Differentiation

Question 1

In one word,
what is
differential calculus
about?

Answer 1

The
instant.

More specifically, the
rate of change at an instant.

Question 2

What type of
line
will tell you the
average speed over time?

Answer 2

Secant line.

A
slope tells you the
ARC of a
vertical variable
with respect to a
horizontal variable.

ARC is

Δy
Δx

which is also the
slope of the secant line.

ex:

Δy
Δx

= y₂ − y₁
x₂ − x₁

= 4 − 0
1 − 0

= 4

Question 3

What type of
line
will tell you
instantaneous speed?

Answer 3

Tangent line.

A
slope tells you the
ARC of a
vertical variable
with respect to a
horizontal variable.

The slope of the secant line will approach that of the tangent line as the distances for the secant line approach zero.

Question 4

g(x) = √(x)

How can you

• *express** the
• *derivative** of √(x)?

Answer 4

A, C, and D.

Question 5

What is a

• *reasonable** estimate of
• *g’(1)**?

A: −2
B: 1/4
C: 2
D: 0
E: −1/4

Answer 5

−2

Question 6

Compare

f(−4) ___ f’(−1).

A: <

B: >

Answer 6

B: >

The slope of the tangent line would be less negative at f(−4).

Question 7

The

• *tangent line** to the graph of
• *function f** at the point
• *(2, 3)** passes through the point
• *(7, 6)**.

What is
f’(2)?

Answer 7

3/5.

The derivative tells you the slope of the tangent.

Δy
Δx

= y₂ − y₁
x₂ − x₁

= 6 − 3
7 − 2

= 3/5

Question 8

What is the

• *formal definition** of the
• *derivative** of a function?

f’(x) = _____?

Answer 8

lim_Δx→0 f(x₀ + Δx) − f(x₀)
Δx

This is basically the slope formula with a limit tacked on:

Δy
Δx

= y₂ − y₁
x₂ − x₁

= lim_Δx→0 f(x₀ + Δx__) − f(x₀)
(x₀ + Δx) − x₀

= lim_Δx→0 f(x₀ + Δx) − f(x₀)
Δx

Question 9

What is the

• *alternate definition** of the
• *derivative** of a function?

f’(x) = _____?

Answer 9

lim_x→a f(x) − f(a)
x − a

This is basically the slope formula with a limit tacked on:

Δy
Δx

= y₂ − y₁
x₂ − x₁

= lim_x→a f(x) − f(a)
x − a

Question 10

How do you write the
equation of a
tangent line using the
formal definition of a
limit of the function below?

f(x) = x² at x = 0.2

Answer 10

1. Find the derivative
   f(x) = x²
   lim_Δx→0 f(x + Δx) − f(x)
   (x + Δx) − x
   f’(x) = lim_Δx→0 (x + Δx)² − x²
   (x + Δx) − x
   = lim_Δx→0 x² + 2x•Δx + Δx² − x²
   Δx
   = lim_Δx→0 2x•Δx + Δx²
   Δx
   = lim_Δx→0 2x + Δx
   f’(x) = 2x
2. Find the slope of the tangent
   f’(x) = 2x
   = 2(0.2)
   f’(0.2) = 0.4
3. Find the right point on the function
   f(x) = x²
   f(0.2) = (0.2)²
   f(0.2) = 0.04
4. Write the equation of the tangent line
   y − y₀ = m(x − x₀)
   y − 0.04 = 0.4(x − 0.2)

Question 11

Given:

• f(−1) = 2
• f(0) = 0
• f(1) = 1
• f(8) = 5

what is the
best estimate of f’(−1/2) we can make
given these values?

Answer 11

−​2

Δy
Δx

= y₂ − y₁
x₂ − x₁

= 2 − 0
−1 − 0

= −2

Question 12

Given:

• f(−1) = 2
• f(0) = 0
• f(1) = 1
• f(8) = 5

what is the
best estimate of f’(8) we can make
given these values?

Answer 12

4/7

Δy
Δx

= y₂ − y₁
x₂ − x₁

= 5 − 1
8 − 1

= 4
7

Question 13

Graphically,
what are
all x-values for which this
function is
not differentiable?

Dashed lines represent asymptotes.

Answer 13

−5, −4, 0

Question 14

Graphically,
what are
all x-values for which this
function is
not differentiable?

There’s a
vertical tangent at x = 1.

Answer 14

1, 5

Question 15

Visually, what
features of a
graph will indicate that
the function is
not differentiable at an x-value?

Answer 15

1. Vertical tangent
   * (the slope of the tangent line is undefined)*
2. Discontinuity
   * (there’s no limit at that location)*
3. “Sharp” turn
   * (the one-sided limits aren’t equivalent, so there’s no limit at that location)*

Question 16

How do you know whether function

• *g(x)** is
• *continuous** at
• *x = a**?

