Exercise 1: The power rule after rewriting: only the power of x crosses the fraction bar

The power rule $\frac{d}{dx}x^n = nx^{n-1}$ holds for every real exponent $n$ , but only on the form $cx^n$ . So every root and every $x$ in a denominator is REWRITTEN first, and this is where most MATH 203 marks are lost, not in the rule itself. Two laws of exponents do all the work: $\frac{1}{x^n} = x^{-n}$ and $\sqrt[q]{x^p} = x^{p/q}$ . A numerical coefficient in the denominator does NOT change sign or position: $\frac{3}{2x^4} = \frac{3}{2}x^{-4}$ , because only $x^4$ is raised to a power.

The sum and constant multiple rules then let you differentiate term by term, and $\frac{d}{dx}e^x = e^x$ . Give every final answer with positive exponents.

a) Differentiate $f(x) = \frac{3}{2x^4} - \frac{x^3}{6} + 5x$ , then compute $f'(2)$ .
b) Differentiate $g(x) = x^2\sqrt x - \frac{4}{\sqrt[3]{x^2}}$ for $x > 0$ , and compute $g'(1)$ .
c) Let $h(x) = \frac{2x - 3}{x\sqrt x}$ for $x > 0$ . Write $h'(x)$ as a single fraction by factoring out the SMALLEST power of $x$ , compute $h'(1)$ , and find where the tangent is horizontal.
d) A student writes: $\frac{1}{2x^3} = 2x^{-3}$ , so its derivative is $-6x^{-4}$ ; and $\frac{d}{dx}x^{-2/3} = -\frac{2}{3}x^{1/3}$ . Correct both lines.
e) Differentiate $p(x) = \frac{e^x}{3} + \frac{e^3}{x}$ and compute $p'(3)$ , exact and to four decimals.

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Answers

a) $f'(x) = -\frac{6}{x^5} - \frac{x^2}{2} + 5$ , $f'(2) = \frac{45}{16}$
b) $g'(x) = \frac{5}{2}x\sqrt x + \frac{8}{3x\sqrt[3]{x^2}}$ , $g'(1) = \frac{31}{6}$
c) $h'(x) = \frac{9 - 2x}{2x^2\sqrt x}$ , $h'(1) = \frac{7}{2}$ , horizontal tangent at $x = \frac{9}{2}$
d) $\frac{1}{2x^3} = \frac{1}{2}x^{-3}$ , derivative $-\frac{3}{2x^4}$ ; $\frac{d}{dx}x^{-2/3} = -\frac{2}{3}x^{-5/3}$
e) $p'(x) = \frac{e^x}{3} - \frac{e^3}{x^2}$ , $p'(3) = \frac{2e^3}{9} \approx 4.4635$

a) Rewrite: $\frac{3}{2x^4} = \frac{3}{2}x^{-4}$ and $\frac{x^3}{6} = \frac{1}{6}x^3$ . Power rule term by term: $\frac{3}{2} \cdot (-4)x^{-5} = -6x^{-5}$ , $\frac{1}{6} \cdot 3x^2 = \frac{1}{2}x^2$ , $(5x)' = 5$ . So $f'(x) = -\frac{6}{x^5} - \frac{x^2}{2} + 5$ . At $x = 2$ : $-\frac{6}{32} - 2 + 5 = 3 - \frac{3}{16} = \frac{45}{16} = 2.8125$ . The classic slip is $\frac{3}{2x^4} = \frac{3}{2}x^{4}$ or $(2x)^{-4}$ : the first forgets that a power in the denominator becomes NEGATIVE, the second raises the $2$ to the power $-4$ as well, and both change every later number.

b) Rewrite with the laws of exponents: $x^2\sqrt x = x^2 \cdot x^{1/2} = x^{5/2}$ (exponents ADD when powers multiply), and $\frac{4}{\sqrt[3]{x^2}} = 4x^{-2/3}$ . Then $g'(x) = \frac{5}{2}x^{3/2} - 4 \cdot \left(-\frac{2}{3}\right)x^{-5/3} = \frac{5}{2}x^{3/2} + \frac{8}{3}x^{-5/3}$ . Back to radicals: $x^{3/2} = x\sqrt x$ and $x^{5/3} = x\sqrt[3]{x^2}$ , so $g'(x) = \frac{5}{2}x\sqrt x + \frac{8}{3x\sqrt[3]{x^2}}$ . At $x = 1$ every power is $1$ : $g'(1) = \frac{5}{2} + \frac{8}{3} = \frac{31}{6}$ . The exponent arithmetic is the whole question: $-\frac{2}{3} - 1 = -\frac{5}{3}$ , since subtracting $1$ moves a negative exponent AWAY from zero.

c) The denominator is one power of $x$ : $x\sqrt x = x^{3/2}$ , so divide each term: $h(x) = 2x \cdot x^{-3/2} - 3x^{-3/2} = 2x^{-1/2} - 3x^{-3/2}$ . Then $h'(x) = -x^{-3/2} + \frac{9}{2}x^{-5/2}$ . The smallest power is $x^{-5/2}$ (the most negative exponent), and factoring it out leaves $x^{-3/2} = x^{-5/2} \cdot x$ : $h'(x) = x^{-5/2}\left(-x + \frac{9}{2}\right) = \frac{9 - 2x}{2x^{5/2}} = \frac{9 - 2x}{2x^2\sqrt x}$ . So $h'(1) = \frac{7}{2}$ . The tangent is horizontal where the numerator is zero, the denominator being positive for $x > 0$ : $x = \frac{9}{2}$ . Factoring out $x^{-3/2}$ instead, the larger power, leaves $x^{-1}$ inside the bracket and a fraction inside a fraction: always factor the SMALLEST power, as with $x^2 + x^3 = x^2(1 + x)$ .

d) First line: in $\frac{1}{2x^3}$ only $x^3$ is in the denominator as a power; the $2$ is a plain number. So $\frac{1}{2x^3} = \frac{1}{2}x^{-3}$ , never $2x^{-3}$ , and the derivative is $\frac{1}{2} \cdot (-3)x^{-4} = -\frac{3}{2x^4}$ . The student's answer is four times too large: at $x = 1$ it gives $-6$ against the true $-\frac{3}{2}$ . Second line: the exponent was increased by $1$ instead of decreased, $-\frac{2}{3} + 1 = \frac{1}{3}$ . The rule subtracts $1$ : $-\frac{2}{3} - 1 = -\frac{5}{3}$ , so $\frac{d}{dx}x^{-2/3} = -\frac{2}{3}x^{-5/3} = -\frac{2}{3x\sqrt[3]{x^2}}$ . A quick sanity check: $x^{-2/3}$ decreases for $x > 0$ , and $-\frac{2}{3}x^{1/3}$ happens to be negative too, so the sign does not catch this one; the scientific calculator does, with $\frac{f(1.001) - f(0.999)}{0.002}$ .

e) $\frac{e^x}{3} = \frac{1}{3}e^x$ has derivative $\frac{1}{3}e^x$ . In $\frac{e^3}{x}$ the numerator $e^3 \approx 20.09$ is a CONSTANT coefficient, so rewrite $\frac{e^3}{x} = e^3x^{-1}$ , whose derivative is $-e^3x^{-2}$ . Hence $p'(x) = \frac{e^x}{3} - \frac{e^3}{x^2}$ , and $p'(3) = \frac{e^3}{3} - \frac{e^3}{9} = \frac{2e^3}{9} \approx 4.4635$ . Treating $e^3$ as a function (writing $3e^2$ , or using the quotient rule with $(e^3)' = e^3$ ) are the two ways to lose this mark; a letter that does not contain $x$ is a number.

