 ## MATH1111 Quizzes

The Product and Quotient Rule Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 14 and corresponds to Section 3.3 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 3. Questions 11, 12 and 14 were illegible on 12/12/05.

There is a film at http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv03.html which goes through the product rule and some examples and the quotient rule is covered at http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv04.html. They use the derivative of trigonometric and logarithmic functions that you haven’t seen before but it is very clearly explained.

There is another written explanation and examples at which only use functions that you know about at http://tutorial.math.lamar.edu/AllBrowsers/2413/ProductQuotientRule.asp

Section 3.3 and 3.4 of The Learning Hub (Mathematics) booklet on differentiation Introduction to Differential Calculus covers differentiating the using the product and quotient rules.

If you are looking for more exercises, with solutions of the product rule try http://archives.math.utk.edu/visual.calculus/2/product˙rule.1/ and for exercises, with solutions of the quotient rule try http://archives.math.utk.edu/visual.calculus/2/quotient˙rule.1/

Which of the following correctly use the product rule to differentiate the given functions?
There may be more than one correct answer. (Zero or more options can be correct)
 a) If $f\left(x\right)=\left(x+6\right)\left({x}^{2}+3\right)$ then ${f}^{\prime }\left(x\right)=3{x}^{2}+12x+3\phantom{\rule{0.3em}{0ex}}.$ b) If $f\left(t\right)=\left({t}^{2}+1\right)\sqrt{t}$ then ${f}^{\prime }\left(t\right)=\frac{5}{2}\sqrt{{t}^{3}}+\frac{1}{2\sqrt{t}}\phantom{\rule{0.3em}{0ex}}.$ c) If $y=\left({x}^{2}-1\right)\left({x}^{2}+6\right)$ then $\frac{dy}{dx}=4{x}^{3}+14x\phantom{\rule{0.3em}{0ex}}.$ d) If $y=\left({t}^{3}+2t\right)\left({t}^{2}+2t+1\right)$ then $\frac{dy}{dt}=5{t}^{4}+8{t}^{3}+9{t}^{2}+8t+2\phantom{\rule{0.3em}{0ex}}.$ e) If $h\left(z\right)=\left({z}^{4}+3z\right)\left(z+{z}^{2}+1\right)$ then ${h}^{\prime }\left(z\right)=5{z}^{5}+6{z}^{4}+4{z}^{3}+9{z}^{2}+6z+3\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be True.
Since $f\left(x\right)=\left(x+6\right)\left({x}^{2}+3\right)⇒{f}^{\prime }\left(x\right)=1\left({x}^{2}+3\right)+\left(x+6\right)2x=3{x}^{2}+12x+3\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
Since $f\left(t\right)=\left({t}^{2}+1\right)\sqrt{t}=\left({t}^{2}+1\right){t}^{\frac{1}{2}}⇒{f}^{\prime }\left(t\right)=\left(2t\right){t}^{\frac{1}{2}}+\left({t}^{2}+1\right)\left(\frac{1}{2}{t}^{-\frac{1}{2}}\right)$
then ${f}^{\prime }\left(t\right)=2{t}^{\frac{3}{2}}+\frac{1}{2}{t}^{\frac{3}{2}}+\frac{1}{2}{t}^{-\frac{1}{2}}=\frac{5}{2}{t}^{\frac{3}{2}}+\frac{1}{2}\frac{1}{\sqrt{t}}=\frac{5}{2}\sqrt{{t}^{3}}+\frac{1}{2\sqrt{t}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be False.
Try again, since $y=\left({x}^{2}-1\right)\left({x}^{2}+6\right)⇒\frac{dy}{dx}=2x\left({x}^{2}+6\right)+\left({x}^{2}-1\right)2x=4{x}^{3}+10x\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be True.
Since $y=\left({t}^{3}+2t\right)\left({t}^{2}+2t+1\right)⇒\frac{dy}{dt}=\left(3{t}^{2}+2\right)\left({t}^{2}+2t+1\right)+\left({t}^{3}+2t\right)\left(2t+2\right)\phantom{\rule{0.3em}{0ex}}.$
Expanding and gathering like terms gives $\frac{dy}{dt}=5{t}^{4}+8{t}^{3}+9{t}^{2}+8t+2\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (e) should be False.
Try again, since $h\left(z\right)=\left({z}^{4}+3z\right)\left(z+{z}^{2}+1\right)⇒{h}^{\prime }\left(z\right)=\left(4{z}^{3}+3\right)\left(z+{z}^{2}+1\right)+\left({z}^{4}+3z\right)\left(1+2z\right)\phantom{\rule{0.3em}{0ex}}.$
