Exercise 0.1: Differentiating with the formulas: twelve direct calculations

For each function, compute the derivative using the differentiation rules, without going back to the limit definition, then evaluate it at the given point. On the last line, it is the SECOND derivative that is asked for.

One method for all twelve lines: first rewrite every root and every fraction as a power, $\sqrt[3]{x}=x^{\frac{1}{3}}$ and $\frac{4}{x}=4x^{-1}$ ; then spot the LAST operation, a sum, a product, a quotient or a power of an expression; apply the matching rule, multiplying by the derivative of the inside if there is one. Substitute the value of $x$ at the very end.

a) $f(x)=2x^{4}-0.5x^{3}+3x-7$ , then $f'(2)$ .
b) $g(x)=\frac{4}{x}+\frac{2}{x^{3}}$ , then $g'(2)$ .
c) $h(x)=6\sqrt[3]{x}-\frac{8}{\sqrt{x}}$ , then $h'(64)$ .
d) $f(x)=(x^{3}-2)(4x+1)$ , then $f'(1)$ .
e) $g(x)=x^{2}\sqrt{x+3}$ , then $g'(1)$ .
f) $h(x)=\frac{2x-1}{x+3}$ , then $h'(1)$ .
g) $f(x)=\frac{x^{2}-1}{x^{2}+x+1}$ , then $f'(2)$ .
h) $g(x)=(x^{2}+x-1)^{4}$ , then $g'(1)$ .
i) $h(x)=\sqrt{x^{2}+9}$ , then $h'(4)$ .
j) $f(x)=\frac{3}{(x^{2}+1)^{2}}$ , then $f'(1)$ .
k) $g(x)=x(1-2x)^{3}$ , then $g'(1)$ .
l) $h(x)=\frac{x-1}{x+2}$ , then the second derivative $h''(1)$ .

Type your answers, the page tells you right or wrong 0/12

a)

f'(2)

b)

g'(2)

c)

h'(64)

d)

f'(1)

e)

g'(1)

f)

h'(1)

g)

f'(2)

h)

g'(1)

i)

h'(4)

j)

f'(1)

k)

g'(1)

l)

h''(1)

Show the solution

Answers

a) $f'(x)=8x^{3}-1.5x^{2}+3$ ; $f'(2)=61$
b) $g'(x)=-\frac{4}{x^{2}}-\frac{6}{x^{4}}$ ; $g'(2)=-\frac{11}{8}$
c) $h'(x)=2x^{-\frac{2}{3}}+4x^{-\frac{3}{2}}$ ; $h'(64)=\frac{17}{128}\approx 0.133$
d) $f'(x)=3x^{2}(4x+1)+4(x^{3}-2)$ ; $f'(1)=11$
e) $g'(x)=2x\sqrt{x+3}+\frac{x^{2}}{2\sqrt{x+3}}$ ; $g'(1)=4.25$
f) $h'(x)=\frac{7}{(x+3)^{2}}$ ; $h'(1)=\frac{7}{16}$
g) $f'(x)=\frac{x^{2}+4x+1}{(x^{2}+x+1)^{2}}$ ; $f'(2)=\frac{13}{49}\approx 0.265$
h) $g'(x)=4(2x+1)(x^{2}+x-1)^{3}$ ; $g'(1)=12$
i) $h'(x)=\frac{x}{\sqrt{x^{2}+9}}$ ; $h'(4)=\frac{4}{5}$
j) $f'(x)=-\frac{12x}{(x^{2}+1)^{3}}$ ; $f'(1)=-\frac{3}{2}$
k) $g'(x)=(1-2x)^{2}(1-8x)$ ; $g'(1)=-7$
l) $h''(x)=-\frac{6}{(x+2)^{3}}$ ; $h''(1)=-\frac{2}{9}\approx-0.222$

The move that decides everything happens BEFORE differentiating: rewrite in powers, then name the last operation. $\frac{2}{x^{3}}$ is the power $2x^{-3}$ and not a quotient to handle with the quotient rule, $(x^{3}-2)(4x+1)$ is a product, and $(x^{2}+x-1)^{4}$ is a power of an expression, hence a chain rule.

a)

Sum of powers: $f'(x)=8x^{3}-1.5x^{2}+3$ , the constant $-7$ disappearing. At $x=2$ : $f'(2)=64-6+3=61$ .

b)

$g(x)=4x^{-1}+2x^{-3}$ , so $g'(x)=-4x^{-2}-6x^{-4}=-\frac{4}{x^{2}}-\frac{6}{x^{4}}$ . At $x=2$ : $g'(2)=-1-\frac{6}{16}=-\frac{11}{8}=-1.375$ . Pitfall: the exponent comes down AND decreases by 1, so $-3$ becomes $-4$ and not $-2$ .

c)

$h(x)=6x^{\frac{1}{3}}-8x^{-\frac{1}{2}}$ , so $h'(x)=2x^{-\frac{2}{3}}+4x^{-\frac{3}{2}}=\frac{2}{\sqrt[3]{x^{2}}}+\frac{4}{x\sqrt{x}}$ . At $x=64$ , $\sqrt[3]{64^{2}}=16$ and $64\sqrt{64}=512$ : $h'(64)=\frac{2}{16}+\frac{4}{512}=\frac{17}{128}\approx 0.133$ . Pitfall: the sign, $-8\times\left(-\frac{1}{2}\right)=+4$ .

d)

Product: $f'(x)=3x^{2}(4x+1)+4(x^{3}-2)$ . At $x=1$ : $f'(1)=3\times 5+4\times(-1)=11$ . Check by expanding: $f(x)=4x^{4}+x^{3}-8x-2$ , $f'(x)=16x^{3}+3x^{2}-8$ , and $f'(1)=11$ . Pitfall: the product of the derivatives, $3x^{2}\times 4=12$ .

e)

Product of a polynomial and a root: $g'(x)=2x\sqrt{x+3}+x^{2}\cdot\frac{1}{2\sqrt{x+3}}$ . At $x=1$ , $\sqrt{4}=2$ : $g'(1)=2\times 2+\frac{1}{4}=4.25$ .

f)

Quotient of two linear expressions: $h'(x)=\frac{2(x+3)-1\cdot(2x-1)}{(x+3)^{2}}=\frac{7}{(x+3)^{2}}$ . At $x=1$ : $h'(1)=\frac{7}{16}=0.4375$ . Pitfall: reversing the order in the numerator, which gives $-\frac{7}{16}$ .

g)