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Exercise 2: Product and quotient rules, then the algebra that follows

The product rule $(fg)' = f'g + fg'$ and the quotient rule $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ give an answer in one line. The rest of the question is algebra: the minus sign of the quotient rule must reach EVERY term of $fg'$ , a common factor such as $e^x$ is taken out, a factor shared by the numerator and the denominator is cancelled, and a fraction inside a fraction is cleared BEFORE differentiating.

Name the rule and the two functions before writing any derivative. The chain rule is not available yet: a squared binomial is expanded.

a) Differentiate $F(x) = \left(\frac{1}{x} + 2\right)(x^2 - 5)$ with the product rule, then check by expanding $F$ first. Compute $F'(1)$ .
b) Differentiate $G(x) = \frac{x^2 - 4}{2x + 1}$ and compute $G'(0)$ .
c) Differentiate $H(x) = \frac{e^x}{x^2 + 1}$ , factor the numerator completely, compute $H'(0)$ , and find where the tangent is horizontal.
d) Let $K(x) = \frac{1 - \frac{2}{x}}{x + \frac{3}{x}}$ for $x \ne 0$ . Clear the complex fraction, then differentiate and compute $K'(1)$ .
e) Let $q(x) = \frac{x}{(x + 1)^2}$ . Without the chain rule, show that $q'(x) = \frac{1 - x}{(x + 1)^3}$ , compute $q'(2)$ , and find where the tangent is horizontal.

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Answers

a) $F'(x) = 4x + 1 + \frac{5}{x^2}$ , $F'(1) = 10$
b) $G'(x) = \frac{2x^2 + 2x + 8}{(2x + 1)^2}$ , $G'(0) = 8$
c) $H'(x) = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$ , $H'(0) = 1$ , horizontal tangent only at $x = 1$
d) $K(x) = \frac{x - 2}{x^2 + 3}$ ( $x \ne 0$ ), $K'(x) = \frac{-x^2 + 4x + 3}{(x^2 + 3)^2}$ , $K'(1) = \frac{3}{8}$
e) $q'(x) = \frac{1 - x}{(x + 1)^3}$ , $q'(2) = -\frac{1}{27}$ , horizontal tangent at $x = 1$ only ( $x = -1$ is excluded)

a) Product rule with $f(x) = x^{-1} + 2$ , $f'(x) = -x^{-2}$ , and $g(x) = x^2 - 5$ , $g'(x) = 2x$ : $F'(x) = -x^{-2}(x^2 - 5) + (x^{-1} + 2)(2x) = -1 + 5x^{-2} + 2 + 4x = 4x + 1 + \frac{5}{x^2}$ . The products $x^{-2} \cdot x^2 = 1$ and $x^{-1} \cdot 2x = 2$ are laws of exponents, and they are where the algebra goes wrong. Check by expanding first: $F(x) = x - 5x^{-1} + 2x^2 - 10$ , so $F'(x) = 1 + 5x^{-2} + 4x$ , the same. $F'(1) = 4 + 1 + 5 = 10$ .

b) Quotient rule with $f(x) = x^2 - 4$ , $f'(x) = 2x$ , and $g(x) = 2x + 1$ , $g'(x) = 2$ : $G'(x) = \frac{2x(2x + 1) - (x^2 - 4) \cdot 2}{(2x + 1)^2}$ . The minus sign multiplies the WHOLE product: $-(x^2 - 4) \cdot 2 = -2x^2 + 8$ . So the numerator is $4x^2 + 2x - 2x^2 + 8 = 2x^2 + 2x + 8$ and $G'(x) = \frac{2x^2 + 2x + 8}{(2x + 1)^2}$ , with $G'(0) = \frac{8}{1} = 8$ . Writing $-2x^2 - 8$ gives $G'(0) = -8$ : one missing pair of parentheses flips the answer. The denominator stays $(2x + 1)^2$ , unexpanded.

c) Quotient rule with top $e^x$ (derivative $e^x$ ) and bottom $x^2 + 1$ (derivative $2x$ ): $H'(x) = \frac{e^x(x^2 + 1) - e^x \cdot 2x}{(x^2 + 1)^2}$ . Factor $e^x$ out of the numerator: $e^x(x^2 - 2x + 1) = e^x(x - 1)^2$ , a perfect square. So $H'(x) = \frac{e^x(x - 1)^2}{(x^2 + 1)^2}$ and $H'(0) = \frac{1 \cdot 1}{1} = 1$ . Since $e^x > 0$ and $(x^2 + 1)^2 > 0$ , $H'(x) = 0$ only when $(x - 1)^2 = 0$ : one horizontal tangent, at $x = 1$ , where $H(1) = \frac{e}{2}$ . Recognising $x^2 - 2x + 1$ as $(x - 1)^2$ is the step that makes the question readable; the factored form also shows that $H'$ is never negative.

d) Multiply the top and the bottom by $x$ , the common denominator of the small fractions: $K(x) = \frac{x - 2}{x^2 + 3}$ , valid for $x \ne 0$ only, since the original expression is undefined at $0$ . Quotient rule: $K'(x) = \frac{1 \cdot (x^2 + 3) - (x - 2) \cdot 2x}{(x^2 + 3)^2} = \frac{x^2 + 3 - 2x^2 + 4x}{(x^2 + 3)^2} = \frac{-x^2 + 4x + 3}{(x^2 + 3)^2}$ , for $x \ne 0$ . At $x = 1$ : $\frac{-1 + 4 + 3}{16} = \frac{3}{8}$ . Differentiating the complex fraction directly is legal but produces fractions of fractions in every term; clearing it first is the standard gesture, and the condition $x \ne 0$ must travel with the simplified form.

e) No chain rule yet, so expand the denominator: $(x + 1)^2 = x^2 + 2x + 1$ , derivative $2x + 2$ . Quotient rule: $q'(x) = \frac{(x^2 + 2x + 1) - x(2x + 2)}{(x + 1)^4} = \frac{1 - x^2}{(x + 1)^4}$ , using $(x^2 + 2x + 1)^2 = (x + 1)^4$ . Now factor the difference of squares, $1 - x^2 = (1 - x)(1 + x)$ , and cancel one factor $x + 1$ : $q'(x) = \frac{1 - x}{(x + 1)^3}$ . Then $q'(2) = \frac{-1}{27}$ . The tangent is horizontal where $1 - x = 0$ , at $x = 1$ (the point $\left(1, \frac{1}{4}\right)$ ). Reading the zeros of $1 - x^2$ BEFORE cancelling gives $x = \pm 1$ , but $x = -1$ is not in the domain of $q$ : it must be rejected, and saying so is part of the answer.

