Unit 3

Differentiation Rules

The chain rule, implicit differentiation, inverse and inverse trig derivatives, higher-order derivatives, and the first uses of all of it: related rates, linear approximation and the Mean Value Theorem.

Topics in this unit

Chain Rule · Derivatives of sin, cos, e^x, ln x · Derivatives of tan, cot, sec, csc · Implicit Differentiation · Derivatives of Inverse Functions · Derivatives of Inverse Trig Functions · Selecting Differentiation Procedures · Higher-Order Derivatives · Straight-Line Motion · Related Rates · Linear Approximation & Differentials · L'Hôpital's Rule · Mean Value Theorem

Chain Rule

Question 1

no calculatormedium

Let

f (x) = (2 x + 1)^{3} e^{x}

. What is

f^{'} (0)

Show the answer

The answer is C.

By the product rule,

f^{'} = \frac{d}{d x} [(2 x + 1)^{3}] e^{x} + (2 x + 1)^{3} e^{x}

. The chain rule gives

\frac{d}{d x} (2 x + 1)^{3} = 3 (2 x + 1)^{2} \cdot 2 = 6 (2 x + 1)^{2}

. So

f^{'} (x) = [6 (2 x + 1)^{2} + (2 x + 1)^{3}] e^{x}

. At

x = 0

[6 (1) + 1] (1) = 7

Why not A:This is $3 (2 x + 1)^{2}$ evaluated naively without the inner factor $2$ or the product rule; the correct value is $7$ .
Why not B:This drops the chain-rule term entirely (just $(2 x + 1)^{3} e^{x}$ at $0$ ); you must also differentiate the cube, giving $6 + 1 = 7$ .
Why not D:This is only the chain-rule piece $6 (2 x + 1)^{2} e^{x}$ at $0$ ; you still must add the second product term $(2 x + 1)^{3} e^{x} = 1$ , giving $7$ .

Question 2

no calculatormedium

The graphs of the functions

f

and

g

are shown below. Let

B (x) = g (f (x))

. What is

B^{'} (2)

Show the answer

The answer is A.

B^{'} (2) = g^{'} (f (2)) f^{'} (2)

. From the graph of

f

f (2) = 3

and

f^{'} (2) = 2

. At

x = 3

g

has slope

- 1

. So

B^{'} (2) = (- 1) (2) = - 2

Why not B:This is $g^{'} (3)$ alone, without the factor $f^{'} (2) = 2$ .
Why not C:This uses the slope of $g$ just left of $x = 2$ , which is $1$ . The outer derivative belongs at $f (2) = 3$ , not at $2$ .
Why not D:This multiplies $f (2) = 3$ by $f^{'} (2) = 2$ . The first factor should be $g^{'} (f (2))$ , the slope of $g$ at $3$ .

Implicit Differentiation

Question 3

no calculatormedium

The point

(2, 3)

lies on the curve

y^{2} = x^{3} + 1

. Which of the following is an equation of the line tangent to the curve at

(2, 3)

Show the answer

The answer is B.

Implicitly,

2 y y^{'} = 3 x^{2}

, so

y^{'} = \frac{3 x ^{2}}{2 y}

. At

(2, 3)

y^{'} = \frac{12}{6} = 2

. The tangent line is

y - 3 = 2 (x - 2)

, i.e.

y = 2 x - 1

Why not A:This uses slope $\frac{1}{2}$ (the reciprocal); the correct slope is $\frac{3 x ^{2}}{2 y} = 2$ .
Why not C:The slope $2$ is right but the line must pass through $(2, 3)$ : $y - 3 = 2 (x - 2)$ gives $y = 2 x - 1$ , not $+ 3$ .
Why not D:The slope is $+ 2$ , not $- 2$ ; $y^{'} = \frac{3 x ^{2}}{2 y} > 0$ at $(2, 3)$ .

Question 4

no calculatormedium

x^{2} + y^{2} = 25

, then

\frac{d y}{d x} =

Show the answer

The answer is A.

