Let $z=e^{x}\sin y,$ where $x=e^{t}$ and $y=\dfrac{\pi }{3}e^{-t}$. Find $\dfrac{dz}{dt}$ when $t=0$.

Solution We begin by finding the partial derivatives of $z$, $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$, and the derivatives $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$. $$\dfrac{\partial z}{\partial x}=e^{x}\sin y\qquad \dfrac{\partial z}{\partial y}=e^{x}\cos y\qquad \dfrac{dx}{dt}=e^{t}\qquad \dfrac{dy}{dt}=-\dfrac{\pi }{3}e^{-t}$$

Then we use Chain Rule I to find $\dfrac{dz}{dt}$. $$\begin{eqnarray*} \frac{dz}{dt}\underset{ \underset{ \color{#0066A7}{\hbox{Chain Rule I}}} {\color{#0066A7}{\uparrow }}} {=}\frac{ \partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{ dt}=(e^{x}\sin y)e^{t}+(e^{x}\cos y)\left( -\frac{\pi }{3}e^{-t}\right)\\[-9pt] \end{eqnarray*}$$

We can stop here and evaluate $\dfrac{dz}{dt}.$ When $t=0$, then $x=e^{0}=1$ and $y=\dfrac{\pi }{3}e^{0}=\dfrac{\pi }{3}$. So, when $t=0$, $$\frac{dz}{dt}=\left( e\sin \frac{\pi }{3}\right) (1)+\left( e\cos \frac{\pi }{3}\right) \left( -\frac{\pi }{3}\right) =\frac{e\sqrt{3}}{2}-\frac{\pi e}{6 }=\frac{e}{6}(3\sqrt{3}-\pi )$$