Let \(z=x^{2}y-y^{2}x,\) where \(x=\sin t\) and \(y=e^{t}\). Find \(\dfrac{dz}{dt}\).

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 \(\dfrac{dy}{dt}.\) \[ \dfrac{\partial z}{\partial x}=2xy-y^{2}\qquad \dfrac{ \partial z}{\partial y}=x^{2}-2xy\qquad \dfrac{dx}{dt} =\cos t\qquad \dfrac{dy}{dt}=e^{t} \]

Then we use Chain Rule I to find \(\dfrac{dz}{dt}.\) \[ \begin{eqnarray*} \frac{dz}{dt}\underset{ \underset{ \color{#0066A7}{\hbox{Chain Rule I}}} {\color{#0066A7}{\uparrow }}} {=}\frac{ \partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{ dt}=(2xy-y^{2})(\cos t)+(x^{2}-2xy)(e^{t}) \tag{1} \end{eqnarray*} \]

Since \(z\) is a function of \(t,\) we express \(\dfrac{dz}{dt}\) in terms of \(t.\) \[ \begin{eqnarray*} \frac{dz}{dt}&&=(2e^{t}\sin t-e^{2t})(\cos t)+(\sin ^{2}t-2e^{t}\sin t)(e^{t})\qquad {\color{#0066A7}{\hbox{\(x=\sin t, y=e^{t}\)}}} \nonumber\\ &&=e^{t}[ \sin ( 2t) -e^{t}\cos t+\sin ^{2}t-2e^{t}\sin t] \qquad {\color{#0066A7}{\hbox{\(2\sin t\cos t =\sin (2t) \)}}} \tag{2} \end{eqnarray*} \]