-1}\right)^{2}}=-\dfrac{1}{2\sqrt{(x-1)}|x-1|}\\ &=\boxed{-\dfrac{1}{2\left(x-1\right)\sqrt{(x-1)}}} \end{align*}

Answer 1

$\left((x-1)^{-1/2}\right)'=-\frac 12(x-1)^{-3/2}$ and this is over.

I think pushing up doing $(x-1)'=1$ step by step is really too much.

Also derivating composition of functions $\frac 1\cdot\circ\sqrt{\cdot}$ is not necessary, go directly with the derivative of $x^\alpha$ in this case. But I admit that considering an highschool level, this may be the only available choice (since fractional exponents may be out of scope).

The point is that when facing more complex functions, the risk of getting confused will increase if one is not familiar with shortcuts. Also the more stuff you write, the more chance a typo WILL occur and ruin it all.

I would keep only terms 1, 2, 3, 7 and 10 (1 being f'(x)) from your calculation, and explain intermediate steps orally if necessary.