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.

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.