Exercise 3: Tangent and normal lines on a curve with an x in the denominator

The tangent line to $y = f(x)$ at $x = a$ is $y = f(a) + f'(a)(x - a)$ , with $f'(a)$ a NUMBER: differentiate first, substitute $a$ afterwards. The normal line at the same point is perpendicular to the tangent, with slope $-\frac{1}{f'(a)}$ when $f'(a) \ne 0$ . A line given in the form $Ax + By = C$ must be solved for $y$ before its slope can be read.

The figure shows the graph of $f(x) = x + \frac{4}{x}$ , defined for $x \ne 0$ .

a) Find $f'(x)$ as a single fraction, then the tangent line at the point of abscissa $1$ , in the form $y = mx + b$ .
b) Find the normal line at the same point.
c) Find the points where the tangent is horizontal.
d) Find the points where the tangent is parallel to the line $3x - 4y = 7$ .
e) The normal line of b) meets the curve at a second point. Find it.

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Answers

a) $f'(x) = \frac{x^2 - 4}{x^2}$ ; tangent $y = -3x + 8$
b) Normal $y = \frac{1}{3}x + \frac{14}{3}$
c) $(2, 4)$ and $(-2, -4)$
d) $(4, 5)$ and $(-4, -5)$
e) $\left(6, \frac{20}{3}\right)$

a) Rewrite $\frac{4}{x} = 4x^{-1}$ : $f'(x) = 1 - 4x^{-2} = 1 - \frac{4}{x^2} = \frac{x^2 - 4}{x^2}$ , over the common denominator $x^2$ . At $a = 1$ : $f(1) = 5$ and $f'(1) = -3$ , so the tangent is $y = 5 - 3(x - 1) = -3x + 8$ . Check: at $x = 1$ the line gives $5$ . The derivative of $\frac{4}{x}$ written as $\frac{0}{1} = 0$ , or as $4$ , are the two slips that make this part worthless; the rewriting $4x^{-1}$ avoids both.

b) The tangent slope is $-3 \ne 0$ , so the normal slope is $-\frac{1}{-3} = \frac{1}{3}$ , and the normal passes through $(1, 5)$ : $y = 5 + \frac{1}{3}(x - 1) = \frac{1}{3}x + \frac{14}{3}$ . The constant needs a common denominator, $5 - \frac{1}{3} = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}$ : this fraction arithmetic is where the question is usually lost, not in the slope. Check: $(-3) \cdot \frac{1}{3} = -1$ .

c) $f'(x) = 0$ when the numerator is zero and the denominator is not: $x^2 - 4 = 0$ , $x = \pm 2$ , both in the domain. $f(2) = 2 + 2 = 4$ and $f(-2) = -2 - 2 = -4$ . The points are $(2, 4)$ and $(-2, -4)$ , the bottom of the right branch and the top of the left one on the figure.

d) Solve the line for $y$ : $-4y = 7 - 3x$ , so $y = \frac{3}{4}x - \frac{7}{4}$ , slope $\frac{3}{4}$ . Reading the slope as $3$ (the coefficient of $x$ before solving) is the classic error. Parallel means equal slopes: $1 - \frac{4}{x^2} = \frac{3}{4}$ , so $\frac{4}{x^2} = \frac{1}{4}$ , $x^2 = 16$ , $x = \pm 4$ . $f(4) = 4 + 1 = 5$ and $f(-4) = -5$ : the points are $(4, 5)$ and $(-4, -5)$ . Note also that $f'(x) = 1 - \frac{4}{x^2} < 1$ for every $x$ : no tangent is parallel to a line of slope $1$ or more.

e) Intersect the normal with the curve: $\frac{1}{3}x + \frac{14}{3} = x + \frac{4}{x}$ . Multiply by $3x$ (allowed, $x \ne 0$ ): $x^2 + 14x = 3x^2 + 12$ , so $2x^2 - 14x + 12 = 0$ , $x^2 - 7x + 6 = 0$ , $(x - 1)(x - 6) = 0$ . The root $x = 1$ is the point we started from, which is a good check; the new point has $x = 6$ and $y = 6 + \frac{4}{6} = \frac{20}{3}$ . On the normal, $\frac{6}{3} + \frac{14}{3} = \frac{20}{3}$ : consistent. The solution figure shows the tangent, the normal, and the two points.

Exercise 4: Prescribed slopes: unknown constants, parallel tangents, a normal in a given direction

Many exam questions give the slope and ask for the point, or give the tangent and ask for the curve. The method never changes: translate each sentence into an equation. The curve passes through a point: $f(a) = b$ . The tangent there has slope $m$ : $f'(a) = m$ . Two lines are parallel: equal slopes. A normal is parallel to a line of slope $m$ : the TANGENT has slope $-\frac{1}{m}$ .

A scientific calculator may be used in d); give decimals to four places.

a) Find $b$ and $c$ so that the parabola $y = x^2 + bx + c$ is tangent to the line $y = x$ at the point where $x = 2$ .
b) Find $a$ and $b$ so that $f(x) = ax + \frac{b}{x}$ has a horizontal tangent at the point $(2, 8)$ .
c) Find the values of $x$ at which the curves $y = x^3 - x$ and $y = x^2$ have parallel tangents.
d) Find the point of $y = e^x$ where the tangent is parallel to $y = 3x - 1$ , and the $y$ -intercept of that tangent.
e) Find the points of $y = 2x^3 - 3x^2$ where the NORMAL line is parallel to the line $x + 12y = 5$ .

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Answers

a) $b = -3$ , $c = 4$ : $y = x^2 - 3x + 4$
b) $a = 2$ , $b = 8$ : $f(x) = 2x + \frac{8}{x}$
c) $x = -\frac{1}{3}$ and $x = 1$
d) $(\ln 3, 3) \approx (1.0986, 3)$ ; intercept $3 - 3\ln 3 \approx -0.2958$
e) $(2, 4)$ and $(-1, -5)$

a) Two conditions, two unknowns. The parabola passes through the point of the line at $x = 2$ , which is $(2, 2)$ : $4 + 2b + c = 2$ . Its slope there equals the slope $1$ of the line: $y' = 2x + b$ , so $4 + b = 1$ and $b = -3$ . Then $c = 2 - 4 + 6 = 4$ , and $y = x^2 - 3x + 4$ . Check: at $x = 2$ , $y = 4 - 6 + 4 = 2$ and $y' = 4 - 3 = 1$ . Using only the slope condition leaves $c$ undetermined: tangency is TWO equations, same point and same slope.

b) Rewrite $\frac{b}{x} = bx^{-1}$ , so $f'(x) = a - \frac{b}{x^2}$ . The point: $f(2) = 2a + \frac{b}{2} = 8$ . The horizontal tangent: $f'(2) = a - \frac{b}{4} = 0$ , so $b = 4a$ . Substituting: $2a + 2a = 8$ , $a = 2$ , $b = 8$ . The function is $f(x) = 2x + \frac{8}{x}$ , and indeed $f(2) = 4 + 4 = 8$ and $f'(2) = 2 - \frac{8}{4} = 0$ . Here $b$ is a constant in a numerator: $\left(\frac{b}{x}\right)' = -\frac{b}{x^2}$ , not $\frac{1}{x}$ or $0$ .

