Evaluating Derivative of $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$ Did I evaluate the following derivative correctly? I know I have not simplified to the utmost extent, but I want to know if this method is correct.
Consider, $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$
Let A = $1+\sqrt{1+\sqrt{1-2x}}$
Let B = $1+\sqrt{1-2x}$
Let C = $1-2x$
$$\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' = \frac{1}{2}A^{-\frac{1}{2}} \cdot \frac{1}{2}B^{-\frac{1}{2}} \cdot \frac{1}{2}C^{-\frac{1}{2}} \cdot -2 = \frac{-1}{4\sqrt{ABC}}$$

Yes, this is valid. Maybe some explanatory steps in between would help to someone reading it though. It strikes me as not being obvious according to standard well-known derivative rules without a few intermediate steps.
$$\begin{align}
\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' &= \left(\sqrt{A}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}A'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\left(1+\sqrt{B}\right))'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}B'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\left(1+\sqrt{C}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}C'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}(-2)\\
&= \frac{-1}{4\sqrt{ABC}}
\end{align}$$