Differentiate term by term:

2 x + 2 y y^{'} = 0

, so

y^{'} = - \frac{x}{y}

Why not B:You inverted the ratio; differentiating gives $2 x + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{y}$ .
Why not C:Solving $2 x + 2 y y^{'} = 0$ for $y^{'}$ gives a negative sign.
Why not D:The $2$ 's from $2 x$ and $2 y y^{'}$ cancel, leaving $- \frac{x}{y}$ .

Question 5

no calculatorhard

For the hyperbola

x^{2} - y^{2} = 9

, what is

\frac{d ^{2} y}{d x ^{2}}

at the point

(5, 4)

Show the answer

The answer is B.

From

2 x - 2 y y^{'} = 0

y^{'} = \frac{x}{y}

. Then

y^{''} = \frac{y - x y ^{'}}{y ^{2}} = \frac{y - x ( x / y )}{y ^{2}} = \frac{y ^{2} - x ^{2}}{y ^{3}} = - \frac{9}{y ^{3}}

. At

y = 4

- \frac{9}{64}

Why not A:This is the first derivative $y^{'} = \frac{x}{y} = \frac{5}{4}$ , not the second derivative.
Why not C:Sign error: $y^{''} = - \frac{9}{y ^{3}}$ is negative (the upper branch is concave down here), so $- \frac{9}{64}$ .
Why not D:This forgets to substitute $y^{'} = \frac{5}{4}$ and to use $x^{2} - y^{2} = 9$ ; the result is $- \frac{9}{64}$ .

Derivatives of Inverse Functions

Question 6

no calculatormedium

The table gives values of a differentiable, increasing function

f

and its derivative.

x 1245 f (x) 3569 f^{'} (x) 2417

g = f^{- 1}

, find

g^{'} (5)

Show the answer

The answer is B.

g^{'} (a) = \frac{1}{f ^{'} ( b )}

where

f (b) = a

. Here

f (2) = 5

, so

g^{'} (5) = \frac{1}{f ^{'} ( 2 )} = \frac{1}{4}

Why not A:Don't forget to take the reciprocal. $f (2) = 5 \Rightarrow g^{'} (5) = \frac{1}{f ^{'} ( 2 )} = \frac{1}{4}$ , not $f^{'} (2)$ .
Why not C:You used the row where $x = 5$ . You need the row where $f (x) = 5$ , i.e. $x = 2$ : $g^{'} (5) = \frac{1}{f ^{'} ( 2 )} = \frac{1}{4}$ .
Why not D:It's not $1$ over the input. Find $x$ with $f (x) = 5$ (that's $x = 2$ ), then $g^{'} (5) = \frac{1}{f ^{'} ( 2 )} = \frac{1}{4}$ .

Derivatives of Inverse Trig Functions

Question 7

no calculatormedium

Let

f (x) = arctan (2 x)

. What is

f^{'} (\frac{1}{2})

Show the answer

The answer is C.

\frac{d}{d x} arctan u = \frac{u ^{'}}{1 + u ^{2}}

with

u = 2 x

u^{'} = 2

. So

f^{'} (x) = \frac{2}{1 + 4 x ^{2}}

, and

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

Why not A:This forgets the chain-rule factor $u^{'} = 2$ ; with it, $\frac{2}{1 + 1} = 1$ .
Why not B:This is the numerator $u^{'} = 2$ alone, ignoring the $1 + 4 x^{2} = 2$ denominator; the value is $1$ .
Why not D:This uses $1 + x^{2}$ instead of $1 + (2 x)^{2}$ in the denominator; correctly $\frac{2}{2} = 1$ .

Selecting Differentiation Procedures

Question 8

calculatormedium

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

, find

f^{'} (2)

to the nearest hundredth.

Show the answer

The answer is D.

f^{'} (x) = \frac{e ^{x} ( x ^{2} + 1 ) - e ^{x} ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{e ^{x} ( x - 1 ) ^{2}}{( x ^{2} + 1 ) ^{2}}

. At

x = 2

\frac{e ^{2}}{25} \approx 0.30

Why not A:That's $\frac{e ^{2}}{5} = f (2)$ , the function value, not its derivative.
Why not B:This is the $uv'$ term $\frac{e ^{x} \cdot 2 x}{( x ^{2} + 1 ) ^{2}}$ at $x = 2$ (that is $\frac{4 e ^{2}}{25} \approx 1.18$ ), the piece you subtract, not the full quotient-rule derivative $\approx 0.30$ .
Why not C:The numerator is $e^{x} (x^{2} + 1) - e^{x} (2 x)$ ; over $(x^{2} + 1)^{2}$ that gives $\approx 0.30$ .