c) The slopes are $3x^2 - 1$ and $2x$ , both at the SAME $x$ (the question compares the two curves above one value of $x$ ). Parallel tangents: $3x^2 - 1 = 2x$ , $3x^2 - 2x - 1 = 0$ , $(3x + 1)(x - 1) = 0$ . So $x = -\frac{1}{3}$ or $x = 1$ . At $x = 1$ both slopes are $2$ ; at $x = -\frac{1}{3}$ both are $-\frac{2}{3}$ . Forgetting to move $2x$ to one side and factoring $3x^2 - 1$ alone is the usual way to lose this part.

d) $(e^x)' = e^x$ , so solve $e^x = 3$ : $x = \ln 3 \approx 1.0986$ , and the point is $(\ln 3, 3)$ , since $e^{\ln 3} = 3$ exactly. The tangent is $y = 3 + 3(x - \ln 3) = 3x + 3 - 3\ln 3$ , with $y$ -intercept $3 - 3\ln 3 \approx -0.2958$ . Keep $\ln 3$ exact until the last line: rounding $x$ to $1.1$ first gives an intercept of $-0.3$ , wrong at the second decimal.

e) Solve the line for $y$ : $y = -\frac{1}{12}x + \frac{5}{12}$ , slope $-\frac{1}{12}$ . A normal with slope $-\frac{1}{12}$ is perpendicular to a tangent of slope $12$ , since $12 \cdot \left(-\frac{1}{12}\right) = -1$ . So solve $y' = 6x^2 - 6x = 12$ : $x^2 - x - 2 = 0$ , $(x - 2)(x + 1) = 0$ , $x = 2$ or $x = -1$ . The points are $(2, 16 - 12) = (2, 4)$ and $(-1, -2 - 3) = (-1, -5)$ . Setting $y' = -\frac{1}{12}$ answers a different question, the one where the TANGENT is parallel to the line.

Exercise 5: Product and quotient rules from a table and from a graph, forward and backward

The product and quotient rules need four numbers at the point: $f(a)$ , $f'(a)$ , $g(a)$ and $g'(a)$ . The functions below are known only through a table; the entry marked ? is unknown.

$\begin{array}{c|cccc} x & f(x) & f'(x) & g(x) & g'(x) \\ \hline 0 & 2 & -1 & 5 & 3 \\ 1 & -3 & 4 & 2 & -5 \\ 3 & 4 & \frac{1}{2} & -1 & ? \end{array}$

The figure shows the graph of another function $k$ and its tangent line at $x = 2$ , which passes through the two marked points $(0, 4)$ and $(4, 2)$ .

a) Let $P = fg$ and $Q = \frac{f}{g}$ . Find $P'(1)$ and $Q'(1)$ .
b) Find $R'(0)$ , $S'(1)$ and $T'(0)$ for $R(x) = e^xf(x)$ , $S(x) = \frac{f(x)}{x^2}$ and $T(x) = \frac{g(x)}{1 + f(x)}$ .
c) The product $H = fg$ satisfies $H'(3) = 5$ . Find the missing value $g'(3)$ .
d) Find the tangent line to the graph of $P = fg$ at $x = 0$ .
e) Read $k(2)$ and $k'(2)$ on the figure, then find $U'(2)$ and $V'(2)$ for $U(x) = x^2k(x)$ and $V(x) = \frac{k(x)}{x}$ .

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Answers

a) $P'(1) = 23$ , $Q'(1) = -\frac{7}{4}$
b) $R'(0) = 1$ , $S'(1) = 10$ , $T'(0) = \frac{14}{9}$
c) $g'(3) = \frac{11}{8}$
d) $y = x + 10$
e) $k(2) = 3$ , $k'(2) = -\frac{1}{2}$ ; $U'(2) = 10$ , $V'(2) = -1$

a) Write the rule with letters, then one number per letter. $P'(1) = f'(1)g(1) + f(1)g'(1) = 4 \cdot 2 + (-3)(-5) = 8 + 15 = 23$ . $Q'(1) = \frac{f'(1)g(1) - f(1)g'(1)}{g(1)^2} = \frac{8 - (-3)(-5)}{4} = \frac{8 - 15}{4} = -\frac{7}{4}$ . The product $(-3)(-5) = +15$ is ADDED in the product rule and SUBTRACTED in the quotient rule: the two rules differ by exactly that sign, and a sign slip on a product of two negatives is the most common error on a table question.

b) $R'(x) = e^xf(x) + e^xf'(x)$ , so $R'(0) = 1 \cdot (2 - 1) = 1$ . For $S$ , rewrite $S(x) = x^{-2}f(x)$ and use the product rule: $S'(x) = -2x^{-3}f(x) + x^{-2}f'(x)$ , so $S'(1) = -2(-3) + 4 = 10$ ; the quotient rule gives the same, $\frac{4 \cdot 1 - (-3) \cdot 2}{1} = 10$ . For $T$ , the denominator $1 + f(x)$ has derivative $f'(x)$ : $T'(0) = \frac{g'(0)(1 + f(0)) - g(0)f'(0)}{(1 + f(0))^2} = \frac{3 \cdot 3 - 5 \cdot (-1)}{9} = \frac{14}{9}$ . The $1$ in $1 + f$ differentiates to $0$ , but it stays in the VALUE $1 + f(0) = 3$ .

c) Work the product rule backward: $H'(3) = f'(3)g(3) + f(3)g'(3) = \frac{1}{2} \cdot (-1) + 4g'(3)$ . Setting this equal to $5$ : $4g'(3) = 5 + \frac{1}{2} = \frac{11}{2}$ , so $g'(3) = \frac{11}{8}$ . The rule is an equation, and an equation can be solved for any one of its letters.

d) The point: $P(0) = f(0)g(0) = 2 \cdot 5 = 10$ , computed, since $P(0)$ is not in the table. The slope: $P'(0) = f'(0)g(0) + f(0)g'(0) = -5 + 6 = 1$ . Tangent: $y = 10 + 1(x - 0) = x + 10$ .

e) The tangent at $x = 2$ goes through $(0, 4)$ and $(4, 2)$ , so its slope is $k'(2) = \frac{2 - 4}{4 - 0} = -\frac{1}{2}$ , and at $x = 2$ it gives $k(2) = 4 - 1 = 3$ , the point of tangency. Product rule: $U'(2) = 2 \cdot 2 \cdot k(2) + 2^2k'(2) = 12 - 2 = 10$ . Quotient rule: $V'(2) = \frac{k'(2) \cdot 2 - k(2) \cdot 1}{2^2} = \frac{-1 - 3}{4} = -1$ . The slope is read between the two MARKED points of the line, never between two values guessed on the curve.

Exercise 6: Higher derivatives: rewrite once, simplify at every step

The second derivative $f'' = (f')'$ is the derivative of the SIMPLIFIED first derivative, and so on: $f'''$ , $f^{(4)}$ , $f^{(n)}$ . Each step starts from the result of the previous one, so an algebra slip at step one is paid again at every later step, and a result left unsimplified makes the next step twice as long. Powers of $x$ are rewritten once, at the start, and stay rewritten.

