## MATH1111 Quizzes

Differentiating Powers and Polynomials Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 12 and corresponds to Section 3.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes again at Wiley, select Section 1. The function referred to in Question 13 is the function in Question 12.

There is a film at http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv03.html which covers this topic well.

There are quizzes and web resources at http://quiz.econ.usyd.edu.au/mathquiz/differentiation/index.php.

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

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
The derivative of a constant is zero.
There is at least one mistake.
For example, choice (c) should be False.
$-3-1=-4$ so ${f}^{\prime }\left(x\right)=-3{x}^{-4}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be True.
Since $y={t}^{-\frac{1}{2}}⇒\frac{dy}{dt}=-\frac{1}{2}{t}^{-\frac{3}{2}}=\frac{-1}{2\sqrt{{t}^{3}}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (e) should be False.
Recall that $h\left(z\right)=-2{z}^{-2}+{z}^{-1}\phantom{\rule{0.3em}{0ex}}.$
${h}^{\prime }\left(z\right)=\frac{4}{{z}^{3}}-\frac{1}{{z}^{2}}\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. True
2. False The derivative of a constant is zero.
3. False $-3-1=-4$ so ${f}^{\prime }\left(x\right)=-3{x}^{-4}\phantom{\rule{0.3em}{0ex}}.$
4. True Since $y={t}^{-\frac{1}{2}}⇒\frac{dy}{dt}=-\frac{1}{2}{t}^{-\frac{3}{2}}=\frac{-1}{2\sqrt{{t}^{3}}}\phantom{\rule{0.3em}{0ex}}.$
5. False Recall that $h\left(z\right)=-2{z}^{-2}+{z}^{-1}\phantom{\rule{0.3em}{0ex}}.$
${h}^{\prime }\left(z\right)=\frac{4}{{z}^{3}}-\frac{1}{{z}^{2}}\phantom{\rule{0.3em}{0ex}}.$
Let $y=4{s}^{3}+3s+7\phantom{\rule{0.3em}{0ex}}.$ Which one of the following statements is correct? Exactly one option must be correct)
 a) $\frac{dy}{ds}=8s+3\phantom{\rule{0.3em}{0ex}}.$ b) $\frac{dy}{ds}=12{s}^{2}+10\phantom{\rule{0.3em}{0ex}}.$ c) $\frac{dy}{ds}=12{s}^{2}+3\phantom{\rule{0.3em}{0ex}}.$ d) None of the above, you cannot differentiate with respect to $s\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you have differentiated $y=4{s}^{2}+3s+7\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, recall that the derivative of a constant is 0.
Choice (c) is correct!
$\frac{dy}{ds}=4×3{x}^{2}+3=12{x}^{2}+3\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, the function can be differentiated, it doesn’t matter what the variable is called.
A particle is moving in a straight line and the distance, in metres, from its starting position at time $t$, in seconds, is given by $s\left(t\right)=27t+7{t}^{2}$ for the first two seconds of its motion. Which of the following gives the formula for the velocity, $v\phantom{\rule{0.3em}{0ex}},$ of the particle, in metres per second for the first two seconds? Exactly one option must be correct)
 a) $v\left(t\right)=27+14t\phantom{\rule{0.3em}{0ex}}.$ b) $v\left(t\right)=27+7t\phantom{\rule{0.3em}{0ex}}.$ c) $v\left(t\right)=34\phantom{\rule{0.3em}{0ex}}.$ d) There is not enough information available.

Choice (a) is correct!
$v\left(t\right)={s}^{\prime }\left(t\right)=27+7{t}^{2}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, $v\left(t\right)={s}^{\prime }\left(t\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, this is the average velocity for the first two seconds.
Choice (d) is incorrect
Try again, recall $v\left(t\right)={s}^{\prime }\left(t\right)\phantom{\rule{0.3em}{0ex}}.$
A particle’s velocity at time $t$ is given by the formula $v\left(t\right)=u+at\phantom{\rule{0.3em}{0ex}}.$
Which of the following could be the function describing the distance of the particle from its starting position. Exactly one option must be correct)
 a) $s\left(t\right)=ut+a{t}^{2}\phantom{\rule{0.3em}{0ex}}.$ b) $s\left(t\right)=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0.3em}{0ex}}.$ c) $s\left(t\right)=u+2a{t}^{2}\phantom{\rule{0.3em}{0ex}}.$ d) None of the above.

Choice (a) is incorrect
Try differentiating the function and see if it gives you $v\left(t\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is correct!
${s}^{\prime }\left(t\right)=u+at=v\left(t\right)$ so this is the correct function.
Choice (c) is incorrect
Try differentiating the function and see if it gives you $v\left(t\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try differentiating the functions and see which one gives you $v\left(t\right)\phantom{\rule{0.3em}{0ex}}.$