 ## MATH1111 Quizzes

Differentiating the Exponential Function Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 13 and corresponds to Section 3.2 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 2.

Section 3.6 of The Mathematics Learning Hub’s booklet on differentiation Introduction to Differential Calculus covers differentiating the exponential function.

Which of the following is the derivative of $g\left(t\right)={5}^{t}\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) ${g}^{\prime }\left(t\right)=\left(ln5\right){5}^{t}\phantom{\rule{0.3em}{0ex}}.$ b) ${g}^{\prime }\left(t\right)=\left(lna\right){5}^{t}\phantom{\rule{0.3em}{0ex}}.$ c) ${g}^{\prime }\left(t\right)=\left(ln5\right){5}^{x}\phantom{\rule{0.3em}{0ex}}.$ d) ${g}^{\prime }\left(t\right)=\frac{1}{ln5}{5}^{t}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
$\frac{d}{dt}\left({a}^{t}\right)=\left(lna\right){a}^{t}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, $\frac{d}{dt}\left({a}^{t}\right)=\left(lna\right){a}^{t}$ and $a=5\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, we are differentiating with respect to $t$, not $x$.
Choice (d) is incorrect
Try again, $\frac{d}{dt}\left({a}^{t}\right)=\left(lna\right){a}^{t}\phantom{\rule{0.3em}{0ex}}.$
Which of the following is the derivative of $y=3{x}^{2}+2{e}^{x}\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) $\frac{dy}{dx}=3{x}^{2}+2{e}^{x}\phantom{\rule{0.3em}{0ex}}.$ b) $\frac{dy}{dx}=5x+2{e}^{x}\phantom{\rule{0.3em}{0ex}}.$ c) $\frac{dy}{dx}=6x+2e\phantom{\rule{0.3em}{0ex}}.$ d) $\frac{dy}{dx}=6x+2{e}^{x}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you have not differentiated the first term.
Choice (b) is incorrect
Try again, you have not differentiated the first term correctly.
Choice (c) is incorrect
Try again, you have not differentiated the last term correctly.
Choice (d) is correct!
The derivative of ${x}^{2}$ is $2x$ and the derivative of ${e}^{x}$ is ${e}^{x}$ so
$\frac{dy}{dx}=3\left(2x\right)+2\left({e}^{x}\right)=6x+2{e}^{x}\phantom{\rule{0.3em}{0ex}}.$
Which of the following is the equation of tangent to $y=2-2{e}^{x}$ at $x=1\phantom{\rule{0.3em}{0ex}}.$ Exactly one option must be correct)
 a) $y=-2x\phantom{\rule{0.3em}{0ex}}.$ b) $y=8.732-5.4336x\phantom{\rule{0.3em}{0ex}}.$ c) $y=2e-2x\phantom{\rule{0.3em}{0ex}}.$ d) $y=2-\left(2e\right)x\phantom{\rule{0.3em}{0ex}}$

Choice (a) is incorrect
Try again, this is equation of the tangent at $x=0\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, you have the correct gradient but the intercept is incorrect.
Choice (c) is incorrect
Try again, you seem to have confused the gradient and the intercept.
Choice (d) is correct!
$\frac{dy}{dx}=-2{e}^{x}$ so ${\frac{dy}{dx}|}_{x=1}=-2e\phantom{\rule{0.3em}{0ex}}.$
The gradient of the tangent is therefore $-2e$ and $y=\left(-2e\right)x+b\phantom{\rule{0.3em}{0ex}}.$
At $x=1\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}y=2-2e$ and the tangent must pass through this point so we substitute these values into the equation of the line to find $b\phantom{\rule{0.3em}{0ex}}.$
$2-2e=-2e\left(1\right)+b⇒b=2$ and $y=\left(-2e\right)x+2=2-\left(2e\right)x\phantom{\rule{0.3em}{0ex}}.$
We saw that if the population of a city increases at a rate which is proportional to the current population and was 2 million in 1980 and 2.5 million in 1990 then the formula for the population $t$ years after 1980 was given by $P\left(t\right)=2{e}^{0.0223t}\phantom{\rule{0.3em}{0ex}}.$ Which of the following correctly gives the population in the form $P\left(t\right)={P}_{0}^{}{a}^{t}\phantom{\rule{0.3em}{0ex}},$ and the growth rate of $t$ years after 1980, to 4 decimal places? Exactly one option must be correct)
 a) $P\left(t\right)=2{\left(1.0226\right)}^{t}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{P}^{\prime }\left(t\right)=2{\left(1.0226\right)}^{t}$ b) $P\left(t\right)=2{\left(1.0226\right)}^{t}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{P}^{\prime }\left(t\right)=0.0446{\left(1.0226\right)}^{t}$ c) $P\left(t\right)={\left(2.0452\right)}^{t}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{P}^{\prime }\left(t\right)=ln\left(2.0452\right){\left(2.0452\right)}^{t}$ d) $P\left(t\right)={\left(2.0452\right)}^{t}\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{P}^{\prime }\left(t\right)=4.094{\left(2.0452\right)}^{t}$

Choice (a) is incorrect
Try again, you do not have the correct growth rate.
Choice (b) is correct!
$P\left(t\right)=2{\left({e}^{0.0223}\right)}^{t}=2{\left(1.0226\right)}^{t}\phantom{\rule{0.3em}{0ex}},$
${P}^{\prime }\left(t\right)=2×ln1.0226{\left(1.0226\right)}^{t}=0.0446{\left(1.0266\right)}^{t}\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you do not have the correct function for the population although you have differentiated it correctly.
Choice (d) is incorrect
Try again, you have both the function for the population and its derivative wrong.