Exact answers are expected; fractions may be given as fractions.

a) Let $f(x) = \sqrt x$ . Find $f'$ , $f''$ and $f'''$ as powers of $x$ , then compute $f''(4)$ and $f'''(1)$ .
b) Find the constants $A$ , $B$ and $C$ such that $y = Ax^2 + Bx + C$ satisfies $y'' + y' - 2y = x^2$ for every $x$ .
c) Let $g(x) = \frac{x - 1}{x + 2}$ . Show that $g'(x) = \frac{3}{(x + 2)^2}$ , then, without the chain rule, find $g''(x)$ in simplified form and compute $g'(1)$ and $g''(1)$ .
d) A polynomial $p$ of degree $3$ satisfies $p(0) = 1$ , $p'(0) = -2$ , $p''(0) = 6$ and $p'''(0) = 12$ . Find $p(x)$ and $p(2)$ .
e) Let $y = e^x\sqrt x$ for $x > 0$ . Find $y''$ and compute $y''(1)$ .

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Answers

a) $f' = \frac{1}{2}x^{-1/2}$ , $f'' = -\frac{1}{4}x^{-3/2}$ , $f''' = \frac{3}{8}x^{-5/2}$ ; $f''(4) = -\frac{1}{32}$ , $f'''(1) = \frac{3}{8}$
b) $A = -\frac{1}{2}$ , $B = -\frac{1}{2}$ , $C = -\frac{3}{4}$
c) $g''(x) = -\frac{6}{(x + 2)^3}$ ; $g'(1) = \frac{1}{3}$ , $g''(1) = -\frac{2}{9}$
d) $p(x) = 2x^3 + 3x^2 - 2x + 1$ , $p(2) = 25$
e) $y'' = e^x\left(x^{1/2} + x^{-1/2} - \frac{1}{4}x^{-3/2}\right) = \frac{e^x(4x^2 + 4x - 1)}{4x\sqrt x}$ , $y''(1) = \frac{7e}{4}$

a) $f(x) = x^{1/2}$ , so $f'(x) = \frac{1}{2}x^{-1/2}$ , $f''(x) = \frac{1}{2} \cdot \left(-\frac{1}{2}\right)x^{-3/2} = -\frac{1}{4}x^{-3/2}$ and $f'''(x) = -\frac{1}{4} \cdot \left(-\frac{3}{2}\right)x^{-5/2} = \frac{3}{8}x^{-5/2}$ . At $x = 4$ : $4^{-3/2} = \frac{1}{4^{3/2}} = \frac{1}{8}$ , so $f''(4) = -\frac{1}{32}$ . At $x = 1$ : $f'''(1) = \frac{3}{8}$ . Writing $f'(x) = \frac{1}{2\sqrt x}$ and then differentiating that fraction with the quotient rule works, but it is exactly the detour that the rewriting avoids; the signs alternate because each new exponent is negative.

b) $y' = 2Ax + B$ and $y'' = 2A$ . Substitute: $2A + (2Ax + B) - 2(Ax^2 + Bx + C) = -2Ax^2 + (2A - 2B)x + (2A + B - 2C)$ . This must equal $x^2 = 1x^2 + 0x + 0$ for EVERY $x$ , so the coefficients match: $-2A = 1$ , $2A - 2B = 0$ , $2A + B - 2C = 0$ . Hence $A = -\frac{1}{2}$ , $B = A = -\frac{1}{2}$ , and $2C = 2A + B = -\frac{3}{2}$ , $C = -\frac{3}{4}$ . Check at $x = 0$ : $y'' + y' - 2y = -1 - \frac{1}{2} + \frac{3}{2} = 0 = 0^2$ . Forgetting to distribute the $-2$ over the three terms of $y$ is where this question is lost.

c) Quotient rule: $g'(x) = \frac{1 \cdot (x + 2) - (x - 1) \cdot 1}{(x + 2)^2} = \frac{3}{(x + 2)^2}$ ; the $x$ terms cancel once the minus sign reaches $-1$ as well, giving $2 + 1 = 3$ . For $g''$ , expand the denominator, $(x + 2)^2 = x^2 + 4x + 4$ with derivative $2x + 4$ , and use the reciprocal rule $\left(\frac{c}{u}\right)' = -\frac{cu'}{u^2}$ : $g''(x) = -\frac{3(2x + 4)}{(x + 2)^4} = -\frac{6(x + 2)}{(x + 2)^4} = -\frac{6}{(x + 2)^3}$ . Factoring $2x + 4 = 2(x + 2)$ and cancelling is the gesture that turns a fourth power into a cube. Then $g'(1) = \frac{3}{9} = \frac{1}{3}$ and $g''(1) = -\frac{6}{27} = -\frac{2}{9}$ .

d) Write $p(x) = a + bx + cx^2 + dx^3$ . Then $p(0) = a$ , $p'(0) = b$ , $p''(x) = 2c + 6dx$ so $p''(0) = 2c$ , and $p'''(x) = 6d$ . The conditions give $a = 1$ , $b = -2$ , $2c = 6$ so $c = 3$ , and $6d = 12$ so $d = 2$ . Hence $p(x) = 2x^3 + 3x^2 - 2x + 1$ and $p(2) = 16 + 12 - 4 + 1 = 25$ . The trap is taking $c = 6$ and $d = 12$ directly: the $k$ -th derivative at $0$ of $x^k$ is $k!$ , so each coefficient is the derivative DIVIDED by $k!$ .

e) Product rule with $e^x$ and $x^{1/2}$ : $y' = e^xx^{1/2} + e^x \cdot \frac{1}{2}x^{-1/2} = e^x\left(x^{1/2} + \frac{1}{2}x^{-1/2}\right)$ , with $e^x$ factored out. Product rule again, on $e^x$ times the bracket: $y'' = e^x\left(x^{1/2} + \frac{1}{2}x^{-1/2}\right) + e^x\left(\frac{1}{2}x^{-1/2} - \frac{1}{4}x^{-3/2}\right) = e^x\left(x^{1/2} + x^{-1/2} - \frac{1}{4}x^{-3/2}\right)$ . Factoring the smallest power $\frac{1}{4}x^{-3/2}$ gives $y'' = \frac{e^x(4x^2 + 4x - 1)}{4x\sqrt x}$ . At $x = 1$ : $y''(1) = e\left(1 + 1 - \frac{1}{4}\right) = \frac{7e}{4} \approx 4.7570$ . Keeping $e^x$ factored after the first step is what keeps the second step to one line.

Exercise 7: Where the rules come from: e to the x, the reciprocal, three factors

Thomas 3.3 does not ask you to accept the rules on faith, and a final exam sometimes asks for one of the short proofs. For the exponential, $\frac{e^{x+h} - e^x}{h} = e^x \cdot \frac{e^h - 1}{h}$ , so everything rests on one limit, $\lim_{h \to 0}\frac{e^h - 1}{h}$ , which the number $e$ is chosen to make equal to $1$ . The other rules are consequences of the product and quotient rules.