Quotient, with $u=x^{2}-1$ , $u'=2x$ , $v=x^{2}+x+1$ , $v'=2x+1$ : $f'(x)=\frac{2x(x^{2}+x+1)-(x^{2}-1)(2x+1)}{(x^{2}+x+1)^{2}}=\frac{x^{2}+4x+1}{(x^{2}+x+1)^{2}}$ . At $x=2$ : $u=3$ , $u'=4$ , $v=7$ , $v'=5$ , so $f'(2)=\frac{4\times 7-3\times 5}{49}=\frac{13}{49}\approx 0.265$ .

h)

Chain: $g'(x)=4(x^{2}+x-1)^{3}\cdot(2x+1)$ . At $x=1$ , the inside equals $1$ : $g'(1)=4\times 1\times 3=12$ . Pitfall: forgetting the factor $(2x+1)$ , which gives $4$ .

i)

$h(x)=(x^{2}+9)^{\frac{1}{2}}$ , so $h'(x)=\frac{1}{2}(x^{2}+9)^{-\frac{1}{2}}\cdot 2x=\frac{x}{\sqrt{x^{2}+9}}$ . At $x=4$ : $h'(4)=\frac{4}{5}=0.8$ . Pitfall: $\sqrt{x^{2}+9}$ is not $x+3$ .

j)

$f(x)=3(x^{2}+1)^{-2}$ , so $f'(x)=-6(x^{2}+1)^{-3}\cdot 2x=-\frac{12x}{(x^{2}+1)^{3}}$ . At $x=1$ : $f'(1)=-\frac{12}{8}=-1.5$ . The rewriting avoids the quotient rule, whose constant numerator would give the same result with more writing.

k)

Product, then chain in the second factor: $g'(x)=1\cdot(1-2x)^{3}+x\cdot 3(1-2x)^{2}\cdot(-2)=(1-2x)^{2}(1-8x)$ . At $x=1$ : $g'(1)=-1-6=-7$ . Pitfall: the factor $-2$ , the derivative of $1-2x$ ; without it you get $-1+3=2$ .

l)

First $h'(x)=\frac{(x+2)-(x-1)}{(x+2)^{2}}=\frac{3}{(x+2)^{2}}=3(x+2)^{-2}$ . Then, differentiating this POWER: $h''(x)=-6(x+2)^{-3}=-\frac{6}{(x+2)^{3}}$ . At $x=1$ : $h''(1)=-\frac{6}{27}=-\frac{2}{9}\approx-0.222$ . Pitfall: replacing $x$ with $1$ in $h'$ before differentiating a second time, which gives the derivative of a constant, $0$ .

Check: for each line, the rate $\frac{f(a+0.001)-f(a-0.001)}{0.002}$ computed on a calculator lands within one thousandth of the value found. For a product, expanding first gives a second route, as in d).

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Exercise 0.2: Writing the equation of the tangent line: ten direct calculations

For each of the first seven functions, write the equation of the tangent line to the curve at the point with $x$ -coordinate $a$ given, in the form $y=mx+b$ , and give the slope $m$ and the $y$ -intercept $b$ . The last three lines reverse the question: the slope is given, and you look for the $x$ -coordinate.

One method: compute $f(a)$ , which gives the point, then $f'(a)$ , which gives the slope; write $y=f'(a)(x-a)+f(a)$ ; expand to read $b$ . When the slope is imposed, solve the equation $f'(x)=m$ .

a) $f(x)=x^{2}-5x+2$ , at $a=4$ .
b) $f(x)=x^{3}-2x^{2}+3$ , at $a=2$ .
c) $f(x)=\frac{8}{x^{2}}$ , at $a=2$ .
d) $f(x)=\sqrt{3x+4}$ , at $a=4$ .
e) $f(x)=\frac{x+4}{x-2}$ , at $a=3$ .
f) $f(x)=(x^{2}-3)^{2}$ , at $a=2$ .
g) $f(x)=\sqrt[3]{x}$ , at $a=8$ .
h) $f(x)=2x^{2}-3x+1$ : the $x$ -coordinate of the point where the tangent has a slope of $5$ .
i) $f(x)=-x^{2}+4x$ : the $x$ -coordinate of the point where the tangent is parallel to the line $y=-2x+7$ , then the $y$ -intercept of that tangent.
j) $f(x)=x^{3}-6x^{2}+9x+2$ : the two $x$ -coordinates where the tangent is horizontal.

Type your answers, the page tells you right or wrong 0/19

a)

Slope

m

y

-intercept

b

b)

Slope

m

y

-intercept

b

c)

Slope

m

y

-intercept

b

d)

Slope

m

y

-intercept

b

e)

Slope

m

y

-intercept

b

f)

Slope

m

y

-intercept

b

g)

Slope

m

y

-intercept

b

h)

x

-coordinate

i)

x

-coordinate

y

-intercept of the tangent

j)

Smaller

x

-coordinate

Larger

x

-coordinate

Show the solution

Answers

a) $y=3x-14$ ; $m=3$ , $b=-14$
b) $y=4x-5$ ; $m=4$ , $b=-5$
c) $y=-2x+6$ ; $m=-2$ , $b=6$
d) $y=\frac{3}{8}x+\frac{5}{2}$ ; $m=0.375$ , $b=2.5$
e) $y=-6x+25$ ; $m=-6$ , $b=25$
f) $y=8x-15$ ; $m=8$ , $b=-15$
g) $y=\frac{1}{12}x+\frac{4}{3}$ ; $m\approx 0.083$ , $b\approx 1.333$
h) $f'(x)=4x-3=5$ ; $x=2$
i) $f'(x)=-2$ at $x=3$ ; tangent $y=-2x+9$ , $b=9$
j) $f'(x)=3(x-1)(x-3)$ ; $x=1$ and $x=3$

The move that decides everything: the slope is read from $f'(a)$ , but the POINT is read from $f(a)$ . The most frequent mistake is to write $y=f'(a)(x-a)+f'(a)$ , or to substitute $a$ into $f'$ before having differentiated.

a)

$f(4)=16-20+2=-2$ and $f'(x)=2x-5$ , so $f'(4)=3$ . The tangent is $y=3(x-4)-2=3x-14$ : slope $3$ , $y$ -intercept $-14$ .

b)

$f(2)=8-8+3=3$ and $f'(x)=3x^{2}-4x$ , so $f'(2)=12-8=4$ . The tangent is $y=4(x-2)+3=4x-5$ : slope $4$ , $y$ -intercept $-5$ .

c)

$f(x)=8x^{-2}$ , so $f'(x)=-16x^{-3}=-\frac{16}{x^{3}}$ . $f(2)=2$ and $f'(2)=-2$ . The tangent is $y=-2(x-2)+2=-2x+6$ : slope $-2$ , $y$ -intercept $6$ . Pitfall: $-2(x-2)=-2x+4$ , the sign of the product.

