Unit 2

Definition of the Derivative

The derivative as a limit and as a rate of change, the basic rules, tangent lines, and where a function fails to be differentiable.

Topics in this unit

Average Rate of Change · Instantaneous Rate of Change & the Derivative at a Point · The Limit Definition of the Derivative · Estimating Derivatives from a Graph or Table · Differentiability & Continuity · Power Rule · Sum, Difference & Constant Multiple Rules · Derivatives of sin, cos, e^x, ln x · Product Rule · Quotient Rule · Product & Quotient Rule from a Table · Derivatives of tan, cot, sec, csc · Tangent & Normal Lines · Horizontal & Parallel Tangents

Average Rate of Change

Question 1

no calculatormedium

The graph of

f

is shown. Over which interval is the average rate of change of

f

the greatest?

Show the answer

The answer is D.

f (x) = \frac{1}{2} x^{2}

curves upward, so secant slopes

\frac{f ( b ) - f ( a )}{b - a}

grow as the interval moves right. Among the choices,

[3, 4]

gives the steepest secant.

Why not A:Closer, but the graph keeps steepening, so $[3, 4]$ has an even greater average rate.
Why not B:For this upward-curving graph, secant slopes increase as $x$ increases. $[0, 1]$ is the flattest, not the steepest.
Why not C:The secant over $[1, 2]$ is steeper than over $[0, 1]$ , but later intervals are steeper still.

Instantaneous Rate of Change & the Derivative at a Point

Question 2

no calculatorhard

Revenue is

R (x) = p (x) n (x)

where

p

is price (dollars) and

n

is units sold, both functions of advertising spend

x

(in thousands of dollars). A table gives

p (3) = 10, p^{'} (3) = - 0.5, n (3) = 200, n^{'} (3) = 20

. Find

R^{'} (3)

and interpret it.

Show the answer

The answer is D.

Product rule:

R^{'} (x) = p^{'} (x) n (x) + p (x) n^{'} (x)

. At

3

R^{'} (3) = (- 0.5) (200) + (10) (20) = - 100 + 200 = 100

. The units are dollars of revenue per (thousand dollars of) advertising, so revenue is increasing at about

100

dollars per unit of

x

Why not A: $2000 = p (3) n (3)$ is $R (3)$ , the revenue value, not its rate of change. The product rule gives $R^{'} (3) = 100$ .
Why not B:That ignores the $p n^{'}$ term. $R^{'} = p^{'} n + p n^{'} = - 100 + 200 = + 100$ , so revenue is rising, not falling.
Why not C:Sign error: $p^{'} (3) n (3) = (- 0.5) (200) = - 100$ , not $+ 100$ . So $R^{'} (3) = - 100 + 200 = 100$ , not $300$ .

The Limit Definition of the Derivative

Question 3

no calculatormedium

For

f (x) = x^{3}

, the limit

x \to 2 lim \frac{x ^{3} - 8}{x - 2}

equals which quantity?

Show the answer

The answer is A.

This is

\frac{f ( x ) - f ( 2 )}{x - 2}

with

f (x) = x^{3}

(since

f (2) = 8

), the

x \to a

form of the derivative. So it is

f^{'} (2)

. With

f^{'} (x) = 3 x^{2}

f^{'} (2) = 12

Why not B:You found the right derivative function $3 x^{2}$ but plugged in $f (2) = 8$ instead of evaluating the slope: $f^{'} (2) = 3 \cdot 4 = 12$ .
Why not C:This limit is the instantaneous rate at $x = 2$ , not an average rate over an interval. It equals $f^{'} (2) = 12$ .
Why not D:That is the function value being subtracted, not the limit. The limit is the slope $f^{'} (2) = 12$ .

Estimating Derivatives from a Graph or Table

Question 4

no calculatormedium

A table gives the temperature

T

(in

^{\circ}

F) of water at times

t

(minutes):

T (2) = 50, T (4) = 55, T (6) = 62

. Which of the following gives the best estimate of

T^{'} (4)

and its correct interpretation?

Show the answer

The answer is D.

The best estimate is the average rate of change over

[2, 6]

, which straddles

t = 4

T^{'} (4) \approx \frac{T ( 6 ) - T ( 2 )}{6 - 2} = \frac{62 - 50}{4} = 3

. The units are

^{\circ}

F per minute, so the water is warming at about

3

degrees per minute.

Why not A:That uses a one-sided slope like $\frac{55 - 50}{2} = 2.5$ . The better estimate straddles $t = 4$ , using $t = 2$ and $t = 6$ : $\frac{62 - 50}{4} = 3$ .
Why not B: $55$ is the value $T (4)$ , not the rate of change. The derivative is a slope with units $^{\circ}$ F per minute.
Why not C:The value $3$ is correct but the units are inverted. $T^{'}$ is $^{\circ}$ F per minute, not minutes per degree.

Differentiability & Continuity

Question 5

no calculatorhard

Let

f (x) = {x^{2} sin (\frac{1}{x}) 0 x \neq = 0 x = 0

. Which statement about

f

x = 0

is true?

Show the answer

The answer is C.

Continuity: since

∣ sin (1/ x) ∣ \leq 1

, we have

∣ f (x) ∣ = ∣ x^{2} sin (1/ x) ∣ \leq x^{2} \to 0 = f (0)

. By the Squeeze Theorem,

lim_{x \to 0} f (x) = 0

, so

f

is continuous. Differentiability:

f^{'} (0) = lim_{h \to 0} \frac{h ^{2} s i n ( 1/ h )}{h} = lim_{h \to 0} h sin (1/ h) = 0

by the Squeeze Theorem (since

∣ h sin (1/ h) ∣ \leq ∣ h ∣ \to 0

). So

f

is differentiable at 0.