Answer 16

g(x) is continuous at x = a if

• *g(a)** and
• *both one-sided limits** are
• *equal**.
• (Which is to say that the lim_x→a g(x) exists and is equal to g(a).)*
• ex:*

g(x) = { x² + 2x , x < 1
{ 4x − 1 , x > 1

g(1) = (1)² + 2(1)
= 3 = lim_x→1⁻ g(x)

lim_x→1⁺ g(x) = 4(1) − 1
= 3

g(1) = lim_x→1⁻ g(x) = lim_x→1⁺ g(x)

So g(x) is continuous at x = 1.

Question 17

How do you know whether function

• *g(x)** is
• *differentiable** at
• *x = a**?

Answer 17

g(x) is differentiable at x = a
if it is
continuous at x = a and
lim_x→a f(x) − f(a)
x − a
exists.

(Which is to say that you can find the slope of a tangent line that intersects g(x) at x = a)

(Once you know it’s continuous, you’re looking for a
vertical tangent or a
“sharp” turn)

ex:

g(x) = { x² + 2x , x < 1
{ 4x − 1 , x > 1

g(1) = (1)² + 2(1)
= 3 = lim_x→1⁻ g(x)

lim_x→1⁺ g(x) = 4(1) − 1
= 3

g(1) = lim_x→1⁻ g(x) = lim_x→1⁺ g(x)

So g(x) is continuous at x = 1.

lim_x→a f(x) − f(a)
x − a

lim_x→1⁻ x² + 2x − (3)
x − 1

= lim_x→1⁻ (x − 1)(x + 3)
x − 1

= lim_x→1⁻ x + 3

= 4

lim_x→1⁺ 4x − 1 − (3)
x − 1

= lim_x→1⁺ 4x − 4
x − 1

= lim_x→1⁺ 4(x − 1__)
x − 1

= 4

Because the one-sided limits are approaching the same value, the limit exists, and g(x) is differentiable at x = 1.

Question 18

Differentiability
implies _____.

Answer 18

Continuity

Proof:

• Differentiability:
  lim_x→c f(x) − f(c) = f’(c)
  x − c
• Continuity:
  lim_x→c (f(x) = f(c))

Assume f is differentiable at x = c

lim_x→c (f(x) − f(c))

= lim_x→c (x − c) • f(x) − f(c)
x − c

= lim_x→c (x − c ) • lim_x→c f(x) − f(c)
x − c

= 0 • lim_x→c f(x) − f(c)
x − c

= 0 • f’(c)

lim_x→c (f(x) − f(c)) = 0

lim_x→c f(x) − lim_x→c f(c) = 0

lim_x→c f(x) − f(c) = 0

lim_x→c f(x) = f(c)
(this is the definition of continuity)

Question 19

Constant Rule:

Answer 19

0.

The derivative of
any constant is 0.

Algebraically:

f(x) = 1

f(x + Δx) = 1

g’(x) = lim_Δx→0 f(x + Δx) − f(x)
Δx

f’(x) = lim_Δx→0 1 − 1
Δx

= lim_Δx→0 0
Δx

= 0

Graphically:
The derivative measures a function’s
instantaneous rate of change at a particular x-value.

Where f(x) is a constant, there is
no change from one x-value to the next, so the derivative is 0.

Question 20

Sum/Difference Rules:

f(x) = g(x) + j(x)

f’(x) = ______

Answer 20

g’(x) + j’(x)

f(x) = g(x) + j(x) ⇒ f’(x) = g’(x) + j’(x)

Algebraically:

f(x) = g(x) + j(x)

f’(x) = lim_Δx→0 g(x + Δx) + j(x + Δx) − (g(x) + j(x))
Δx

= lim_Δx→0 g(x + Δx) − g(x) + j(x + Δx) − j(x)
Δx

= lim_Δx→0 g(x + Δx) − g(x) + j(x + Δx) − j(x)
Δx Δx

= lim_Δx→0 g(x + Δx) − g(x) + lim_Δx→0 j(x + Δx) − j(x)
Δx Δx

= g’(x) + j’(x)

The difference rule is identical.