d)

$f(4)=\sqrt{16}=4$ and $f'(x)=\frac{3}{2\sqrt{3x+4}}$ , so $f'(4)=\frac{3}{8}=0.375$ . The tangent is $y=\frac{3}{8}(x-4)+4=\frac{3}{8}x+\frac{5}{2}$ : slope $0.375$ , $y$ -intercept $2.5$ . Pitfall: forgetting the factor $3$ , the derivative of $3x+4$ , which gives a slope of $\frac{1}{8}$ .

e)

$f(3)=\frac{7}{1}=7$ and $f'(x)=\frac{(x-2)-(x+4)}{(x-2)^{2}}=\frac{-6}{(x-2)^{2}}$ , so $f'(3)=-6$ . The tangent is $y=-6(x-3)+7=-6x+25$ : slope $-6$ , $y$ -intercept $25$ .

f)

$f(2)=(4-3)^{2}=1$ and $f'(x)=2(x^{2}-3)\cdot 2x=4x(x^{2}-3)$ , so $f'(2)=8$ . The tangent is $y=8(x-2)+1=8x-15$ : slope $8$ , $y$ -intercept $-15$ .

g)

$f(8)=2$ and $f'(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^{2}}}$ , so $f'(8)=\frac{1}{3\times 4}=\frac{1}{12}$ . The tangent is $y=\frac{1}{12}(x-8)+2=\frac{1}{12}x+\frac{4}{3}$ : slope $\frac{1}{12}\approx 0.083$ , $y$ -intercept $\frac{4}{3}\approx 1.333$ . This is the line that will be used, in the chapter on the differential, to estimate a cube root close to $2$ .

h)

$f'(x)=4x-3$ . Solve $4x-3=5$ : $x=2$ . The point is $(2, 3)$ . Pitfall: solving $f(x)=5$ , which looks for a HEIGHT and not a slope.

i)

Two parallel lines have the same slope, here $-2$ . $f'(x)=-2x+4=-2$ gives $x=3$ , and $f(3)=-9+12=3$ . The tangent is $y=-2(x-3)+3=-2x+9$ : $y$ -intercept $9$ , and not $7$ , which is that of the given line.

j)

Horizontal means zero slope: $f'(x)=3x^{2}-12x+9=3(x^{2}-4x+3)=3(x-1)(x-3)=0$ , so $x=1$ or $x=3$ . The points are $(1, 6)$ and $(3, 2)$ .

Check: the tangent passes through the point, so $ma+b=f(a)$ ; and it hugs the curve, so the difference $f(a+0.01)-\left(m(a+0.01)+b\right)$ does not exceed a few ten-thousandths, whereas it does with a wrong slope.

Exercise 1: From the average rate to the instantaneous rate

An AVERAGE rate of change is computed between two dates; an INSTANTANEOUS rate is computed at one date. The second exists only as a limit of the first, and that is the whole definition of the derivative.

The expenses of a program are $D(t)=0.4t^{2}+6t+120$ millions of dollars, $t$ years after 2015.

a) Compute the average rate of change of the expenses between $t=2$ and $t=6$ .
b) Write the average rate of change between $t=4$ and $t=4+h$ , then simplify it.
c) Deduce $D'(4)$ by taking the limit.
d) Compare your answers to a) and c). Explain the coincidence.
e) Give the units of $D$ and of $D'$ , then interpret $D'(4)$ in one sentence stating a rate.

Show the solution

Answers

a) $\dfrac{170.4-133.6}{4}=9.2$ million dollars per year.
b) $\dfrac{D(4+h)-D(4)}{h}=9.2+0.4h$ for $h\ne 0$ .
c) $D'(4)=9.2$ million dollars per year.
d) Both equal $9.2$ : 4 is the midpoint of $[2, 6]$ , a property reserved for quadratics.
e) Millions of dollars for $D$ , millions of dollars per year for $D'$ : in 2019, expenses were increasing by 9.2 million per year.

a)

$D(2)=0.4\times 4+12+120=1.6+12+120=133.6$ and $D(6)=0.4\times 36+36+120=14.4+36+120=170.4$ . The average rate is $\dfrac{170.4-133.6}{6-2}=\dfrac{36.8}{4}=9.2$ million dollars per year. Geometrically, it is the slope of the SECANT line joining the two points of the curve.

b)

$D(4)=0.4\times 16+24+120=6.4+24+120=150.4$ . Then $D(4+h)=0.4(16+8h+h^{2})+6(4+h)+120=6.4+3.2h+0.4h^{2}+24+6h+120=150.4+9.2h+0.4h^{2}$ . The average rate is therefore $\dfrac{D(4+h)-D(4)}{h}=\dfrac{9.2h+0.4h^{2}}{h}=9.2+0.4h$ for $h\ne 0$ . This is the figure: a line in $h$ , with a hole at $h=0$ , since an average rate over an interval of length zero makes no sense.

c)

$D'(4)=\lim_{h\to 0}(9.2+0.4h)=9.2$ million dollars per year. We have just filled the hole: the limit exists even though the expression is not defined at $h=0$ , and this is exactly the removable discontinuity of the previous chapter, repaired. Every derivative is this repair.

d)

Both equal 9.2, and it is no accident: 4 is the MIDPOINT of the interval $[2, 6]$ . For a quadratic function, the average rate over an interval always equals the instantaneous rate at the midpoint of that interval. Direct check: $D'(t)=0.8t+6$ , so $D'(4)=3.2+6=9.2$ , while $D'(2)=7.6$ and $D'(6)=10.8$ , whose mean is indeed 9.2. The property fails as soon as the degree exceeds 2, and invoking it on an arbitrary function is a mistake.

e)

$D$ is measured in millions of dollars, $t$ in years, so $D'$ is measured in MILLIONS OF DOLLARS PER YEAR: the unit of a derivative is always the unit of the output divided by the unit of the input. Sentence stating a rate: in 2019, the program's expenses were increasing at a rate of 9.2 million dollars per year. The tense matters: it is a rate at that instant, not an amount over the year.

Exercise 2: The basic rules: power, product, quotient

Once the definition is understood, it is no longer used: the rules do the work. Three of them cover most of the course, and two writing traps cost more marks than the rules themselves.

The first trap is to reverse the order of the numerator in the quotient rule; the second is to believe that the derivative of a product is the product of the derivatives.

a) Differentiate $f(x)=5x^{4}-\dfrac{3}{x^{2}}+7\sqrt{x}-9$ .
b) Differentiate $g(x)=(2x+5)(x^{2}-3x)$ with the product rule, then check by expanding first.
c) Differentiate $h(x)=\dfrac{3x+1}{x^{2}+4}$ with the quotient rule.
d) Compute $h'(0)$ .
e) Show on the example $u(x)=v(x)=x$ that the derivative of a product is not the product of the derivatives.