A scientific calculator is used in a); give four decimals.

a) Compute $\frac{e^h - 1}{h}$ for $h = 0.1$ , $0.01$ , $0.001$ and $h = -0.01$ . What do the values suggest, and what does it give for $\frac{d}{dx}e^x$ ?
b) Deduce from the quotient rule the reciprocal rule $\left(\frac{1}{g}\right)' = -\frac{g'}{g^2}$ , then prove $\frac{d}{dx}x^{-n} = -nx^{-n-1}$ for a positive integer $n$ . Use the reciprocal rule to find the derivative of $\frac{1}{x^2 + x}$ at $x = 1$ .
c) Prove that $(uvw)' = u'vw + uv'w + uvw'$ for three differentiable functions. Apply it to $y = (x - 1)(x + 2)(x + 3)$ at $x = 1$ , and check by expanding.
d) Taking $u = v = w = f$ in c), show that $(f^3)' = 3f^2f'$ . Use it to compute the derivative of $(x^2 + 1)^3$ at $x = 1$ and of $(e^x + 1)^3$ at $x = 0$ .
e) A student writes $\frac{d}{dx}(x^2 + 1)^3 = 3(x^2 + 1)^2$ . Compare with d) at $x = 1$ and at $x = 2$ , and name the missing factor.

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Answers

a) $1.0517$ , $1.0050$ , $1.0005$ , $0.9950$ : the limit is $1$ , so $(e^x)' = e^x$
b) $\left(\frac{1}{g}\right)' = \frac{0 \cdot g - 1 \cdot g'}{g^2}$ ; $(x^{-n})' = -\frac{nx^{n-1}}{x^{2n}} = -nx^{-n-1}$ ; value $-\frac{3}{4}$
c) $y'(1) = 12$ , and $y = x^3 + 4x^2 + x - 6$ gives $y'(1) = 3 + 8 + 1 = 12$
d) $(f^3)' = f'ff + ff'f + fff' = 3f^2f'$ ; values $24$ and $12$
e) Student: $12$ at $x = 1$ and $75$ at $x = 2$ ; true: $24$ and $300$ ; missing factor $2x$

a) With the calculator: $h = 0.1$ gives $\frac{e^{0.1} - 1}{0.1} \approx 1.0517$ , $h = 0.01$ gives $1.0050$ , $h = 0.001$ gives $1.0005$ , and $h = -0.01$ gives $0.9950$ . The values approach $1$ from both sides, which suggests $\lim_{h \to 0}\frac{e^h - 1}{h} = 1$ (a table suggests a limit, it does not prove it; Thomas takes this limit as the defining property of $e$ ). Then $\frac{d}{dx}e^x = \lim_{h \to 0}e^x \cdot \frac{e^h - 1}{h} = e^x \cdot 1 = e^x$ , because $e^x$ does not depend on $h$ and factors out of the limit. The step $e^{x+h} = e^xe^h$ is a law of exponents, and it is the whole proof.

b) Quotient rule with top $1$ (derivative $0$ ) and bottom $g$ : $\left(\frac{1}{g}\right)' = \frac{0 \cdot g - 1 \cdot g'}{g^2} = -\frac{g'}{g^2}$ . With $g(x) = x^n$ , $g'(x) = nx^{n-1}$ : $\frac{d}{dx}x^{-n} = -\frac{nx^{n-1}}{x^{2n}} = -nx^{n-1-2n} = -nx^{-n-1}$ , which is the power rule for the exponent $-n$ . For $\frac{1}{x^2 + x}$ : $g(x) = x^2 + x$ , $g'(x) = 2x + 1$ , so the derivative is $-\frac{2x + 1}{(x^2 + x)^2}$ , equal to $-\frac{3}{4}$ at $x = 1$ . The exponent arithmetic $n - 1 - 2n = -n - 1$ is the step to write out.

c) Group two factors: $uvw = (uv)w$ . Product rule: $(uvw)' = (uv)'w + (uv)w' = (u'v + uv')w + uvw' = u'vw + uv'w + uvw'$ . Each term differentiates ONE factor and keeps the other two. For $y$ : $u = x - 1$ , $v = x + 2$ , $w = x + 3$ , all with derivative $1$ . At $x = 1$ , $u(1) = 0$ , so the two terms containing $u$ vanish: $y'(1) = 1 \cdot 3 \cdot 4 + 0 + 0 = 12$ . Expanding: $(x - 1)(x^2 + 5x + 6) = x^3 + 4x^2 + x - 6$ , so $y' = 3x^2 + 8x + 1$ and $y'(1) = 12$ . A factor that is zero at the point kills every term where it is kept, which is a real shortcut on an exam.

d) With $u = v = w = f$ : $(f^3)' = f'ff + ff'f + fff' = 3f^2f'$ . For $f(x) = x^2 + 1$ , $f'(x) = 2x$ : the derivative is $3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2$ , equal to $6 \cdot 4 = 24$ at $x = 1$ . Check by expanding $(x^2 + 1)^3 = x^6 + 3x^4 + 3x^2 + 1$ : derivative $6x^5 + 12x^3 + 6x$ , which is $24$ at $x = 1$ . For $f(x) = e^x + 1$ , $f'(x) = e^x$ : the derivative is $3(e^x + 1)^2e^x$ , equal to $3 \cdot 4 \cdot 1 = 12$ at $x = 0$ . This is a first case of the chain rule, obtained with the product rule alone.

e) The student's formula gives $3 \cdot 2^2 = 12$ at $x = 1$ and $3 \cdot 5^2 = 75$ at $x = 2$ , while the true derivative $6x(x^2 + 1)^2$ gives $24$ and $6 \cdot 2 \cdot 25 = 300$ . The ratio is $2$ , then $4$ : it is $2x$ , the derivative $f'$ of the inside, and it is exactly the factor the student dropped. The power rule applies to $x^n$ , a power of $x$ itself; for a power of a FUNCTION, the formula of d) is $3f^2f'$ , and forgetting $f'$ is the error that the chain rule chapter will name.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample with numbers (a calculator check is acceptable as evidence, not as a proof), and write the correct statement.

a) $\frac{d}{dx}x^{-3} = -3x^{-2}$ .
b) The derivative of $f(x) = \frac{x^2 + 3x}{x}$ is $\frac{2x + 3}{1} = 2x + 3$ .
c) If $f'(a) = g'(a)$ , then $f$ and $g$ have the same tangent line at $x = a$ .
d) The only function equal to its own derivative is $e^x$ .
e) The tangent line to $y = f(x)$ at $x = a$ is $y = f'(a)x + f(a)$ .