Question 21

Constant Multiple Rule:

• *d [k•f(x)]** = _____?
• *dx**

Answer 21

k• d [f(x)]
dx

Also:
f(x) = k•g(x) ⇒ f’(x) = k•g’(x)

Algebraically:

h’(x) = lim_Δx→0 f(x + Δx) − f(x)
Δx

f(x) = k•g(x)

f’(x) = lim_Δx→0 k•g(x + Δx) − k•g(x)
Δx

= lim_Δx→0 k•g(x + Δx) − k•g(x)
Δx

= lim_Δx→0 k • g(x + Δx) − g(x)
Δx

= k•lim_Δx→0 g(x + Δx) − g(x)
Δx

= k•g’(x)

Question 22

Constant Rule:

f(x) = 2x³ − √x + 1/x + 2

f’(x) = _____

Answer 22

6x² − ½x^−1/2 − x⁻²

*f(x) = x<sup>n</sup>
f'(x) = nx<sup>n−1​</sup>*

f(x) = 2x³ − √x + 1/x + 2

= 2x³ − x^1/2 + x⁻¹ + 2

f’(x) = d/dx(2x³) − d/dx(x^1/2) + d/dx(x⁻¹) + d/dx(2)

= (3)2x³⁻¹ − (½)x^1/2−1 + (−1)x⁻¹⁻² + 0

= 6x² − ½x^−1/2 − x⁻²

Question 23

Power Rule:

f(x) = xⁿ, n ≠ 0

f’(x) = _____

Answer 23

f’(x) = nxⁿ⁻¹

Proof:
(for positive integers)

Question 24

f(x) = 2x³ − √x + 1/x + 2

f’(x) = _____

Answer 24

6x² − ½x^−1/2 − x⁻²

*f(x) = x<sup>n</sup>
f'(x) = nx<sup>n−1​</sup>*

f(x) = 2x³ − √x + 1/x + 2

= 2x³ − x^1/2 + x⁻¹ + 2

f’(x) = d/dx(2x³) − d/dx(x^1/2) + d/dx(x⁻¹) + d/dx(2)

= (3)2x³⁻¹ − (½)x^1/2−1 + (−1)x⁻¹⁻² + 0

= 6x² − ½x^−1/2 − x⁻²

Question 25

How would you write the * *equation** of a * *line** * *tangent** to **f(x) = x³ − 6x² + x − 5** at **x = 1**?

Answer 25

1. **Find the point the tangent touches** *(this is just f(1))* f(x) = x³ − 6x² + x − 5 f(1) = (1)³ − 6(1) + (1) − 5 = −9 _(1, −9)_ 2. **Find the slope of the tangent** at that point *(this is just the derivative)* f'(x) = 3x² − 12x² + 1 f'(1) = 3(1)² − 12(1)² + 1 _= −8_ 3. **Write the equation of the line** y − y₀ = m(x − x₀) y + 9 = −8(x − 1) *(for point/slope form)* y = −8x + 8 − 9 y = −8x − 1 *(for slope/intercept form)*

Question 26

**f(x) = sin(x)** **f'(x)** = \_\_\_\_\_

Answer 26

**cos(x)** ## Footnote Green: slope of 1 Orange: slope of 0 Red: slope of −1

Question 27

**f(x) = cos(x)** **f'(x)** = \_\_\_\_\_

Answer 27

**−sin(x)** ## Footnote Green: slope of 1 Orange: slope of 0 Red: slope of −1

Question 28

**f(x) = e^x** **f'(x)** = \_\_\_\_\_

Answer 28

**e^x** * This is one of the "magical" things about e.* * In fact, e can be defined this way.*

Question 29

**f(x) = ln(x)** **f'(x)** = \_\_\_\_\_

Answer 29

**_1_ x**

Question 30

_Product Rule_: ## Footnote **p(x) = [f₁(x) • f₂(x)]** **p'(x) = \_\_\_\_\_**

Answer 30

**f₁'(x)** **• f₂(x) + f₁(x) • f₂'(x)** ## Footnote p(x) = [ln(x) • cos(x)] p'(x) = d/dx(ln(x)) • cos(x) + ln(x) • d/dx(cos(x)) = 1/x • cos(x) + ln(x) • (−sin(x)) = _cos(x)_ − ln(x) • sin(x) x

Question 31

_Quotient Rule_: ## Footnote **q(x) = [_n(x)_] [d(x)]** **q'(x) = \_\_\_\_\_**

Answer 31

**_n'(x)_** **_• d(x) − n(x) • d'(x)_ ( d(x) )²** ## Footnote q(x) = [_sin(x)_] [x²] q'(x) = _d/dx(sin(x)) • x² + sin(x) • d/dx(x²)_ (x²)² = _x² • cos(x) − sin(x) • 2x_ x⁴ = _x (x•cos(x) − 2•sin(x))_ x⁴​ = _x•cos(x) − 2•sin(x)_ x³​

Question 32

**f(x) = tan(x)** **f'(x)** = \_\_\_\_\_

Answer 32

**_1_ = sec²(x) cos²(x)** ## Footnote f(x) = tan(x) = _sin(x)_ cos(x) f'(x) = _d_ [_sin(x)_] dx [cos(x)] = _cos(x) • cos(x) + sin(x) • sin(x)_ cos²(x) = _cos²(x) + sin²(x)_ cos²(x) = _1_ cos²(x) = sec²(x)