Why not A:The $sin (1/ x)$ part oscillates, but the $x^{2}$ in front squeezes it: $∣ f (x) ∣ \leq x^{2} \to 0 = f (0)$ . So $f$ is continuous at $0$ .
Why not B:Continuity is right, but check the derivative too. The difference quotient is $h sin (1/ h)$ , which the same squeeze sends to $0$ , so $f^{'} (0) = 0$ exists.
Why not D:The oscillation alone would kill the limit, but here it's multiplied by $x^{2}$ . Since $- x^{2} \leq f (x) \leq x^{2}$ , the limit is $0$ .

Question 6

no calculatormedium

Let

f (x) = {a x^{2} + 3 b x - 1 x \leq 2 x > 2

. For what values of

a

and

b

f

differentiable at

x = 2

Show the answer

The answer is A.

Differentiability at

x = 2

requires both continuity and matching derivatives. Continuity:

4 a + 3 = 2 b - 1 \Rightarrow 2 b = 4 a + 4 \Rightarrow b = 2 a + 2

. Derivative match:

f^{'} (x) = 2 a x

on the left gives

4 a

x = 2

;

f^{'} (x) = b

on the right. So

4 a = b

. Solving the system

b = 2 a + 2

and

b = 4 a

4 a = 2 a + 2 \Rightarrow a = 1

b = 4

Why not B: $a = - 1, b = - 4$ satisfies the slope condition $4 a = b$ but fails continuity: the left piece gives $a (2)^{2} + 3 = - 1$ while the right gives $b (2) - 1 = - 9$ . Both conditions must hold, so $a = 1, b = 4$ .
Why not C:With $a = 1$ the left value at $x = 2$ is $4 (1) + 3 = 7$ , so continuity needs $2 b - 1 = 7 \Rightarrow b = 4$ , not $2$ . And $b = 2$ would leave a corner (left slope $4 a = 4 \neq = b = 2$ ). The match is $a = 1, b = 4$ .
Why not D:A solution does exist. Differentiability gives two equations (one from continuity, one from matching slopes), and the system has a unique solution: $a = 1, b = 4$ .

Product Rule

Question 7

no calculatormedium

f (x) = x sin x

, what is

f^{'} (\frac{π}{2})

Show the answer

The answer is B.

Product rule:

f^{'} (x) = \frac{1}{2 x} sin x + x cos x

. At

x = \frac{π}{2}

sin \frac{π}{2} = 1

cos \frac{π}{2} = 0

, so

f^{'} (\frac{π}{2}) = \frac{1}{2 π /2} = \frac{2}{2 π}

Why not A:You forgot the $\frac{1}{2}$ from $\frac{d}{d x} x = \frac{1}{2 x}$ (not $\frac{1}{x}$ ); that halves it to $\frac{2}{2 π}$ .
Why not C:Sign: $sin \frac{π}{2} = + 1$ , so the term is positive, $\frac{2}{2 π}$ .
Why not D:That's the $x cos x$ term with $cos \frac{π}{2}$ taken as $1$ . But $cos \frac{π}{2} = 0$ , so that term vanishes.

Tangent & Normal Lines

Question 8

no calculatormedium

Find an equation of the normal line to

f (x) = x^{3} - x

x = 2

Show the answer

The answer is D.

f^{'} (x) = 3 x^{2} - 1

, so

f^{'} (2) = 11

and

f (2) = 6

. The normal is perpendicular, slope

- 1/ f^{'} (2) = - \frac{1}{11}

, through

(2, 6)

y - 6 = - \frac{1}{11} (x - 2) \Rightarrow y = \frac{68}{11} - \frac{x}{11}

Why not A:That is the $tangent$ line (slope $f^{'} (2) = 11$ ). The normal slope is the negative reciprocal, $- \frac{1}{11}$ .
Why not B:The normal slope is the negative $reciprocal$ $- 1/ f^{'} (2) = - \frac{1}{11}$ , not $- f^{'} (2) = - 11$ .
Why not C:Take the $negative$ reciprocal: $- 1/ f^{'} (2) = - \frac{1}{11}$ , not $+ \frac{1}{11}$ .

Horizontal & Parallel Tangents

Question 9

no calculatorhard

The function

f

is defined by

f (x) = \frac{x}{x + 2}

. At which point(s)

(x, y)

on the graph of

f

does the tangent line have slope

\frac{1}{2}

Show the answer

The answer is A.

f^{'} (x) = \frac{( x + 2 ) - x}{( x + 2 ) ^{2}} = \frac{2}{( x + 2 ) ^{2}}

. Set

\frac{2}{( x + 2 ) ^{2}} = \frac{1}{2} \Rightarrow (x + 2)^{2} = 4 \Rightarrow x = 0

x = - 4

. Then

f (0) = 0

and

f (- 4) = \frac{- 4}{- 2} = 2

, giving

(0, 0)

and

(- 4, 2)

Why not B:Solving $f^{'} (x) = \frac{2}{( x + 2 ) ^{2}} = \frac{1}{2}$ gives $(x + 2)^{2} = 4$ , so $x = 0$ AND $x = - 4$ . You kept only one root; $x = - 4$ gives the point $(- 4, 2)$ .
Why not C:There are two solutions of $(x + 2)^{2} = 4$ : $x = - 4$ and $x = 0$ . The root $x = 0$ gives the point $(0, 0)$ .
Why not D:The $x$ -values are right, but $f (- 4) = \frac{- 4}{- 4 + 2} = \frac{- 4}{- 2} = 2$ , not $- 2$ . The point is $(- 4, 2)$ .

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