Differentiating a Function of Two Variables Where Each Variable Is a Function of \(t\)

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 \(t=0\), \[ \frac{dz}{dt}=\left( e\sin \frac{\pi }{3}\right) (1)+\left( e\cos \frac{\pi }{3}\right) \left( -\frac{\pi }{3}\right) =\frac{e\sqrt{3}}{2}-\frac{\pi e}{6 }=\frac{e}{6}(3\sqrt{3}-\pi ) \]