Partial differential equation of a function with multiple dependencies 
*

*Given the following functions and their dependencies:

$$
u(r,v,w) \\
v(r,t) \\
w(r,t)
$$
What is the chain rule expansion of $$\frac{\partial u}{\partial r}$$


*Given the following functions and their dependencies:

$$
y(x,\mu,\sigma) \\
\mu(x) \\
\sigma(x,\mu)
$$
What is the chain rule expansion of $$\frac{\partial y}{\partial x}$$
I am stuck at writing down the first term itself. If I write $\frac{\partial u}{\partial r} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial r} + \dots$, then I get $\frac{\partial u}{\partial r}$ on both sides of the equation, which is incorrect.
 A: \begin{equation}
u=u(r,v(r,t),w(r,t)) \\
\end{equation}
Because $v$ and $w$ depend on $r$ you can't just take $u$'s partial derivative . Instead you need to take its total derivative w.r.t $r$:
$$\dfrac{\mathrm{du}}{\mathrm{dr}}=\frac{\partial u}{\partial r}\underbrace{\frac{\mathrm{dr}}{\mathrm{dr}}}_{1}+\frac{\partial u}{\partial v}\frac{\mathrm{dv}}{\mathrm{dr}}+\frac{\partial u}{\partial w}\frac{\mathrm{dw}}{\mathrm{dr}}$$
$$\dfrac{\mathrm{du}}{\mathrm{dr}}=\frac{\partial u}{\partial r}+\frac{\partial u}{\partial v}\frac{\mathrm{dv}}{\mathrm{dr}}+\frac{\partial u}{\partial w}\frac{\mathrm{dw}}{\mathrm{dr}}$$
For $t$ it's a bit different since $u$ does not depend on $t$ directly:
$$\dfrac{\mathrm{du}}{\mathrm{dt}}=\frac{\partial u}{\partial r}\underbrace{{\frac{\mathrm{dr}}{\mathrm{dt}}}}_{0}+\frac{\partial u}{\partial v}{\frac{\mathrm{du}}{\mathrm{dt}}}+\frac{\partial u}{\partial w}{\frac{\mathrm{dw}}{\mathrm{dt}}}$$
$$\dfrac{\mathrm{du}}{\mathrm{dt}}=\frac{\partial u}{\partial v}{\frac{\mathrm{dv}}{\mathrm{dt}}}+\frac{\partial u}{\partial w}{\frac{\mathrm{dw}}{\mathrm{dt}}}$$
