Let $p(x,y,z) = q(q(x^2, xy), q(xyz, sin(x^2y^2z^3)))$ where $q$ is a function of 2-variables. Find all partial derivatives.
What I know/tried: Chain rule needs to be applied but the main issue is how the function is written. I have tried to take the derivative w.r.t $x$ but it is a landscape page wide.
Bonus Question: Does $\frac{\partial p}{\partial z}$ exist?
Using the comments I think I have figured it out: Let $a = q(x^2, xy)$ and $b = q(xyz, sin(x^2y^2z^3))$.
Quickly take the partials w.r.t $x$ now...
$\frac{\partial a}{\partial x} = \frac{\partial q}{\partial x^2}(2x) + \frac{\partial q}{\partial xy}(y)$
and
$\frac{\partial b}{\partial x} = \frac{\partial q}{\partial (xyz)}(yz) + \frac{\partial q}{\partial (sin(x^2y^2z^3))}(cos(x^2y^2z^3)(2xy^2z^3))$
So then rewrite and apply chain rule...
$p(x,y,z) = q(a, b)$
$\frac{\partial p}{\partial x} = \frac{\partial q}{\partial a}\frac{\partial a}{\partial x} + \frac{\partial q}{\partial b}\frac{\partial b}{\partial x}$
After this, we substitute the work we did and since $q$ is arbitrary, we leave the ones with $q$.
$\frac{\partial p}{\partial x} = \frac{\partial q}{\partial a}[\frac{\partial q}{\partial x^2}(2x) + \frac{\partial q}{\partial xy}(y)]+ \frac{\partial q}{\partial b}[\frac{\partial q}{\partial (xyz)}(yz) + \frac{\partial q}{\partial (sin(x^2y^2z^3))}(cos(x^2y^2z^3)(2xy^2z^3))]$
Rinse and repeat for y and z.
Bonus question: Yes this derivative does exist.