## MATH1111 Quizzes

The Chain Rule Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 15 and corresponds to Section 3.4 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 4.

There is a film at http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv05.html which covers the chain rule, although it uses the derivative of trigonometric functions which you haven’t covered yet, but don’t let that stop you.

Section 3.5 of The Learning Hub (Mathematics) booklet on differentiation Introduction to Differential Calculus covers differentiating using the chain rule.

There is an applet at http://www.scottsarra.org/applets/calculus/FunctionComposition.html which shows you what the composition of two functions looks like and shows the tangent at each point. There are plenty of exercises with solutions at the following sites but you may not know how to do some of them yet as they involve trigonometric and logarithmic functions.
http://www.math.ucdavis.edu/ kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html
http://archives.math.utk.edu/visual.calculus/2/chain_ rule.2/

The chain rule is applied to composite functions $h\left(x\right)=f\left(g\left(x\right)\right)$ where we describe $f\left(x\right)$ as the outside function and $g\left(x\right)$ as the inside function.
Each of the following functions can be composed of two simple functions which can be differentiated without using the chain rule.
Which of the following statements are true and satisfy the above condition?
(Zero or more options can be correct)
 a) If $h\left(x\right)={e}^{{x}^{2}+5}$ then the inside function is $g\left(x\right)={e}^{x}\phantom{\rule{0.3em}{0ex}}.$ b) If $h\left(x\right)=\sqrt{{x}^{2}+5}$ then the inside function is $g\left(x\right)={x}^{2}+5\phantom{\rule{0.3em}{0ex}}.$ c) If $h\left(x\right)=\frac{1}{{\left({x}^{3}+2x+1\right)}^{2}}$ then the inside function is $g\left(x\right)={\left({x}^{3}+2x+1\right)}^{2}\phantom{\rule{0.3em}{0ex}}.$ d) If $h\left(x\right)={\left(2{x}^{4}+{e}^{x}\right)}^{3}$ then the outside function is $f\left(x\right)={x}^{3}\phantom{\rule{0.3em}{0ex}}.$ e) If $h\left(x\right)=\sqrt[3]{{\left(2{x}^{2}+3x+1\right)}^{2}}$ then the outside function is $f\left(x\right)={x}^{\frac{2}{3}}$