Expanding and gathering like terms gives ${h}^{\prime }\left(z\right)=6{z}^{5}+5{z}^{4}+4{z}^{3}+9{z}^{2}+6z+3\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. True Since $f\left(x\right)=\left(x+6\right)\left({x}^{2}+3\right)⇒{f}^{\prime }\left(x\right)=1\left({x}^{2}+3\right)+\left(x+6\right)2x=3{x}^{2}+12x+3\phantom{\rule{0.3em}{0ex}}.$
2. True Since $f\left(t\right)=\left({t}^{2}+1\right)\sqrt{t}=\left({t}^{2}+1\right){t}^{\frac{1}{2}}⇒{f}^{\prime }\left(t\right)=\left(2t\right){t}^{\frac{1}{2}}+\left({t}^{2}+1\right)\left(\frac{1}{2}{t}^{-\frac{1}{2}}\right)$
then ${f}^{\prime }\left(t\right)=2{t}^{\frac{3}{2}}+\frac{1}{2}{t}^{\frac{3}{2}}+\frac{1}{2}{t}^{-\frac{1}{2}}=\frac{5}{2}{t}^{\frac{3}{2}}+\frac{1}{2}\frac{1}{\sqrt{t}}=\frac{5}{2}\sqrt{{t}^{3}}+\frac{1}{2\sqrt{t}}\phantom{\rule{0.3em}{0ex}}.$
3. False Try again, since $y=\left({x}^{2}-1\right)\left({x}^{2}+6\right)⇒\frac{dy}{dx}=2x\left({x}^{2}+6\right)+\left({x}^{2}-1\right)2x=4{x}^{3}+10x\phantom{\rule{0.3em}{0ex}}.$
4. True Since $y=\left({t}^{3}+2t\right)\left({t}^{2}+2t+1\right)⇒\frac{dy}{dt}=\left(3{t}^{2}+2\right)\left({t}^{2}+2t+1\right)+\left({t}^{3}+2t\right)\left(2t+2\right)\phantom{\rule{0.3em}{0ex}}.$
Expanding and gathering like terms gives $\frac{dy}{dt}=5{t}^{4}+8{t}^{3}+9{t}^{2}+8t+2\phantom{\rule{0.3em}{0ex}}.$
5. False Try again, since $h\left(z\right)=\left({z}^{4}+3z\right)\left(z+{z}^{2}+1\right)⇒{h}^{\prime }\left(z\right)=\left(4{z}^{3}+3\right)\left(z+{z}^{2}+1\right)+\left({z}^{4}+3z\right)\left(1+2z\right)\phantom{\rule{0.3em}{0ex}}.$
Expanding and gathering like terms gives ${h}^{\prime }\left(z\right)=6{z}^{5}+5{z}^{4}+4{z}^{3}+9{z}^{2}+6z+3\phantom{\rule{0.3em}{0ex}}.$
Which of the following correctly use the quotient rule to differentiate the given functions?
There may be more than one correct answer. (Zero or more options can be correct)
 a) If $y=\frac{{x}^{2}-5}{4x}$ then $\frac{dy}{dx}=-\frac{4{x}^{2}-8x+20}{16{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$ b) If $f\left(t\right)=\frac{2t+1}{2t-1}$ then ${f}^{\prime }\left(t\right)=\frac{-4}{{\left(2t-1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ c) If $f\left(x\right)=\frac{{x}^{3}-1}{{x}^{2}+3x+1}$ then ${f}^{\prime }\left(x\right)=\frac{{x}^{4}+6{x}^{3}+3{x}^{2}+2x+3}{{\left({x}^{2}+3x+1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ d) If $y=\frac{5t-2}{3t+5}$ then $\frac{dy}{dt}=\frac{30t+19}{{\left(3t+5\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ e) If $h\left(z\right)=\frac{{z}^{2}}{{2}^{z}}$ then ${h}^{\prime }\left(z\right)=\frac{2z-{z}^{2}ln2}{{2}^{z}}\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be False.
Try again, you may have missed an sign change.
$\frac{dy}{dx}=\frac{2x\left(4x\right)-\left({x}^{2}-5\right)4}{{\left(4{x}^{2}\right)}^{2}}=\frac{8x-4{x}^{2}+20}{16{x}^{2}}=-\frac{4{x}^{2}-8x-20}{16{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
${f}^{\prime }\left(t\right)=\frac{2\left(2t-1\right)-\left(2t+1\right)\left(2\right)}{{\left(2t-1\right)}^{2}}=\frac{4t-2-4t-2}{{\left(2t-1\right)}^{2}}=\frac{-4}{{\left(2t-1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be True.
${f}^{\prime }\left(x\right)=\frac{3{x}^{2}\left({x}^{2}+3x+1\right)-\left({x}^{3}-1\right)\left(2x+3\right)}{{\left({x}^{2}+3x+1\right)}^{2}}=\frac{{x}^{4}+6{x}^{3}+3{x}^{2}+2x+3}{{\left({x}^{2}+3x+1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be False.
Try again, you may have missed a couple of sign changes.
$\frac{dy}{dt}=\frac{5\left(3t+5\right)-\left(5t-2\right)3}{{\left(3t+5\right)}^{2}}=\frac{15t+25-15t+6}{{\left(3t+5\right)}^{2}}=\frac{31}{{\left(3t+5\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (e) should be True.
${h}^{\prime }\left(z\right)=\frac{2z\left({2}^{z}\right)-{z}^{2}\left(ln2\left({2}^{z}\right)\right)}{{\left({2}^{z}\right)}^{2}}=\frac{{2}^{z}\left(2z-{z}^{2}ln2\right)}{{\left({2}^{z}\right)}^{2}}=\frac{2z-{z}^{2}ln2}{{2}^{z}}\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. False Try again, you may have missed an sign change.
