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 = $1-2x$
$$\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' = \frac{1}{2}A^{-\frac{1}{2}} \cdot \frac{1}{2}B^{-\frac{1}{2}} \cdot \frac{1}{2}C^{-\frac{1}{2}} \cdot -2 = \frac{-1}{4\sqrt{ABC}}$$
 A: 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.
$$\begin{align}
\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' &= \left(\sqrt{A}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}A'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\left(1+\sqrt{B}\right))'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}B'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\left(1+\sqrt{C}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}C'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}(-2)\\
&= \frac{-1}{4\sqrt{ABC}}
\end{align}$$
A: Another way to see it is that with $f(x)=\sqrt{1+x}$, and $g(x)=-2x$, you are calculating the derivative of
$$
f\circ f\circ f\circ g(x)
$$
then you apply the chain rule.
A: Another way of thinking about it that makes the use of chain rule on nested functions is like so:
Define $f(x) = \sqrt{1+x}$ and $g(x) = -2x$. Then the original expression $y(x) = f(f(f(g(x))))$, which can be more compactly represented as $y = f^3g$ if we understand that the "exponent" is denoting repeated composition and that the independent variable $x$ that we are differentiating with respect to, is left implicit.
Then application of chain rule immediately gets us:
$y' = f'(f^2g)\cdot f'fg \cdot f'g \cdot g'$
So $y'(x) = \frac 12 (1+ \sqrt{1+\sqrt{1-2x}})^{-\frac 12} \cdot \frac 12(1+\sqrt{1-2x})^{-\frac 12} \cdot \frac 12(1-2x)^{-\frac 12}\cdot (-2)$.
The results should be the same after simplification.