Higher-Order Derivatives

Question 9

no calculatormedium

Let

f (x) = sin (3 x)

. What is

f^{''} (\frac{π}{6})

Show the answer

The answer is A.

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

and

f^{''} (x) = - 9 sin (3 x)

(chain rule each time). At

x = \frac{π}{6}

sin (\frac{π}{2}) = 1

, so

f^{''} = - 9

Why not B: $f^{''} (x) = - 9 sin (3 x)$ carries a minus sign; at $\frac{π}{6}$ it is $- 9$ .
Why not C:This used only one factor of $3$ ; the second derivative brings down $3$ twice, giving $- 9 sin (3 x)$ .
Why not D:This forgets the $3^{2} = 9$ from differentiating the inside twice; the value is $- 9$ , not $- 1$ .

Linear Approximation & Differentials

Question 11

no calculatormedium

Let

f

be the function given by

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

. The tangent line to the graph of

f

x = 2

is used to approximate values of

f (x)

. Which of the following is the greatest value of

x

for which the error resulting from this tangent line approximation is less than

0.5

Show the answer

The answer is D.

The tangent at

x = 2

L (x) = - 3 + 2 (x - 2)

, and the error is

∣ f (x) - L (x) ∣ = ∣ x^{2} - 4 x + 4∣ = (x - 2)^{2}

. We need

(x - 2)^{2} < 0.5

, i.e.

∣ x - 2∣ < 0.5 \approx 0.707

. Among the choices,

x = 2.7

gives error

0.49 < 0.5

while

x = 2.8

gives

0.64

Why not A:The error is $(x - 2)^{2}$ ; at $x = 2.8$ it is $0.64 > 0.5$ , too large.
Why not B:The error $(2.6 - 2)^{2} = 0.36 < 0.5$ , but $2.7$ also satisfies it and is greater.
Why not C:The error $(2.4 - 2)^{2} = 0.16 < 0.5$ , but this is not the greatest such $x$ .

Mean Value Theorem

Question 12

no calculatorhard

Let

g

be differentiable on the closed interval

[0, 10]

with

g (0) = 7

g (4) = - 1

, and

g (10) = 7

. Which of the following must be true? I.

g

has at least 2 zeros in

(0, 10)

.
II. The graph of

g

has at least one horizontal tangent in

(0, 10)

.
III.

g^{'} (c) = - 3

for some

c

(0, 4)

Show the answer

The answer is C.

g

changes sign

7 \to - 1

[0, 4]

and

- 1 \to 7

[4, 10]

, so the IVT gives a zero in each subinterval, at least 2 zeros. II:

g (0) = g (10) = 7

, so Rolle's Theorem gives a horizontal tangent. III: on

(0, 4)

the MVT only forces

g^{'} (c) = \frac{- 1 - 7}{4 - 0} = - 2

, not

- 3

; the asked slope differs from the average rate, so it need not occur. Hence I and II only.

Why not A:III is not guaranteed: on $(0, 4)$ the MVT forces only $g^{'} (c) = \frac{- 1 - 7}{4 - 0} = - 2$ , and $- 3$ is not that average rate, so it need not occur.
Why not B:II also holds: $g (0) = g (10) = 7$ , so Rolle's Theorem gives a horizontal tangent in $(0, 10)$ .
Why not D:I also holds: $g$ goes $7 \to - 1 \to 7$ , so the IVT yields at least two zeros.

The full bank has 3,000+ more, every one with a worked explanation. Bought once, yours forever.

Get the full AP Calculus bank for $10

AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this product.

Differentiation Rules

Chain Rule

Question 1

Question 2

Implicit Differentiation

Question 3

Question 4

Question 5

Derivatives of Inverse Functions

Question 6

Derivatives of Inverse Trig Functions

Question 7

Selecting Differentiation Procedures

Question 8

Higher-Order Derivatives

Question 9

Related Rates

Question 10

Linear Approximation & Differentials

Question 11

Mean Value Theorem

Question 12