Q.If $z(x,y)=x\tan^{-1}(xy),\ x=t^2,\ y=se^t,\ s,t\in\mathbb R$. Find $\dfrac{\partial z}{\partial s}$ and $\dfrac{\partial z}{\partial t}$ at $s=t=1$.

When the two variables $x,y$ of a function $W(x,y)$ are themselves both functions of a single parameter $t$, the composite $W(x(t),y(t))$ depends ultimately only on $t$, and its ordinary derivative $\dfrac{dW}{dt}$ can be found without first eliminating $x,y$ in favour of $t$.

Theorem 8.2 (Function of Function / Chain Rule, one parameter). If $W(x,y)$ has partial derivatives $\dfrac{\partial W}{\partial x},\dfrac{\partial W}{\partial y}$, and $x=x(t),\,y=y(t)$ are both differentiable, then $W$ is a differentiable function of $t$ and

The tree diagram is the standard memory aid: $W$ branches to $x$ and $y$ (via $\partial W/\partial x,\ \partial W/\partial y$), and each of $x,y$ branches down to $t$ (via $dx/dt,\ dy/dt$); multiply along each branch and add over the two paths.

Theorem 8.3 (Chain Rule, two parameters). If $W(x,y)$ has partial derivatives, and $x=x(s,t),\,y=y(s,t)$ both have partial derivatives with respect to $s$ and $t$, then

This is the tool that converts a function between coordinate systems — e.g. Cartesian $(x,y)$ to polar $(r,\theta)$ via $x=r\cos\theta,\,y=r\sin\theta$ — without re-deriving from scratch, and it generalizes to any number of intermediate and parameter variables (three-variable $W(x,y,z)$ with $x,y,z$ each functions of $s,t$, and so on). …