Yes chains inside chains.
One way to think about it is to just keep pushing out one more nested function and differentiating, working it like so.
$\frac {d}{dx} \sin^3\sqrt{x^3 + 2x^2}\\
(3\sin^2\sqrt{x^3 + 2x^2})\frac {d}{dx}\sin\sqrt{x^3 + 2x^2}\\
(3\sin^2\sqrt{x^3 + 2x^2})(\cos\sqrt{x^3 + 2x^2})\frac {d}{dx}\sqrt{x^3 + 2x^2}\\
(3\sin^2\sqrt{x^3 + 2x^2})(\cos\sqrt{x^3 + 2x^2})(\frac {1}{2\sqrt{x^3+2x^2}})\frac {d}{dx}(x^3 + 2x^2)\\
(3\sin^2\sqrt{x^3 + 2x^2})(\cos\sqrt{x^3 + 2x^2})(\frac {1}{2\sqrt{x^3+2x^2}})(3x^2 + 4x)$
The other is to think of the composition of functions beforehand.
$u = x^3 + 2x^2\\
v = \sqrt u\\
w = \sin v\\
y = w^3$
And then apply the chain rule:
$\frac {dy}{dx} = \frac {dy}{dw}\frac {dw}{dv}\frac {dv}{du}\frac {du}{dx}\\
\frac {dy}{dx} = (3w^2)(\cos v)(\frac {1}{2\sqrt{u}})(3x^2 + 4x)\\
\frac {dy}{dx} = (3\sin^2 \sqrt{x^3+2x^2})(\cos \sqrt{x^3+2x^2})(\frac {1}{2\sqrt{x^3+2x^2}})(3x^2 + 4x)\\$
I would say the second approach is conceptually more difficult, but easier to keep organized.
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signs. $\endgroup$ – saulspatz Oct 14 '20 at 17:43