Show the solution

Answers

a) $f'(x)=20x^{3}+\dfrac{6}{x^{3}}+\dfrac{7}{2\sqrt{x}}$ .
b) $g'(x)=6x^{2}-2x-15$ , confirmed by the expansion $g(x)=2x^{3}-x^{2}-15x$ .
c) $h'(x)=\dfrac{-3x^{2}-2x+12}{(x^{2}+4)^{2}}$ .
d) $h'(0)=\dfrac{12}{16}=0.75$ .
e) $(x\cdot x)'=2x$ , whereas the product of the derivatives is $1$ : the two differ everywhere except at $x=0.5$ .

a)

First rewrite each term as a power: $f(x)=5x^{4}-3x^{-2}+7x^{1/2}-9$ . The rule $\left(x^{n}\right)'=nx^{n-1}$ then applies to all of them, negative and fractional exponents included. $f'(x)=20x^{3}+6x^{-3}+\dfrac{7}{2}x^{-1/2}$ , that is $f'(x)=20x^{3}+\dfrac{6}{x^{3}}+\dfrac{7}{2\sqrt{x}}$ . Two signs to watch: $-3x^{-2}$ gives $-3\times(-2)x^{-3}=+6x^{-3}$ , the minus becomes a plus, and the constant $-9$ disappears.

b)

Product rule, $(uv)'=u'v+uv'$ , with $u=2x+5$ and $v=x^{2}-3x$ : $g'(x)=2(x^{2}-3x)+(2x+5)(2x-3)=2x^{2}-6x+4x^{2}-6x+10x-15=6x^{2}-2x-15$ . Check by expanding first: $g(x)=2x^{3}-6x^{2}+5x^{2}-15x=2x^{3}-x^{2}-15x$ , so $g'(x)=6x^{2}-2x-15$ . The two routes agree, and this is the cheapest check on a product of polynomials.

c)

Quotient rule, $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}$ , with $u=3x+1$ and $v=x^{2}+4$ : $h'(x)=\dfrac{3(x^{2}+4)-(3x+1)(2x)}{(x^{2}+4)^{2}}=\dfrac{3x^{2}+12-6x^{2}-2x}{(x^{2}+4)^{2}}=\dfrac{-3x^{2}-2x+12}{(x^{2}+4)^{2}}$ . The ORDER of the numerator is not negotiable: $u'v-uv'$ and not the reverse, otherwise the whole sign of the derivative is wrong, and with it the entire table of variation that follows.

d)

$h'(0)=\dfrac{-0-0+12}{(0+4)^{2}}=\dfrac{12}{16}=0.75$ . Rough check: $h(0)=\dfrac{1}{4}=0.25$ and $h(0.1)=\dfrac{1.3}{4.01}\approx 0.324$ , so the average rate is about $0.74$ over this small interval. This is indeed the expected order of magnitude, and this kind of numerical check catches most sign errors.

e)

With $u(x)=v(x)=x$ , the product is $x^{2}$ , whose derivative is $2x$ . The product of the derivatives would be $1\times 1=1$ . The two agree only at $x=0.5$ , so the “product of the derivatives” rule is false in general. The correct rule, $u'v+uv'$ , gives here $1\times x+x\times 1=2x$ . The same counterexample settles the quotient question: $\dfrac{x}{x}=1$ has derivative 0, whereas the quotient of the derivatives would be 1.

Exercise 3: The chain rule: differentiating a nested model

In the social sciences, models are nested: the utilization rate of a service determines the waiting time, which determines satisfaction. The chain rule says how the rates multiply along this chain.

In Leibniz notation, it reads $\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}$ , and the units cancel as in a fraction.

a) Differentiate $u(x)=(3x^{2}-5)^{4}$ .
b) Differentiate $v(x)=\sqrt{7x+2}$ .
c) Differentiate $w(x)=\dfrac{1}{(2x+9)^{3}}$ .
d) The satisfaction of users is $S=100-8W$ points, where $W=\dfrac{\rho}{1-\rho}$ is the waiting time from the previous chapter. Compute $\dfrac{dS}{d\rho}$ at $\rho=0.8$ .
e) State the chain rule in words, relying on the units of question d).

Show the solution

Answers

a) $u'(x)=24x(3x^{2}-5)^{3}$ .
b) $v'(x)=\dfrac{7}{2\sqrt{7x+2}}$ .
c) $w'(x)=\dfrac{-6}{(2x+9)^{4}}$ .
d) $\dfrac{dS}{d\rho}=-8\times 25=-200$ points per unit of $\rho$ , that is 2 points per percentage point.
e) Multiply the derivatives of each level; the intermediate units cancel as in a fraction.

a)

The outer layer is the fourth power, the inner layer is $3x^{2}-5$ . Differentiate the outer layer keeping the inner one intact, then MULTIPLY by the derivative of the inner layer: $u'(x)=4(3x^{2}-5)^{3}\times 6x=24x(3x^{2}-5)^{3}$ . Forgetting the factor $6x$ is the most frequent mistake of the course; you can spot it by the degree: the function has degree 8, so its derivative must have degree 7, and without that factor you only reach degree 6.

b)

Write $v(x)=(7x+2)^{1/2}$ , so $v'(x)=\dfrac{1}{2}(7x+2)^{-1/2}\times 7=\dfrac{7}{2\sqrt{7x+2}}$ . The domain narrows when passing to the derivative: $v$ exists for $7x+2\ge 0$ , whereas $v'$ requires $7x+2>0$ strictly, the root moving to the denominator. At $x=-\dfrac{2}{7}$ , the function exists and the derivative does not.

c)

Write $w(x)=(2x+9)^{-3}$ , so $w'(x)=-3(2x+9)^{-4}\times 2=\dfrac{-6}{(2x+9)^{4}}$ . The derivative is negative wherever it exists, which is consistent: $w$ is a negative power of an increasing function, so it decreases.

d)

$\dfrac{dS}{dW}=-8$ points per service time. $\dfrac{dW}{d\rho}=\dfrac{(1)(1-\rho)-\rho(-1)}{(1-\rho)^{2}}=\dfrac{1}{(1-\rho)^{2}}$ , so at $\rho=0.8$ : $\dfrac{1}{0.04}=25$ service times per unit of $\rho$ . The product is $-8\times 25=-200$ points per unit of $\rho$ , that is, 2 satisfaction points lost per additional percentage point of utilization. At $\rho=0.5$ , the same calculation would give $-8\times 4=-32$ , six times less: it is still the asymptote speaking.

e)

The derivative of the composite is the PRODUCT of the derivatives taken at their own level. In units: $\dfrac{dS}{d\rho}$ is expressed in points per unit of $\rho$ , and the calculation multiplied points per service time by service times per unit of $\rho$ ; the service time cancels as in an ordinary fraction. This cancellation is not a convenient picture, it is the surest check on a chain of models: if the units do not cancel, a link is upside down.