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Answers

a) False: $\frac{d}{dx}x^{-3} = -3x^{-4}$ , which is $-\frac{3}{16}$ at $x = 2$ , not $-\frac{3}{4}$
b) False: $f(x) = x + 3$ for $x \ne 0$ , so $f'(x) = 1$
c) False: the tangents are only parallel; same line only if also $f(a) = g(a)$
d) False: $ce^x$ works for every constant $c$ , for instance $5e^x$ and $0$
e) False: $y = f(a) + f'(a)(x - a)$ ; for $x^3$ at $a = 2$ , $y = 12x - 16$

a) FALSE. The student added $1$ to the exponent: $-3 + 1 = -2$ . The power rule SUBTRACTS $1$ , and for a negative exponent that moves it further from zero: $-3 - 1 = -4$ . Correct: $\frac{d}{dx}x^{-3} = -3x^{-4} = -\frac{3}{x^4}$ . At $x = 2$ the true slope is $-\frac{3}{16} = -0.1875$ and the student's formula gives $-\frac{3}{4}$ ; the calculator quotient $\frac{2.001^{-3} - 1.999^{-3}}{0.002} \approx -0.1875$ settles it.

b) FALSE. The student took the quotient of the derivatives, a rule that does not exist. For $x \ne 0$ , $\frac{x^2 + 3x}{x} = x + 3$ (divide each term by $x$ ), so $f'(x) = 1$ on the domain $x \ne 0$ . At $x = 5$ the student's formula gives $13$ against the true $1$ . Correct: simplify first, $f(x) = x + 3$ for $x \ne 0$ , and $f'(x) = 1$ ; the quotient rule would give $\frac{(2x + 3)x - (x^2 + 3x)}{x^2} = \frac{x^2}{x^2} = 1$ as well.

c) FALSE. Equal derivatives give equal SLOPES, so the tangent lines are parallel, but they pass through $(a, f(a))$ and $(a, g(a))$ , which may differ. With $f(x) = x^2$ , $g(x) = x^2 + 1$ and $a = 0$ : $f'(0) = g'(0) = 0$ , and the tangents are $y = 0$ and $y = 1$ , two different lines. Correct: $f$ and $g$ have the same tangent line at $a$ when $f(a) = g(a)$ AND $f'(a) = g'(a)$ , the two conditions of a tangency.

d) FALSE. For any constant $c$ , $(ce^x)' = ce^x$ by the constant multiple rule: $5e^x$ equals its own derivative, and so does the zero function. At $x = 0$ , $(5e^x)' = 5 \ne e^0 = 1$ , so $5e^x$ is a different function from $e^x$ with the same property. Correct: the functions equal to their own derivative are the functions $ce^x$ (that there are no others is proved later in the course, with a tool that is not in this chapter).

e) FALSE. The formula forgets to shift by $a$ : the line $y = f'(a)x + f(a)$ has the right slope but passes through $(0, f(a))$ , not through $(a, f(a))$ . For $f(x) = x^3$ at $a = 2$ : $f(2) = 8$ , $f'(2) = 12$ ; the student gets $y = 12x + 8$ , which gives $32$ at $x = 2$ instead of $8$ . Correct: $y = f(a) + f'(a)(x - a) = 8 + 12(x - 2) = 12x - 16$ , and the check at $x = 2$ gives $8$ . Plugging $x = a$ into the tangent line is the five-second check that catches this error every time.

Exercise 9: A camera dolly that stops without turning back

On a film set, a camera dolly moves along a straight rail. Its position, measured from its starting mark, is $s(t) = t^4 - 8t^3 + 18t^2$ centimetres at time $t$ seconds, for $0 \le t \le 5$ . Its velocity is $v = s'$ and its acceleration is $a = v' = s''$ . The sign of $v$ gives the direction of motion; the dolly speeds up when $v$ and $a$ have the same sign and slows down when they have opposite signs.

The figure shows the position. The acceleration is not constant, so the formulas of constant-acceleration kinematics do not apply: everything comes from $s'$ and $s''$ , FACTORED.

a) Find $v(t)$ and $a(t)$ , and factor both completely. Compute $v(2)$ and $a(2)$ .
b) When is the dolly at rest? Does it change direction at those instants?
c) Find $s(1)$ , $s(3)$ and $s(5)$ , the displacement and the total distance over $[0, 5]$ , and the average velocity.
d) Draw a sign chart of $v$ and $a$ , and find when the dolly speeds up and when it slows down.
e) At what times is the velocity equal to $16$ cm/s? What is the acceleration at each of them?

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Answers

a) $v(t) = 4t(t - 3)^2$ , $a(t) = 12(t - 1)(t - 3)$ ; $v(2) = 8$ cm/s, $a(2) = -12$ cm/s $^2$
b) At $t = 0$ and $t = 3$ s; no change of direction, $v \ge 0$ throughout
c) $s(1) = 11$ , $s(3) = 27$ , $s(5) = 75$ cm; displacement $=$ distance $= 75$ cm; average velocity $15$ cm/s
d) Speeds up on $(0, 1)$ and $(3, 5]$ , slows down on $(1, 3)$
e) $t = 1$ s ( $a = 0$ ) and $t = 4$ s ( $a = 36$ cm/s $^2$ )

a) Power rule: $v(t) = 4t^3 - 24t^2 + 36t$ cm/s and $a(t) = 12t^2 - 48t + 36$ cm/s $^2$ . Factor out the common factor FIRST: $v(t) = 4t(t^2 - 6t + 9) = 4t(t - 3)^2$ , a perfect square in the bracket; $a(t) = 12(t^2 - 4t + 3) = 12(t - 1)(t - 3)$ . Then $v(2) = 4 \cdot 2 \cdot 1 = 8$ cm/s and $a(2) = 12 \cdot 1 \cdot (-1) = -12$ cm/s $^2$ . Trying to factor $4t^3 - 24t^2 + 36t$ as a trinomial without taking out $4t$ is where most attempts stall.

b) $v(t) = 0$ when $t = 0$ or $(t - 3)^2 = 0$ : the dolly is at rest at $t = 0$ and $t = 3$ s. But $4t \ge 0$ and $(t - 3)^2 \ge 0$ on $[0, 5]$ , so $v(t) \ge 0$ throughout: the velocity touches zero at $t = 3$ without changing sign. The dolly stops for an instant and then continues FORWARD; it never changes direction. On the figure the position graph flattens at $(3, 27)$ and keeps rising. A zero of $v$ is a candidate for a change of direction, never a proof of one: the sign on both sides decides, and a SQUARED factor does not change sign.

c) $s(1) = 1 - 8 + 18 = 11$ , $s(3) = 81 - 216 + 162 = 27$ and $s(5) = 625 - 1000 + 450 = 75$ cm. The displacement is $s(5) - s(0) = 75$ cm. Since the dolly never moves backward, the distance equals the displacement: $75$ cm. Splitting at $t = 3$ gives $27 + 48 = 75$ , the same, which is the check. The average velocity is $\frac{75}{5} = 15$ cm/s. Subtracting a backward leg that does not exist is the error that the sign study of b) prevents.

d) On $(0, 1)$ : $v > 0$ and $a > 0$ (both factors of $a$ negative), same sign, the dolly speeds up. On $(1, 3)$ : $v > 0$ and $a < 0$ , it slows down, all the way to its stop at $t = 3$ . On $(3, 5]$ : $v > 0$ and $a > 0$ , it speeds up again. The solution figure gathers the signs. At $t = 3$ both $v$ and $a$ are zero; at $t = 1$ only $a$ is zero, which is the instant where the dolly stops gaining speed and starts losing it.

e) Solve $4t(t - 3)^2 = 16$ , that is $t(t - 3)^2 = 4$ , or $t^3 - 6t^2 + 9t - 4 = 0$ . Test the divisors of $4$ : $t = 1$ gives $1 - 6 + 9 - 4 = 0$ , so $t - 1$ is a factor, and dividing gives $(t - 1)(t^2 - 5t + 4) = (t - 1)^2(t - 4)$ . The velocity is $16$ cm/s at $t = 1$ s and $t = 4$ s. Then $a(1) = 0$ and $a(4) = 12 \cdot 3 \cdot 1 = 36$ cm/s $^2$ . The double root $t = 1$ is not a coincidence: it is where the velocity stops increasing, so the horizontal line $v = 16$ only touches the velocity graph there.

