Let $z=e-x(3xy-4x+y2),x=\sin t$ and $y=\cos t$

(a) Find $\frac{dz}{dt}$

by using the Chain Rule, the Theorem $12.32$

(b) Find $\frac{dz}{dt}$by evaluating $f\left(x\right(t),y(t\left)\right)=f(\sin t,\cos t)$and taking the derivative of the resulting function.

(c) Show that your answers from parts (a) and (b) are the same. Which method was easier?

Let $z=e^{-x}\left(3xy-4x+y^2\right),x=\sin t$ and $y=\cos t$

The objective is to find $\frac{dz}{dt}$using chain rule.

According to the chain rule

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

where $z=f(x,y),x=u\left(t\right)$and $y=v\left(t\right)$

First, find $\frac{\partial z}{\partial x}$

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left\{e^{-x}\left(3xy-4x+y^2\right)\right\} \\ =e^{-x}\frac{\partial }{\partial x}\left(3xy-4x+y^2\right)+\left(3xy-4x+y^2\right)\frac{\partial }{\partial x}{e}^{-x} \\ =e^{-x}\left(3y\frac{\partial }{\partial x}x-4\frac{\partial }{\partial x}x+\frac{\partial }{\partial x}{y}^2\right)+\left(3xy-4x+y^2\right)e^{-x}\frac{\partial }{\partial x}(-x) \\ =e^{-x}(3y-4·1+0)+\left(3xy-4x+y^2\right)e^{-x}(-1) \\ =e^{-x}(3y-4)-\left(3xy-4x+y^2\right)e^{-x} \end{gathered}$$

Next, find $\frac{\partial z}{\partial y}$

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left\{e^{-x}\left(3xy-4x+y^2\right)\right\} \\ =e^{-x}\frac{\partial }{\partial y}\left(3xy-4x+y^2\right) \\ =e^{-x}\left(3x\frac{\partial }{\partial y}y-4\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}{y}^2\right) \\ =e^{-x}(3x·1-4·0+2y) \\ =e^{-x}(3x+2y) \end{gathered}$$

Again,

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}\sin t \\ =\cos t \end{gathered}$$

Also,

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\cos t \\ =-\sin t \end{gathered}$$

Thus,

$$\begin{gathered} \frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt} \\ =\left\{e^{-x}(3y-4)-\left(3xy-4x+y^2\right)e^{-x}\right\}\cos t+e^{-x}(3x+2y)(-\sin t) \\ =e^{-x}\left[\left\{(3y-4)-\left(3xy-4x+y^2\right)\right\}\cos t+(3x+2y)(-\sin t)\right] \\ =e^{-x}\left[\left\{(3\cos t-4)-\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\right\}\cos t\right. \\ +(3\sin t+2\cos t)(-\sin t)] \end{gathered}$$

$$\begin{gathered} =e^{-\sin t}\left(3{\cos }^{2}t-4\cos t-3\sin t{\cos }^{2}t+4\sin t\cos t-{\cos }^{3}t\right. \\ \left.-3{\sin }^{2}t-2\sin t\cos t\right) \\ =e^{-\sin t}\left(3{\cos }^{2}t-4\cos t-3\sin t{\cos }^{2}t+2\sin t\cos t-{\cos }^{3}t-3{\sin }^{2}t\right) \\ =e^{-\sin t}\left(2\sin t\cos t-3\sin t{\cos }^{2}t-{\cos }^{3}t+3{\cos }^{2}t-3{\sin }^{2}t-4\cos t\right) \end{gathered}$$

The objective is to find $\frac{dz}{dt}$

Rephrase the function as follows:

$z=e^{-x\sin t}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\dots \dots \text{(1)}$

Differentiate ( 1 ) with respect to $t$

$$\begin{gathered} \frac{dz}{dt}=\frac{d}{dt}\left\{e^{-\sin t}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\right\} \\ =e^{-\sin t}\frac{d}{dt}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right) \\ +\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\frac{d}{dt}{e}^{-\sin t} \\ =e^{-\sin t}\left(3\frac{d}{dt}\sin t\cos t-4\frac{d}{dt}\sin t+\frac{d}{dt}{\cos }^{2}t\right) \\ +\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)e^{-\sin t}\frac{d}{dt}(-\sin t) \end{gathered}$$

Parts (a) and (b) show that the outcome is the same. The method based on the chain rule is less difficult than the method used in part (b).