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.

*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.

*There is at least one mistake.*

For example, choice (c) should be False.

*There is at least one mistake.*

For example, choice (d) should be True.

*There is at least one mistake.*

For example, choice (e) should be False.

${h}^{\prime}\left(z\right)=\frac{4}{{z}^{3}}-\frac{1}{{z}^{2}}\phantom{\rule{0.3em}{0ex}}.$

*Correct!*

*True**False*The derivative of a constant is zero.*False*$-3-1=-4$ so ${f}^{\prime}\left(x\right)=-3{x}^{-4}\phantom{\rule{0.3em}{0ex}}.$*True*Since $y={t}^{-\frac{1}{2}}\Rightarrow \frac{dy}{dt}=-\frac{1}{2}{t}^{-\frac{3}{2}}=\frac{-1}{2\sqrt{{t}^{3}}}\phantom{\rule{0.3em}{0ex}}.$*False*Recall that $h\left(z\right)=-2{z}^{-2}+{z}^{-1}\phantom{\rule{0.3em}{0ex}}.$

${h}^{\prime}\left(z\right)=\frac{4}{{z}^{3}}-\frac{1}{{z}^{2}}\phantom{\rule{0.3em}{0ex}}.$

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

*Choice (d) is incorrect*

*Choice (a) is correct!*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is incorrect*

Which of the following could be the function describing the distance of the particle from its starting position. Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

*Choice (c) is incorrect*

*Choice (d) is incorrect*