Exercise 4: The tangent line, its slope and its units

The derivative at a point is the slope of the tangent line. That sentence is geometric; in the social sciences, it is the INTERPRETATION of this slope, with its units, that earns the marks.

The cumulative enrollments in an assistance program are $N(t)=\dfrac{1{,}500t}{t+8}$ people, $t$ months after opening. The figure below represents the situation.

a) Compute $N'(t)$ with the quotient rule and simplify.
b) Compute $N(4)$ and $N'(4)$ .
c) Write the equation of the tangent line to the curve at $t=4$ .
d) Compare $N'(4)$ and $N'(20)$ , then say what the decrease of $N'$ means for the program.
e) Give the units of $N$ , $N'$ and $N''$ .

Show the solution

Answers

a) $N'(t)=\dfrac{12{,}000}{(t+8)^{2}}$ .
b) $N(4)=500$ people and $N'(4)=\dfrac{250}{3}\approx 83.33$ people per month.
c) $y=500+\dfrac{250}{3}(t-4)$ .
d) $N'(20)\approx 15.31$ : the rate is divided by more than five, and the model levels off at 1,500 enrollees.
e) People, people per month, people per month squared.

a)

With $u=1{,}500t$ and $v=t+8$ : $N'(t)=\dfrac{1{,}500(t+8)-1{,}500t\times 1}{(t+8)^{2}}=\dfrac{1{,}500t+12{,}000-1{,}500t}{(t+8)^{2}}=\dfrac{12{,}000}{(t+8)^{2}}$ . The $t$ disappears from the numerator, a sign that the simplification has been carried all the way. The derivative is strictly positive: cumulative enrollments can only grow, which a cumulative total always does.

b)

$N(4)=\dfrac{6{,}000}{12}=500$ people enrolled after four months. $N'(4)=\dfrac{12{,}000}{144}=\dfrac{250}{3}\approx 83.33$ people per month.

c)

The equation of the tangent line is $y=N(4)+N'(4)(t-4)$ , so $y=500+\dfrac{250}{3}(t-4)$ , that is $y\approx 500+83.33(t-4)$ . The form $y=f(a)+f'(a)(x-a)$ is preferable to $y=mx+b$ : it reads directly, the point first, the slope second, and it avoids computing a $y$ -intercept that nobody needs.

d)

$N'(20)=\dfrac{12{,}000}{784}\approx 15.31$ people per month, against 83.33 in the fourth month: the enrollment rate is divided by more than five. The program keeps recruiting, but more and more slowly; this is the signature of a model that levels off, and indeed $\lim_{t\to+\infty}N(t)=1{,}500$ , the program's ceiling. Enrollments will not stop, they will become negligible, and the management decision is about the moment when the recruiting effort stops being worth it.

e)

$N$ is counted in people. $N'$ is counted in PEOPLE PER MONTH, the unit of the output divided by that of the input. $N''$ is counted in people per month squared, that is, in monthly change of the monthly rate. This last unit looks strange, yet it is the one used by every report that writes “the increase is slowing down”: slowing down is a statement about $N''$ , not about $N'$ or $N$ .

Exercise 5: The second derivative: speeding up or slowing down

Three sentences, three functions: “the total is rising” is about the function, “the rate is high” is about the derivative, “the increase is slowing down” is about the second derivative. A report that mixes them up says nothing sensible.

The cumulative number of requests filed with a help service is $C(t)=-0.5t^{3}+9t^{2}+40t+300$ , $t$ weeks after opening, for $0\le t\le 14$ .

a) Compute $C'(t)$ and $C''(t)$ .
b) In which week is the RATE of filing at its maximum? What is its value?
c) Give the sign of $C''$ before and after that week, and translate it into plain language.
d) In which week does the cumulative total stop increasing?
e) A newspaper headlines “requests are falling” in week 8. Is it right? What should it have written?

Show the solution

Answers

a) $C'(t)=-1.5t^{2}+18t+40$ and $C''(t)=-3t+18$ .
b) Week 6, with $C'(6)=94$ requests per week.
c) $C''>0$ before week 6, the rate speeds up; $C''<0$ after, it slows down without becoming negative.
d) $t=\dfrac{18+\sqrt{564}}{3}\approx 13.92$ weeks.
e) It is wrong: $C'(8)=88>0$ . It should have written that the increase is SLOWING.

a)

$C'(t)=-1.5t^{2}+18t+40$ , in requests per week. $C''(t)=-3t+18$ , in requests per week squared. The first derivative is the weekly RATE of filing, the second is the change in that rate.

b)

The rate $C'$ is a downward-opening parabola, so its maximum is at its vertex, where $C''$ vanishes: $-3t+18=0$ gives $t=6$ . The maximum rate is $C'(6)=-1.5\times 36+108+40=-54+108+40=94$ requests per week. Note that here we look for the maximum of $C'$ , so we set $C''$ to zero and not $C'$ : this is the costliest confusion of the chapter.

c)

For $t<6$ , $C''(t)>0$ : the rate INCREASES, the service receives more requests each week than the week before. For $t>6$ , $C''(t)<0$ : the rate DECREASES, the service still receives new requests every week, but fewer and fewer. Week 6 is the turning point, and it is the one when the most staff is needed at the counter.

d)

The total stops increasing when $C'(t)=0$ : $-1.5t^{2}+18t+40=0$ , that is $1.5t^{2}-18t-40=0$ . The discriminant is $324+240=564$ , so $t=\dfrac{18+\sqrt{564}}{3}\approx\dfrac{18+23.75}{3}\approx 13.92$ weeks, the negative root being rejected. From about the fourteenth week on, the model predicts a cumulative total that decreases, which is impossible for a CUMULATIVE count: it is the domain of validity of the model that ends there, as the statement suggested by restricting $t$ to 14.

e)

It is wrong. In week 8, $C'(8)=-96+144+40=88$ requests per week, a positive and high number: the service still receives 88 new requests every week. What is falling is the RATE, since $C''(8)=-6<0$ and 88 is below the maximum of 94. It should have written “the increase in requests is slowing down”, or “the service receives fewer new requests than the week before”. Confusing $C'<0$ with $C''<0$ is the most widespread interpretation error in the press, and it is exactly what differential calculus is there to settle.