There is at least one mistake.
For example, choice (a) should be False.
If we have to raise all of ${x}^{2}+5$ to the power $e$ so ${x}^{2}+5$ is the inside function.
There is at least one mistake.
For example, choice (b) should be True.
There is at least one mistake.
For example, choice (c) should be False.
$g\left(x\right)={\left({x}^{3}+2x+1\right)}^{2}$ needs to be differentiated using the chain rule.
The inside function should be $g\left(x\right)=\left({x}^{3}+2x+1\right)\phantom{\rule{0.3em}{0ex}},$ and the outside function should be $f\left(x\right)=\frac{1}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$
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 True.
Since $h\left(x\right)=\sqrt[3]{{\left(2{x}^{2}+3x+1\right)}^{2}}={\left(2{x}^{2}+3x+1\right)}^{\frac{2}{3}}$ and the statement is correct.
Correct!
1. False If we have to raise all of ${x}^{2}+5$ to the power $e$ so ${x}^{2}+5$ is the inside function.
2. True
3. False $g\left(x\right)={\left({x}^{3}+2x+1\right)}^{2}$ needs to be differentiated using the chain rule.
The inside function should be $g\left(x\right)=\left({x}^{3}+2x+1\right)\phantom{\rule{0.3em}{0ex}},$ and the outside function should be $f\left(x\right)=\frac{1}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$
4. True
5. True Since $h\left(x\right)=\sqrt[3]{{\left(2{x}^{2}+3x+1\right)}^{2}}={\left(2{x}^{2}+3x+1\right)}^{\frac{2}{3}}$ and the statement is correct.
Which of the following have been differentiated correctly? (Zero or more options can be correct)
 a) If $h\left(x\right)={e}^{{x}^{2}+5}$  then  ${h}^{\prime }\left(x\right)=2x{e}^{{x}^{2}+5}\phantom{\rule{0.3em}{0ex}}.$ b) If $h\left(x\right)=\sqrt{{x}^{2}+5}$  then  ${h}^{\prime }\left(x\right)=\frac{x}{\sqrt{{x}^{2}+5}}\phantom{\rule{0.3em}{0ex}}.$ c) If $h\left(x\right)=\frac{1}{{\left({x}^{3}+2x+1\right)}^{2}}$  then ${h}^{\prime }\left(x\right)=\frac{3{x}^{2}+2}{{\left({x}^{3}+2x+1\right)}^{3}}\phantom{\rule{0.3em}{0ex}}.$ d) If $h\left(x\right)={\left(2{x}^{4}+{e}^{x}\right)}^{3}$  then  ${h}^{\prime }\left(x\right)=3{\left(8x+{e}^{2}\right)}^{2}\left(2{x}^{4}+{e}^{x}\right)\phantom{\rule{0.3em}{0ex}}.$ e) If $h\left(x\right)=\sqrt[3]{{\left(2{x}^{2}+3x+1\right)}^{2}}$  then  ${h}^{\prime }\left(x\right)=\frac{2\left(2x+3\right)}{3\sqrt[3]{2{x}^{2}+3x+1}}\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be True.
This is a straight forward application of the chain rule.
There is at least one mistake.
For example, choice (b) should be True.
Since $h\left(x\right)={\left({x}^{2}+5\right)}^{\frac{1}{2}}$ we have
${h}^{\prime }\left(x\right)=2x\frac{1}{2}{\left({x}^{2}+5\right)}^{-\frac{1}{2}}=\frac{x}{\sqrt{{x}^{2}+5}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be False.
Since $h\left(x\right)={\left({x}^{3}+2x+1\right)}^{-2}$ we have
${h}^{\prime }\left(x\right)=\left(3{x}^{2}+2\right)\left(-2\right){\left({x}^{3}+2x+1\right)}^{-3}=\frac{-2\left(3{x}^{2}+2\right)}{{\left({x}^{3}+2x+1\right)}^{3}}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be False.
${h}^{\prime }\left(x\right)=\left(8x+{e}^{2}\right)3{\left(2{x}^{4}+{e}^{x}\right)}^{2}=3\left(8x+{e}^{2}\right){\left(2{x}^{4}+{e}^{x}\right)}^{2}\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (e) should be True.
Since $h\left(x\right)={\left(2{x}^{2}+3x+1\right)}^{\frac{2}{3}}$ we have
${h}^{\prime }\left(x\right)=\left(4x+3\right)\frac{2}{3}{\left(2{x}^{2}+3x+1\right)}^{-\frac{1}{3}}=\frac{2\left(2x+3\right)}{3\sqrt[3]{2{x}^{2}+3x+1}}\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. True This is a straight forward application of the chain rule.
2. True Since $h\left(x\right)={\left({x}^{2}+5\right)}^{\frac{1}{2}}$ we have
${h}^{\prime }\left(x\right)=2x\frac{1}{2}{\left({x}^{2}+5\right)}^{-\frac{1}{2}}=\frac{x}{\sqrt{{x}^{2}+5}}\phantom{\rule{0.3em}{0ex}}.$
3. False Since $h\left(x\right)={\left({x}^{3}+2x+1\right)}^{-2}$ we have
${h}^{\prime }\left(x\right)=\left(3{x}^{2}+2\right)\left(-2\right){\left({x}^{3}+2x+1\right)}^{-3}=\frac{-2\left(3{x}^{2}+2\right)}{{\left({x}^{3}+2x+1\right)}^{3}}\phantom{\rule{0.3em}{0ex}}.$
4. False ${h}^{\prime }\left(x\right)=\left(8x+{e}^{2}\right)3{\left(2{x}^{4}+{e}^{x}\right)}^{2}=3\left(8x+{e}^{2}\right){\left(2{x}^{4}+{e}^{x}\right)}^{2}\phantom{\rule{0.3em}{0ex}}.$
5. True Since $h\left(x\right)={\left(2{x}^{2}+3x+1\right)}^{\frac{2}{3}}$ we have
${h}^{\prime }\left(x\right)=\left(4x+3\right)\frac{2}{3}{\left(2{x}^{2}+3x+1\right)}^{-\frac{1}{3}}=\frac{2\left(2x+3\right)}{3\sqrt[3]{2{x}^{2}+3x+1}}\phantom{\rule{0.3em}{0ex}}.$
At which of the following points is the derivative of $y=t{e}^{-2t}$ zero? Exactly one option must be correct)
 a) The derivative is never zero. b) $\left(\frac{1}{2}\phantom{\rule{0.3em}{0ex}},-\frac{1}{2e}\right)$ c) $\left(-\frac{1}{2}\phantom{\rule{0.3em}{0ex}},-\frac{e}{2}\right)$ d) $\left(\frac{1}{2}\phantom{\rule{0.3em}{0ex}},\frac{1}{2e}\right)$

Choice (a) is incorrect
Try again, there is a solution.
Choice (b) is incorrect
Try again, you have not substituted into the formula for $y$ correctly.
Choice (c) is incorrect
Try again, you do not have the correct value for $t\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is correct!
$\frac{dy}{dt}={e}^{-2t}-t\left(2\right){e}^{-2t}={e}^{-2t}\left(1-2t\right)$
Hence $\frac{dy}{dt}=0$ when $t=\frac{1}{2}$ and $y=\frac{1}{2}{e}^{-1}=\frac{1}{2e}\phantom{\rule{0.3em}{0ex}}.$
On which of the following intervals is $y=t{e}^{-2t}$ concave up? Exactly one option must be correct)
 a) No interval, $y=t{e}^{-2t}$ is always concave down. b) $\left(-\infty \phantom{\rule{0.3em}{0ex}},\infty \right)\phantom{\rule{0.3em}{0ex}},$ $y=t{e}^{-2t}$ is always concave up. c) $\left(1\phantom{\rule{0.3em}{0ex}},\infty \right)\phantom{\rule{0.3em}{0ex}},$ $y=t{e}^{-2t}$ is concave up when $t>1\phantom{\rule{0.3em}{0ex}}.$ d) $\left(-\infty \phantom{\rule{0.3em}{0ex}},1\right)\phantom{\rule{0.3em}{0ex}},$ $y=t{e}^{-2t}$ is concave up when $t<1\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you need to find the second derivative and find out when it is positive.
Choice (b) is incorrect
Try again, you need to find the second derivative and find out when it is positive.
Choice (c) is correct!
From Question 3 we know that $\frac{dy}{dt}={e}^{-2t}-2t{e}^{-2t}$ so we have
$\frac{{d}^{2}y}{d{t}^{2}}=-2{e}^{-2t}-2\left({e}^{-2t}-2t{e}^{-2t}\right)={e}^{-2t}\left(-2-2+4t\right)={e}^{-2t}\left(4t-4\right)\phantom{\rule{0.3em}{0ex}}.$
The second derivative is zero at $t=1$ and it is positive when $t>1$ and negative when $t<1\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, you have found the correct place where the concavity changes.
You need the second derivative to be positive.