Chain rule with triple composition We are supposed to apply the chain rule on the following function $f$:
$$
f(x) = \sqrt{x+\sqrt{2x+\sqrt{3x}}}  
$$
I assumed we could rewrite this as $$ f(x) = g(h(j(x))) $$
However, I was not sure how to define the functions $$g(x), h(x), j(x) $$
Any help would be appreciated.
 A: I would simply do something like this:
\begin{align}
g(x) &= \sqrt{x}\\
h(x) &= x+\sqrt{2x+\sqrt{3x}}
\end{align}
so that $f(x) = g(h(x))$. Then by the chain rule
$$f'(x) = g'(h(x))h'(x) = \frac{1}{2\sqrt{h(x)}}h'(x)$$
Then we apply the chain rule to find $h'(x)$.
$$h'(x) = 1+\frac{1}{2\sqrt{2x+\sqrt{3x}}}(2x+\sqrt{3x})' $$
Then
$$(2x+\sqrt{3x})' = 2+\frac{1}{2\sqrt{3x}}(3x)'=2+\frac{3}{2\sqrt{3x}}. $$ 
Finally we get that
$$f'(x)=\frac{1}{2\sqrt{x+\sqrt{2x+\sqrt{3x}}}}\left(1+\frac{1}{2\sqrt{2x+\sqrt{3x}}}\left(2+\frac{3}{2\sqrt{3x}}\right)\right) $$
Which I'm sure can be simplified somewhat, please tell me anyone if I made any errors in this computation.
A: Do it in steps.
Let $j = \sqrt{3x}$
Then $f = \sqrt{x + \sqrt{2x + j}}$
Let $h = {\sqrt {2x + j}}$
And so on.
$\displaystyle \frac {\mathrm df}{\mathrm dx} = \frac {\mathrm df}{\mathrm dg}\frac {\mathrm dg}{\mathrm dh} \frac {\mathrm dh}{\mathrm dj}\frac {\mathrm dj}{\mathrm dx}$
edit: of course, the other answers are just as correct, but I find it easier to introduce more variables and then substitute them in when you're done.
A: It seems more realistic to write your function as a composition of two functions, instead of three. However, if you do successfully rewrite your function as the composition of three functions then $$ [g(h(j(x)))]'=g'(h(j(x))\cdot h'(j(x)) \cdot j'(x) $$ or for a finite collection of differentiable functions $\{f_i \}_{i=1}^n$, the derivative of their composition is $$\left[f_1 \circ f_2 \circ \dots \circ f_n\right]' =\prod_{i=1}^n \left(f'_i\circ f_{i+1} \circ \dots \circ f_n\right)$$
A: 1) $j(x) = 2x + \sqrt{3x}$
$j'(x) = 2 + \frac{\sqrt{3}}{2\sqrt{x}}$
2) $h(x) = x + \sqrt{2x + \sqrt{3x}} = x + \sqrt{j(x)}$ 
$h'(x) = 1 + \frac{1}{2*\sqrt{j(x)}} * j'(x)$ $=1 + \frac{1}{2*\sqrt{2x + \sqrt{3x}}} * (2 + \frac{\sqrt{3}}{2\sqrt{x}})$
3) $g(x) = \sqrt{x + \sqrt{2x + \sqrt{3x}}} = \sqrt{h(x)}$
$g'(x) = \frac{1}{2*\sqrt{h(x)}} * h'(x)$ $=\frac{1}{2*\sqrt{x + \sqrt{2x + \sqrt{3x}}}} * (1 + \frac{1}{2*\sqrt{2x + \sqrt{3x}}} * (2 + \frac{\sqrt{3}}{2\sqrt{x}}))$
A: Taking @Eff awesome start point
$$
f = \sqrt{x + \sqrt{2x + \sqrt{3x}}}
$$
as
$$
f^2 - x = \sqrt{2x + \sqrt{3x}}\implies \left(f^2-x\right)^2 - 2x = \sqrt{3x}
$$
and then do implicit differentiation to get $f'$? maybe