Exercise 6: The three ways to lose a derivative

A function can be perfectly defined, perfectly continuous, and have no derivative at a point. Three situations produce this defect, and two of them are manufactured by administrative decisions.

We study the contribution of a user $T(x)=0.2x$ for $0\le x\le 40$ and $T(x)=8+0.5(x-40)$ for $x>40$ , where $x$ is the number of service hours consumed.

a) Show that $T$ is continuous at $x=40$ .
b) Compute the two one-sided derivatives at $x=40$ and conclude.
c) The data plan of chapter 1 jumps from 255 to 250 dollars at $x=50$ . Is it differentiable at 50? State the general rule that lets you answer without any calculation.
d) Let $g(x)=\sqrt[3]{x-5}$ . Compute $\dfrac{g(x)-g(5)}{x-5}$ for $x=5.001$ then $x=5.000001$ , and conclude.
e) Why does a marginal-rate schedule always produce a corner at the bracket boundaries?

Type your answers, the page tells you right or wrong 0/10

a)

Common value at

x=40

dollars

b)

Left derivative

Right derivative

The curve has

c)

Differentiable at 50

The rule is

d)

Quotient at

x=5.001

Quotient at

x=5.000001

The curve has

e)

A progressive schedule is

Show the solution

Answers

a) Both branches and the value give 8 dollars: $T$ is continuous at 40.
b) $0.2$ on the left and $0.5$ on the right: not differentiable, the curve has a corner.
c) No: differentiable implies continuous, and the plan is discontinuous at 50.
d) $100$ then $10{,}000$ : the quotient tends to $+\infty$ , vertical tangent at $x=5$ .
e) The marginal rate IS the slope: continuity can be matched, the slope cannot.

a)

To the left of 40, $T(x)=0.2x$ tends to $0.2\times 40=8$ . To the right, $T(x)=8+0.5(x-40)$ tends to $8+0=8$ . The value at the point is $T(40)=8$ . All three coincide, so $T$ is continuous at 40: the schedule does not make anyone pay abruptly more for one more hour.

b)

The LEFT derivative is the slope of the first branch, $0.2$ dollar per hour. The RIGHT derivative is that of the second, $0.5$ dollar per hour. They differ, so $T$ is NOT differentiable at $x=40$ : the curve has a CORNER there. The marginal rate is ambiguous at that point, which is why every official document writes “up to 40 hours” and “beyond 40 hours”, never “at 40 hours”.

c)

No, it is not differentiable at 50, and no calculation is needed: DIFFERENTIABLE IMPLIES CONTINUOUS. By the contrapositive, a function that is discontinuous at a point is not differentiable there. The plan is discontinuous at 50, so the question is settled. Mind the direction of the implication: the converse is false, and question b) has just given the counterexample, $T$ continuous and not differentiable at 40.

d)

$g(5)=0$ , so the quotient equals $\dfrac{(x-5)^{1/3}}{x-5}=(x-5)^{-2/3}$ . For $x=5.001$ : $(10^{-3})^{-2/3}=10^{2}=100$ . For $x=5.000001$ : $(10^{-6})^{-2/3}=10^{4}=10{,}000$ . The quotient tends to no number, it tends to $+\infty$ : the curve has a VERTICAL TANGENT at $x=5$ . The function is continuous there, it equals 0, and its derivative does not exist. This is the third way to lose a derivative, after the corner and the jump.

e)

Because a marginal rate is exactly a SLOPE, and a bracket schedule changes rate from one bracket to the next. Continuity is imposed, as we saw in chapter 2, by choosing the parameter that joins the two formulas; but nothing can join the SLOPES, since changing the rate is precisely the point of the bracket. A progressive schedule is therefore, by construction, continuous everywhere and not differentiable at the bracket boundaries. This is not a design flaw: it is the mathematical trace of a political decision, that of taxing differently above a threshold.

Exercise 7: The relative rate of change, in percent per year

In demography and economics, the derivative is almost never published alone: what is published is the RELATIVE rate of change, that is, the derivative divided by the value, expressed as a percentage. The two numbers tell different stories about the same phenomenon.

Two municipalities: $P_{A}(t)=50{,}000+1{,}200t$ and $P_{B}(t)=12{,}000+500t$ inhabitants, $t$ years after 2020.

a) Compute $P_{A}'(t)$ and $P_{B}'(t)$ . Which one grows faster in absolute terms?
b) The relative rate is $r(t)=\dfrac{P'(t)}{P(t)}\times 100$ . Compute $r_{A}(0)$ , $r_{A}(10)$ and $r_{A}(25)$ .
c) Explain why a constant absolute growth always gives a decreasing relative rate.
d) Compute $r_{B}(0)$ , then show that $r_{B}(t)>r_{A}(t)$ for all $t\ge 0$ .
e) A daily newspaper must pick only one of the two figures for its headline. Which one, and under what condition?

Type your answers, the page tells you right or wrong 0/12

a)

P_{A}'

res./yr

P_{B}'

res./yr

In absolute terms, the faster one is

b)

r_{A}(0)

%

r_{A}(10)

%

r_{A}(25)

%

c)

The relative rate decreases because

Its limit as t tends to infinity

d)

r_{B}(0)

%

After cross-multiplying, what remains is

e)

To size schools, you need

To compare two cities of different sizes

Show the solution

Answers

a) $P_{A}'=1{,}200$ and $P_{B}'=500$ inhabitants per year: A grows faster in absolute terms.
b) $2.4$ %, about $1.94$ % and $1.5$ %.
c) Constant numerator, growing denominator: the quotient decreases toward 0.
d) $r_{B}(0)\approx 4.17$ %; cross-multiplying leaves $25{,}000{,}000>14{,}400{,}000$ , true for every $t$ .
e) Absolute to size services, relative to compare cities of different sizes.

a)

$P_{A}'(t)=1{,}200$ and $P_{B}'(t)=500$ inhabitants per year, both constant. In ABSOLUTE terms, A grows more than twice as fast: it gains 1,200 inhabitants a year against 500.

b)

$r_{A}(0)=\dfrac{1{,}200}{50{,}000}\times 100=2.4$ %. $r_{A}(10)=\dfrac{1{,}200}{62{,}000}\times 100\approx 1.94$ %. $r_{A}(25)=\dfrac{1{,}200}{80{,}000}\times 100=1.5$ %. The relative rate falls from year to year, while the municipality still gains exactly 1,200 inhabitants.

c)