$\frac{dy}{dx}=\frac{2x\left(4x\right)-\left({x}^{2}-5\right)4}{{\left(4{x}^{2}\right)}^{2}}=\frac{8x-4{x}^{2}+20}{16{x}^{2}}=-\frac{4{x}^{2}-8x-20}{16{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$
2. True ${f}^{\prime }\left(t\right)=\frac{2\left(2t-1\right)-\left(2t+1\right)\left(2\right)}{{\left(2t-1\right)}^{2}}=\frac{4t-2-4t-2}{{\left(2t-1\right)}^{2}}=\frac{-4}{{\left(2t-1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
3. True ${f}^{\prime }\left(x\right)=\frac{3{x}^{2}\left({x}^{2}+3x+1\right)-\left({x}^{3}-1\right)\left(2x+3\right)}{{\left({x}^{2}+3x+1\right)}^{2}}=\frac{{x}^{4}+6{x}^{3}+3{x}^{2}+2x+3}{{\left({x}^{2}+3x+1\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
4. False Try again, you may have missed a couple of sign changes.
$\frac{dy}{dt}=\frac{5\left(3t+5\right)-\left(5t-2\right)3}{{\left(3t+5\right)}^{2}}=\frac{15t+25-15t+6}{{\left(3t+5\right)}^{2}}=\frac{31}{{\left(3t+5\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$
5. True ${h}^{\prime }\left(z\right)=\frac{2z\left({2}^{z}\right)-{z}^{2}\left(ln2\left({2}^{z}\right)\right)}{{\left({2}^{z}\right)}^{2}}=\frac{{2}^{z}\left(2z-{z}^{2}ln2\right)}{{\left({2}^{z}\right)}^{2}}=\frac{2z-{z}^{2}ln2}{{2}^{z}}\phantom{\rule{0.3em}{0ex}}.$
Which of the following correctly differentiate $g\left(x\right)=3{x}^{2}\left(2x+3\right)+\frac{{x}^{3}}{{x}^{2}+2}\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) ${g}^{\prime }\left(x\right)=18x\left(x+1\right)+\frac{{x}^{2}\left({x}^{2}+6\right)}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ b) ${g}^{\prime }\left(x\right)=17{x}^{2}+18x+\frac{2{x}^{4}+6{x}^{2}}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ c) ${g}^{\prime }\left(x\right)=18{x}^{2}+18x+\frac{5{x}^{4}+6{x}^{2}}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$ d) ${g}^{\prime }\left(x\right)=12x+\frac{3{x}^{2}}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
$\begin{array}{llll}\hfill {g}^{\prime }\left(x\right)& =6x\left(2x+3\right)+3{x}^{2}\left(2\right)+\frac{3{x}^{2}\left({x}^{2}+2\right)-{x}^{3}\left(2x\right)}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =18{x}^{2}+18x+\frac{{x}^{4}+6{x}^{2}}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =18x\left(x+1\right)+\frac{{x}^{2}\left({x}^{2}+6\right)}{{\left({x}^{2}+2\right)}^{2}}\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Choice (b) is incorrect
Try again, you seem to have made a few errors gathering like terms.
Choice (c) is incorrect
Try again, you seem to have made a mistake with the signs.
Choice (d) is incorrect
Try again, you have not used the product or quotient rule correctly.
The concentration of a certain drug in the blood at time $t$ hours after taking the does is $x$ units, where $x=0.3t{\left(0.3\right)}^{t}\phantom{\rule{0.3em}{0ex}}.$ Which of the following is the rate of change of the concentration? Exactly one option must be correct)
 a) $\frac{dx}{dt}={\left(0.3\right)}^{t}\left(1+\left(ln0.3\right)t\right)\phantom{\rule{0.3em}{0ex}}.$ b) $\frac{dx}{dt}=\left(0.3\right)\left(ln0.3\right)\left({\left(0.3\right)}^{t}\right)\phantom{\rule{0.3em}{0ex}}.$ c) $\frac{dx}{dt}={\left(0.3\right)}^{t+1}\left(1+t\right)\phantom{\rule{0.3em}{0ex}}.$ d) $\frac{dx}{dt}={\left(0.3\right)}^{t+1}\left(1+\left(ln0.3\right)t\right)\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you do not seem to have factorised correctly.
Choice (b) is incorrect
Try again, you have not used the product rule correctly.
Choice (c) is incorrect
Try again, you have not differentiated ${\left(0.3\right)}^{t}$ correctly.
Choice (d) is correct!
$\frac{dx}{dt}=0.3{\left(0.3\right)}^{t}+0.3tln0.3{\left(0.3\right)}^{t}={\left(0.3\right)}^{t+1}\left(1+\left(ln0.3\right)t\right)\phantom{\rule{0.3em}{0ex}}.$