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*}$$