Because the numerator of the quotient is constant and its denominator grows. $r(t)=\dfrac{1{,}200}{50{,}000+1{,}200t}\times 100$ : the fraction has a value that decreases as soon as $t$ increases, and its limit at $+\infty$ is 0. Said without calculation: adding 1,200 people to 50,000 is 2.4%; adding them to 80,000 is 1.5%. The same absolute effort weighs less and less as the base grows, and that is the only reason a rich country always shows a lower percentage growth than a poor country that progresses less in absolute terms.

d)

$r_{B}(0)=\dfrac{500}{12{,}000}\times 100\approx 4.17$ %, almost double $r_{A}(0)$ . To compare at every moment, write $r_{B}(t)>r_{A}(t)$ as $\dfrac{500}{12{,}000+500t}>\dfrac{1{,}200}{50{,}000+1{,}200t}$ . Both denominators are positive, so we cross-multiply without changing the direction: $500\,(50{,}000+1{,}200t)>1{,}200\,(12{,}000+500t)$ , that is $25{,}000{,}000+600{,}000t>14{,}400{,}000+600{,}000t$ . The terms in $t$ CANCEL and what remains is $25{,}000{,}000>14{,}400{,}000$ , true for every $t$ . The small municipality therefore keeps the relative advantage forever, even though the absolute gap keeps widening in favour of the large one.

e)

Neither one lies, and each answers a different question. The ABSOLUTE rate, 1,200 against 500, is the right figure for sizing schools, buses and housing: these are real people to accommodate. The RELATIVE rate, 4.17% against 2.4%, is the right figure for comparing dynamics of different sizes, or for speaking of pressure on an existing urban fabric. The honest headline therefore names the question before the figure: “B is growing twice as fast in proportion” and “A is welcoming twice as many people” are both accurate. The only dishonest headline is the one that writes “grows faster” without saying in what sense.

Exercise 8: Five statements to correct

Each of these five sentences has been heard in class. Say whether it is true or false, and if it is false, write the correct statement.

A correction that merely says “false” earns no marks: it is the correct sentence that is asked for.

a) The derivative of a product is the product of the derivatives.
b) If $f$ is continuous at $a$ , then $f$ is differentiable at $a$ .
c) $f'(a)$ is the slope of the secant line passing through the point with $x$ -coordinate $a$ .
d) If $f''(t)<0$ , then $f$ is decreasing.
e) A derivative is a number, so it has no unit.

Type your answers, the page tells you right or wrong 0/9

a)

This statement is

The correct rule is

b)

This statement is

The correct direction of the implication is

c)

This statement is

It is the slope of

d)

This statement is

Whether a function increases or decreases is read from

e)

This statement is

Show the solution

Answers

a) False: $(uv)'=u'v+uv'$ , and $(x\cdot x)'=2x\ne 1$ .
b) False: it is differentiable that implies continuous.
c) False: it is the slope of the tangent line, the secant line giving an average rate.
d) False: $f''<0$ says that the increase is slowing down; whether the function rises or falls is read from $f'$ .
e) False: the unit of the output divided by that of the input.

a)

FALSE. The rule is $(uv)'=u'v+uv'$ . The counterexample fits on one line: $(x\times x)'=(x^{2})'=2x$ , whereas the product of the derivatives would be $1\times 1=1$ . Correct statement: the derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.

b)

FALSE, it is the converse that is true. Differentiable implies continuous, never the reverse. The contribution $T$ of exercise 6 is continuous at 40 and not differentiable there, and the cube root is continuous at 5 and not differentiable there. Correct statement: if $f$ is differentiable at $a$ , then it is continuous there.

c)

FALSE. $f'(a)$ is the slope of the TANGENT line at the point with $x$ -coordinate $a$ . The secant line joins TWO points and its slope is an AVERAGE rate of change; the tangent line touches only one and its slope is the instantaneous rate. Correct statement: $f'(a)$ is the limit of the slopes of the secant lines through $a$ , that is, the slope of the tangent line at $a$ .

d)

FALSE, and it is the costliest interpretation error of the course. $f''<0$ says that the GROWTH is slowing down, not that the function is decreasing. Exercise 5 shows it: at $t=8$ , $C''(8)=-6<0$ and yet $C'(8)=88>0$ , so the total is still rising by 88 units a week. Correct statement: $f''<0$ means that $f'$ is decreasing, so the curve is concave down; whether $f$ is increasing or decreasing is read from the sign of $f'$ , never from that of $f''$ .

e)

FALSE. A derivative always carries the unit of the OUTPUT divided by that of the INPUT: millions of dollars per year, people per month, satisfaction points per unit of utilization rate. It is even the fastest check on a chain of models, since the intermediate units must cancel. Correct statement: the unit of a derivative is the quotient of the unit of the function by that of the variable, and a second derivative carries that unit divided a second time.

Exercise 9: Problem: the graduation rate, from the table to the derivative

Real data never arrive as a function: they arrive as a table. Here is the graduation rate of a college network, recorded every two years.

$t=0$ : $71.2$ %; $t=2$ : $73.0$ %; $t=4$ : $74.4$ %; $t=6$ : $75.4$ %; $t=8$ : $76.0$ %, where $t$ counts the years since 2014.

a) Compute the four successive average rates of change, over each of the two-year intervals.
b) These four rates decrease by a constant amount. Check that $D(t)=71.2+t-0.05t^{2}$ reproduces the five data points exactly.
c) Compute $D'(t)$ , then $D'(0)$ and $D'(8)$ .
d) In which year does the graduation rate stop increasing, and what value does it reach then?
e) Estimate $D'(4)$ with the central difference $\dfrac{D(6)-D(2)}{4}$ and compare it with the exact value. Explain the coincidence.

Show the solution

Answers

a) $0.90$ ; $0.70$ ; $0.50$ ; $0.30$ point per year.
b) $D$ gives back $71.2$ ; $73.0$ ; $74.4$ ; $75.4$ ; $76.0$ : the fit is exact.
c) $D'(t)=1-0.1t$ , so $D'(0)=1$ and $D'(8)=0.2$ point per year.
d) $t=10$ , that is 2024, with $D(10)=76.2$ %.
e) $0.60$ point per year in both cases: the central difference is exact on a parabola.

a)

From 0 to 2: $\dfrac{73.0-71.2}{2}=0.90$ point per year. From 2 to 4: $\dfrac{74.4-73.0}{2}=0.70$ . From 4 to 6: $\dfrac{75.4-74.4}{2}=0.50$ . From 6 to 8: $\dfrac{76.0-75.4}{2}=0.30$ . The rate keeps rising, but more and more slowly, and the drop in the rate is regular, $-0.20$ point per year at each interval.

b)

