Tree diagram Chain Rule

Question

Let $$ \begin{align} w &= f(x,y,z)\\ z &= z(x,y)\\ x &= x(t)\\ y &= y(t)\\ \end{align} $$ be differentiable functions.
Find the formula for the derivative of $w$ with respect to $t$.

My answer: $w = f(x,y,z) \implies \dfrac{dw}{dt} = \dfrac{dw}{dx} \cdot \dfrac{dw}{dy} \cdot\dfrac{dw}{dz}$ $$ \begin{align} z &= z(x,y) \\ x &= x(t) \\ y &= y(t) \\ \end{align} \implies \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y}\cdot\frac{dy}{dt} $$ It is getting confusing to me. I used to make a tree diagram to make it a bit easier but this time, it is out of my reach. Can someone give me a hint?

Answer 1

Note that \begin{eqnarray*} w\to \left\{\begin{aligned} \color{blue}{x \to x=(t)} \\ \color{red}{ y \to y=(t)} \\ \color{green}{ z \to z= z(x,y) \to \left\{\begin{aligned} \color{blue}{x\to x(t)} \\ \color{red}{y \to y(t)}\end{aligned} \right.} \\ \end{aligned}\right. \end{eqnarray*} Now, by chain rule, you have $$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\color{blue}{\frac{dx}{dt}}+\frac{\partial w}{\partial y}\color{red}{\frac{dy}{dt}}+\frac{\partial w}{\partial z}\color{\green}{\frac{\partial z}{\partial x}}\color{blue}{\frac{dx}{dt}}+\frac{\partial w}{\partial z}\color{\green}{\frac{\partial z}{\partial y}}\color{red}{\frac{dy}{dt}}$$