## MATH1111 Quizzes

The Chain Rule Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on the chain rule and corresponds to Section 14.6 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There is a useful explanation and extra examples at http://www.math.hmc.edu/calculus/tutorials/multichainrule/. There are interactive exercises at http://www.math.temple.edu/ cow/ but you need to go to Book III, Functions, Chain Rule to find the questions. Pressing help gives you an explanation of how to use the chain rule.

There are more web quizzes at Wiley, select Section 6. This quiz has 14 questions.

Suppose $z=2xy$ where $x={t}^{2}+1$ and $y=3-t$. Use the chain rule to determine which of the following is $\frac{dz}{dt}$. Exactly one option must be correct)
 a) $\frac{dz}{dt}=6t-4{t}^{2}-2$ b) $\frac{dz}{dt}=4{t}^{3}+6t-6$ c) $\frac{dz}{dt}=6t-2$ d) $\frac{dz}{dt}=12t-6{t}^{2}-2$

Choice (a) is incorrect
Try again, you may have lost a $t$.
Choice (b) is incorrect
Try again, check your partial derivatives.
Choice (c) is incorrect
Try again, carefully watch your signs.
Choice (d) is correct!
We compute $\frac{\partial z}{\partial x}=2y$, $\frac{dx}{dt}=2t$ $\frac{\partial z}{\partial y}=2x$ and $\frac{dy}{dt}=-1$, so
 $\frac{dz}{dt}$ = $\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$ = $2y\cdot 2t-2x$ = $2\left(3-t\right)\cdot 2t-2\left({t}^{2}+1\right)$ = $12t-4{t}^{2}-2{t}^{2}-2=12t-6{t}^{2}-2$.
Suppose $z=xy-2{y}^{2}$ where $x=3t+1$ and $y=2t$. Use the chain rule to determine which of the following is $\frac{dz}{dt}$. Exactly one option must be correct)
 a) $\frac{dz}{dt}=-4t+2$ b) $\frac{dz}{dt}=-16t-t$ c) $\frac{dz}{dt}=18t+2$ d) $\frac{dz}{dt}=-10{t}^{2}+8t+1$

Choice (a) is correct!
We compute $\frac{\partial z}{\partial x}=y$, $\frac{dx}{dt}=3$ $\frac{\partial z}{\partial y}=x-4y$ and $\frac{dy}{dt}=2$, so
 $\frac{dz}{dt}$ = $\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$ = $y\cdot 3+\left(x-4y\right)\cdot 2$ = $2x-5y$ = $2\left(3t+1\right)-10t=-4t+2$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Try again, you may have not differentiated $x$ and $y$ with respect to $t$.
Which of the following is the derivative, $\frac{dz}{dt}$, written in terms of $t$ where $z={x}^{2}y+{y}^{2}x$ where $x=cost$ and $y={t}^{2}?$ Exactly one option must be correct)
 a) $\frac{dz}{dt}=2{t}^{2}costsint+{t}^{4}sint+2t{cos}^{2}t+4{t}^{3}cost$ b) $\frac{dz}{dt}=-2{t}^{2}costsint-2{t}^{2}sint+4tcost+4{t}^{3}$ c) $\frac{dz}{dt}=-2{t}^{2}costsint-{t}^{4}sint+2t{cos}^{2}t+4{t}^{3}cost$ d) $\frac{dz}{dt}=\left(2xy+{y}^{2}\right)\left(-sint\right)+\left({x}^{2}+2xy\right)2t$

Choice (a) is incorrect
Try again, you may not have differentiated $cost$ correctly, check your signs.
Choice (b) is incorrect
Try again, you may not have found the correct partial derivatives.
Choice (c) is correct!
We compute $\frac{\partial z}{\partial x}=2xy+{y}^{2}$, $\frac{dx}{dt}=-sint$ $\frac{\partial z}{\partial y}={x}^{2}+2xy$ and $\frac{dy}{dt}=2t$. Therefore, $\begin{array}{llll}\hfill \frac{dz}{dt}& =\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(2xy+{y}^{2}\right)\cdot sint+\left({x}^{2}+2xy\right)\cdot 2t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(2{t}^{2}cost+{t}^{4}\right)\cdot \left(-sint\right)+\left({cos}^{2}t+2{t}^{2}cost\cdot 2t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-2{t}^{2}costsint-{t}^{4}sint+2t{cos}^{2}t+4{t}^{3}cost.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (d) is incorrect
Try again, this is $\frac{dz}{dt}$ but not in terms of $t$.
Suppose $z=lnxy$ where $x=\sqrt{t}$ and $y={t}^{2}$. Use the chain rule to determine which of the following is $\frac{dz}{dt}$. Exactly one option must be correct)
 a) $\frac{dz}{dt}=\frac{1}{{t}^{2}}+\frac{2{t}^{3}}{\sqrt{t}}$ b) $\frac{dz}{dt}=\frac{3}{2t}$ c) $\frac{dz}{dt}=\frac{2}{t}$ d) $\frac{dz}{dt}=\frac{1}{2t}$

Choice (a) is incorrect
Try again, you may have not have differentiated $lnxy$ correctly.
Choice (b) is correct!
We compute $\frac{\partial z}{\partial x}=\frac{1}{x}$, $\frac{dx}{dt}=\frac{1}{2\sqrt{t}}$ $\frac{\partial z}{\partial y}=\frac{1}{y}$ and $\frac{dy}{dt}=2t$. Therefore, $\begin{array}{llll}\hfill \frac{dz}{dt}& =\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{x}\cdot \frac{1}{2\sqrt{t}}+\frac{1}{y}\cdot 2t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{\sqrt{t}}\cdot \frac{1}{2\sqrt{t}}+\frac{1}{{t}^{2}}\cdot 2t\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2t}+\frac{1}{t}=\frac{3}{2t}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (c) is incorrect