Exercise 10: Marginal cost and marginal revenue: the next unit, estimated by a derivative

A small workshop makes and sells $x$ units of a product per week. Its weekly cost is $C(x) = 1200 + 15x - 0.1x^2 + 0.0005x^3$ dollars, and market research shows that it can sell $x$ units at the price $p(x) = 60 - 2\sqrt x$ dollars each, so its weekly revenue is $R(x) = x\,p(x)$ . The MARGINAL cost $C'(x)$ and the marginal revenue $R'(x)$ , in dollars per unit, estimate the extra cost and the extra revenue of producing one more unit.

The figure shows the revenue $R$ . A scientific calculator is allowed; amounts of money are given to the cent.

a) Find $C'(x)$ and the marginal cost at $x = 100$ . Compare with the exact extra cost $C(101) - C(100)$ .
b) Find $R'(x)$ with the product rule, in simplified form, and compare the marginal revenue $R'(100)$ with the price $p(100)$ .
c) Find the marginal profit $R'(x) - C'(x)$ at $x = 100$ and at $x = 400$ , and interpret the signs.
d) The workshop estimates that with $a$ thousand dollars of advertising it sells $S(a) = \frac{200a}{a + 5}$ extra units per week. Find $S'(a)$ , then $S'(5)$ and $S'(15)$ , with their units.
e) A student claims that the extra revenue from the $101$ st unit is the price, $p(100) = 40$ dollars. Compute $R(101) - R(100)$ and explain.

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Answers

a) $C'(x) = 15 - 0.2x + 0.0015x^2$ ; $C'(100) = 10$ dollars per unit; $C(101) - C(100) = 10.05$ dollars
b) $R'(x) = 60 - 3\sqrt x$ ; $R'(100) = 30$ dollars per unit, less than $p(100) = 40$ dollars
c) $20$ dollars per unit at $x = 100$ ; $-175$ dollars per unit at $x = 400$
d) $S'(a) = \frac{1000}{(a + 5)^2}$ ; $S'(5) = 10$ and $S'(15) = 2.5$ units per thousand dollars
e) $R(101) - R(100) \approx 29.93$ dollars, close to $R'(100) = 30$ , not $40$

a) Power rule term by term: $C'(x) = 15 - 0.2x + 0.0015x^2$ dollars per unit, the fixed cost $1200$ disappearing as every constant does. $C'(100) = 15 - 20 + 15 = 10$ dollars per unit. The exact extra cost is $C(101) - C(100) = 15 - 0.1(10201 - 10000) + 0.0005(1030301 - 1000000) = 15 - 20.10 + 15.1505 \approx 10.05$ dollars. The derivative is the slope of the tangent, the difference is the slope of a secant over one unit; they differ by five cents, which is why economists use $C'(100)$ as the cost of the $101$ st unit.

b) Product rule with $x$ (derivative $1$ ) and $p(x) = 60 - 2x^{1/2}$ (derivative $-x^{-1/2}$ ): $R'(x) = 1 \cdot (60 - 2\sqrt x) + x \cdot \left(-\frac{1}{\sqrt x}\right) = 60 - 2\sqrt x - \sqrt x = 60 - 3\sqrt x$ , using $\frac{x}{\sqrt x} = \sqrt x$ . So $R'(100) = 60 - 30 = 30$ dollars per unit, while $p(100) = 60 - 20 = 40$ dollars. The marginal revenue is smaller than the price because selling one more unit forces the price down on ALL the units sold, and the term $x\,p'(x) = -10$ counts that loss. Expanding first, $R(x) = 60x - 2x^{3/2}$ , gives the same $R'(x) = 60 - 3x^{1/2}$ .

c) At $x = 100$ : $R'(100) - C'(100) = 30 - 10 = 20$ dollars per unit, positive, so the $101$ st unit adds about $20$ dollars of profit. At $x = 400$ : $R'(400) = 60 - 3 \cdot 20 = 0$ and $C'(400) = 15 - 80 + 240 = 175$ , so the marginal profit is $-175$ dollars per unit: at that level, one more unit brings no extra revenue and costs about $175$ dollars, so it LOSES money. The sign of a marginal profit says whether the next unit helps or hurts; it says nothing yet about the best production level, which is a question for a later chapter.

d) Quotient rule with top $200a$ (derivative $200$ ) and bottom $a + 5$ (derivative $1$ ): $S'(a) = \frac{200(a + 5) - 200a}{(a + 5)^2} = \frac{1000}{(a + 5)^2}$ , the $200a$ terms cancelling. Units: extra units sold per week, per thousand dollars of advertising. $S'(5) = \frac{1000}{100} = 10$ and $S'(15) = \frac{1000}{400} = 2.5$ units per thousand dollars: the fifth thousand dollars of advertising brings about $10$ extra units, the fifteenth only about $2.5$ . Forgetting the parentheses around $200(a + 5)$ gives $200a + 5 - 200a = 5$ in the numerator, a derivative two hundred times too small.

e) $R(100) = 100 \cdot 40 = 4000$ dollars and $R(101) = 101(60 - 2\sqrt{101}) \approx 101 \cdot 39.9002 \approx 4029.93$ dollars, so $R(101) - R(100) \approx 29.93$ dollars. The student's $40$ dollars counts the price of the new unit but not the price cut on the $100$ units already sold. The derivative $R'(100) = 30$ predicted the right amount; the price $p(100)$ did not. The solution figure shows both lines through $(100, 4000)$ : the tangent of slope $30$ follows the revenue curve, the line of slope $40$ leaves it at once.

Corrected exercises: Differentiation rules (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: The power rule after rewriting: only the power of x crosses the fraction bar

Exercise 2: Product and quotient rules, then the algebra that follows

Exercise 3: Tangent and normal lines on a curve with an x in the denominator

Exercise 4: Prescribed slopes: unknown constants, parallel tangents, a normal in a given direction

Exercise 5: Product and quotient rules from a table and from a graph, forward and backward

Part B: problems and reasoning (/50)

Exercise 6: Higher derivatives: rewrite once, simplify at every step

Exercise 7: Where the rules come from: e to the x, the reciprocal, three factors

Exercise 8: Five statements to correct

Exercise 9: A camera dolly that stops without turning back

Exercise 10: Marginal cost and marginal revenue: the next unit, estimated by a derivative

See also

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