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/

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)

*There is at least one mistake.*

For example, choice (a) should be False.

*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.

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.

*Correct!*

*False*If we have to raise all of ${x}^{2}+5$ to the power $e$ so ${x}^{2}+5$ is the inside function.*True**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 $f\left(x\right)=\frac{1}{{x}^{2}}\phantom{\rule{0.3em}{0ex}}.$*True**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.

*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 True.

${h}^{\prime}\left(x\right)=2x\frac{1}{2}{\left({x}^{2}+5\right)}^{-\frac{1}{2}}=\frac{x}{\sqrt{{x}^{2}+5}}\phantom{\rule{0.3em}{0ex}}.$

*There is at least one mistake.*

For example, choice (c) should be False.

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

*There is at least one mistake.*

For example, choice (d) should be False.

*There is at least one mistake.*

For example, choice (e) should be True.

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

*Correct!*

*True*This is a straight forward application of the chain rule.*True*Since $h\left(x\right)={\left({x}^{2}+5\right)}^{\frac{1}{2}}$ we have

${h}^{\prime}\left(x\right)=2x\frac{1}{2}{\left({x}^{2}+5\right)}^{-\frac{1}{2}}=\frac{x}{\sqrt{{x}^{2}+5}}\phantom{\rule{0.3em}{0ex}}.$*False*Since $h\left(x\right)={\left({x}^{3}+2x+1\right)}^{-2}$ we have

${h}^{\prime}\left(x\right)=\left(3{x}^{2}+2\right)\left(-2\right){\left({x}^{3}+2x+1\right)}^{-3}=\frac{-2\left(3{x}^{2}+2\right)}{{\left({x}^{3}+2x+1\right)}^{3}}\phantom{\rule{0.3em}{0ex}}.$*False*${h}^{\prime}\left(x\right)=\left(8x+{e}^{2}\right)3{\left(2{x}^{4}+{e}^{x}\right)}^{2}=3\left(8x+{e}^{2}\right){\left(2{x}^{4}+{e}^{x}\right)}^{2}\phantom{\rule{0.3em}{0ex}}.$*True*Since $h\left(x\right)={\left(2{x}^{2}+3x+1\right)}^{\frac{2}{3}}$ we have

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

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

Hence $\frac{dy}{dt}=0$ when $t=\frac{1}{2}$ and $y=\frac{1}{2}{e}^{-1}=\frac{1}{2e}\phantom{\rule{0.3em}{0ex}}.$

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is correct!*

$\frac{{d}^{2}y}{d{t}^{2}}=-2{e}^{-2t}-2\left({e}^{-2t}-2t{e}^{-2t}\right)={e}^{-2t}\left(-2-2+4t\right)={e}^{-2t}\left(4t-4\right)\phantom{\rule{0.3em}{0ex}}.$

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*

You need the second derivative to be positive.