$D(0)=71.2$ . $D(2)=71.2+2-0.05\times 4=71.2+2-0.2=73.0$ . $D(4)=71.2+4-0.8=74.4$ . $D(6)=71.2+6-1.8=75.4$ . $D(8)=71.2+8-3.2=76.0$ . The five values are reproduced EXACTLY, and it is no luck: average rates of change that decrease by a constant step are the signature of a quadratic model, since $D''$ is constant there.

c)

$D'(t)=1-0.1t$ , in percentage points per year. $D'(0)=1$ point per year in 2014, $D'(8)=1-0.8=0.2$ point per year in 2022. The rate of improvement has been divided by five in eight years.

d)

The rate stops increasing when $D'(t)=0$ , that is $1-0.1t=0$ , so $t=10$ : in 2024. The maximum is $D(10)=71.2+10-0.05\times 100=71.2+10-5=76.2$ %. The model therefore announces a ceiling at 76.2%, then a decline, and this decline is the point where we must ask whether the quadratic model remains valid: nothing in the data proves that the rate WILL fall. A second-degree polynomial fitted on a slowdown phase always predicts a reversal, whether the phenomenon wants it or not.

e)

The central difference gives $\dfrac{75.4-73.0}{4}=\dfrac{2.4}{4}=0.60$ point per year, and the exact value is $D'(4)=1-0.4=0.60$ . The estimate is EXACT, to the last decimal. The reason is that of exercise 1: for a quadratic function, the average rate over an interval is the instantaneous rate at the midpoint, and 4 is the midpoint of $[2, 6]$ . This is why statistical agencies estimate a derivative by central difference rather than by forward difference: on a curved trend, the central difference is markedly more accurate, and it is exact if the curvature is constant.

Exercise 10: Problem: a fundraising campaign

The amount raised by a campaign is a PRODUCT: the number of donors times the average gift. When the two factors move in opposite directions, only the derivative of the product settles the matter.

A campaign has $n(t)=200+30t$ donors and collects an average gift of $m(t)=150-4t$ dollars, $t$ weeks after launch.

a) Write the total amount $M(t)=n(t)\,m(t)$ in expanded form, and give the domain of validity of the model.
b) Compute $M'(t)$ with the product rule, then check from the expanded form.
c) Compute $M'(0)$ and $M'(20)$ , and interpret each of the two signs.
d) In which week is the amount raised at its maximum? What is it then?
e) The average gift is also $\dfrac{M(t)}{n(t)}$ . Differentiate this quotient and recover $m'(t)$ .

Show the solution

Answers

a) $M(t)=30{,}000+3{,}700t-120t^{2}$ , valid on $[0, 37.5)$ .
b) $M'(t)=3{,}700-240t$ dollars per week, by both routes.
c) $M'(0)=3{,}700$ and $M'(20)=-1{,}100$ : the campaign rises, then falls.
d) $t=\dfrac{3{,}700}{240}\approx 15.42$ weeks, for about $58{,}521$ dollars.
e) The derivative of the quotient is $-4$ , exactly $m'(t)$ .

a)

$M(t)=(200+30t)(150-4t)=30{,}000-800t+4{,}500t-120t^{2}=30{,}000+3{,}700t-120t^{2}$ dollars. The domain of validity requires a positive average gift: $150-4t>0$ gives $t<37.5$ . The model therefore holds on $[0, 37.5)$ , and beyond it would predict negative gifts.

b)

Product rule: $M'(t)=n'(t)m(t)+n(t)m'(t)=30(150-4t)+(200+30t)(-4)=4{,}500-120t-800-120t=3{,}700-240t$ dollars per week. Check from the expanded form: $\bigl(30{,}000+3{,}700t-120t^{2}\bigr)'=3{,}700-240t$ . The two routes agree. The term $-240t$ contains the TWO opposing effects, each weighing $-120t$ : that of donors who are more numerous but less generous, and that of an average gift applied to a growing base.

c)

$M'(0)=3{,}700$ dollars per week: at launch, the campaign gains more by recruiting than it loses in average generosity. $M'(20)=3{,}700-4{,}800=-1{,}100$ dollars per week: in the twentieth week, the erosion of the average gift wins out, and each additional week makes the amount raised FALL, even though the number of donors keeps growing. This is the kind of situation where a dashboard that tracks only the number of donors gives a signal exactly opposite to reality.

d)

Set $M'$ to zero: $3{,}700-240t=0$ gives $t=\dfrac{3{,}700}{240}\approx 15.42$ weeks. Since $M''(t)=-240<0$ , it is indeed a maximum. The value is $M(15.42)\approx 30{,}000+3{,}700\times 15.4167-120\times 237.67\approx 58{,}521$ dollars. In practice, the campaign should stop around the fifteenth week, or change method to raise the average gift.

e)

$\dfrac{M(t)}{n(t)}=\dfrac{30{,}000+3{,}700t-120t^{2}}{200+30t}$ . By the quotient rule, the numerator of the derivative is $(3{,}700-240t)(200+30t)-(30{,}000+3{,}700t-120t^{2})(30)$ . Expanding: $740{,}000+111{,}000t-48{,}000t-7{,}200t^{2}-900{,}000-111{,}000t+3{,}600t^{2}$ , that is $-160{,}000-48{,}000t-3{,}600t^{2}$ . We recognize $-4\,(200+30t)^{2}=-4\,(40{,}000+12{,}000t+900t^{2})=-160{,}000-48{,}000t-3{,}600t^{2}$ . The derivative is therefore $\dfrac{-4(200+30t)^{2}}{(200+30t)^{2}}=-4$ , that is, exactly $m'(t)$ . The check is long but worth doing once: it shows that the quotient rule and the product rule tell the same story, and that the average gift falls by 4 dollars a week, whatever happens to the number of donors.

Corrected exercises: the derivative and differentiation rules (201-103)

Course recap

Calculations: derivatives to compute

Exercise 0.1: Differentiating with the formulas: twelve direct calculations

Calculations: tangent lines to write

Exercise 0.2: Writing the equation of the tangent line: ten direct calculations

Part A: the basics (/50)

Exercise 1: From the average rate to the instantaneous rate

Exercise 2: The basic rules: power, product, quotient

Exercise 3: The chain rule: differentiating a nested model

Exercise 4: The tangent line, its slope and its units

Exercise 5: The second derivative: speeding up or slowing down

Part B: problems and reasoning (/50)

Exercise 6: The three ways to lose a derivative

Exercise 7: The relative rate of change, in percent per year

Exercise 8: Five statements to correct

Exercise 9: Problem: the graduation rate, from the table to the derivative

Exercise 10: Problem: a fundraising campaign

See also

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