Question 33

**f(x) = cot(x)** **f'(x)** = \_\_\_\_\_

Answer 33

**− _1_ = −csc²(x) sin²(x)** ## Footnote f(x) = cot(x) = _cos(x)_ sin(x) f'(x) = _d_ [_cos(x)_] dx [sin(x)] = _−sin(x) • sin(x) − cos(x) • cos(x)_ sin²(x) = _−sin²(x) − cos²(x)_ sin²(x) = − _1_ sin²(x) = −csc²(x)

Question 34

**f(x) = sec(x)** **f'(x)** = \_\_\_\_\_

Answer 34

**tan(x) • sec(x)** ## Footnote f(x) = sec(x) = _1_ cos(x) = _0 • cos(x) + sin(x) • 1_ cos²(x) = _sin(x)_ cos²(x) = _sin(x)_ • _1_ cos(x) cos(x) = tan(x) • sec(x)

Question 35

**f(x) = csc(x)** **f'(x)** = \_\_\_\_\_

Answer 35

**−cot(x) • csc(x)** ## Footnote f(x) = csc(x) = _1_ sin(x) = _0 • sin(x) − cos(x) • 1_ sin²(x) = _−cos(x)_ sin²(x) = _−cos(x)_ • _1_ sin(x) sin(x) = −cot(x) • csc(x)

Question 36

_Chain Rule_: ## Footnote **f(x) = o(i(x))** **f'(x) = \_\_\_\_\_**

Answer 36

**o'( i(x)) • i'(x)** *ex:* * h(x) = (5 − 6x)⁵ = o(i(x)) * i(x) = 5 − 6x i'(x) = −6 * o(x) = x⁵ o'(x) = 5x⁴ * h'(x) = o'(i(x)) • i'(x) * = 5 ( 5 − 6x)⁴ • −6 * −30( 5 − 6x)⁴

Question 37

Is **f(x) = cos²(x)** a _composite function_?

Answer 37

**Yes**, so the **chain rule applies**. *A function is composite if you can write it as _f(g(x))_.* *In other words, it is a _function within a function_, or a _function of a function_.* cos²(x) = (cos(x))² = o(i(x)) i(x) = cos(x) o(x) = x²

Question 38

**f(x) = b^x**, where b is _any positive number_ ## Footnote **_d_ [b^x] = \_\_\_\_\_ dx**

Answer 38

**b^x • ln (b)** ## Footnote d/dx [e^x] = e^x​ b = e^{ln (b)} *(remember: log_b a = c ⇔ b^a = c) (ln (b) is the number that you have to raise e to in order to get b, so if you raise e to that number, you get b)* d/dx [b^x] = d/dx [(e^{ln (b)})^x] *(substitution)* = e^{ln (b) • x} • ln(b) (*chain rule)* = (ln (b)) • (e^{ln (b)})^x = b^x • ln (b)

Question 39

**f(x) = log_b a**, where b is _any positive number_ ## Footnote **d/dx [log_b a] = \_\_\_\_\_**

Answer 39

**_1_ (ln b) • a** ## Footnote d/dx [ln x] = 1/x​

Question 40

* *_d_ [y]** = \_\_\_\_\_? * *dx**

Answer 40

**_dy_ dx** *These are simply equivalent statements.*

Question 41

* *_d_ [y²]** = \_\_\_\_\_? * *dx**

Answer 41

**2y • _dy_ dx** *Just the chain rule:* c(x) = o(i(x)) c'(x) = o'(i(x) • i'(x) * First, we assume that y does change with respect to x — it's not a constant. * Then you can set up y as a function of x: y(x) = _d_ [y²] dx = _d(y²)_ • _dy_ dy dx ↑ This part is **just the chain rule**: *The **derivative of \*something\* squared** **with respect to that \*something\*** times the **derivative of that \*something\* with respect to x**.* = 2y • _dy_ dx

Question 42

* *_d_ [xy]** = \_\_\_\_\_? * *dx**

Answer 42

**y + x • _dy_ dx** *Apply the product rule: p(x) = f₁(x) • f₂(x) p'(x) = f₁'(x) • f₂(x) + f₁(x) • f₂'(x)* * *_d_ [xy]** * *dx** = _dx_ • y + x • _dy_ dx dx = y + x • _dy_ dx

Question 43

Given **x² + y² = 1**, how do you obtain * *_dy_**? * *dx**

Answer 43

**Implicit differentiation**: *Treat the _variable you're looking for_ as a _function of the other_ and _solve_.* ## Footnote x² + y² = 1 _d_ (x² + y²) = _d_ (1) dx dx _d_ (x²) + _d_ (y²) = 0 dx dx 2x + 2y • _dy_ = 0 dx 2y • _dy_ = −2x dx _dy_ = − _x_ dx y

Question 44

**_d_ sin⁻¹ (x) = \_\_\_\_\_ dx**

Answer 44

**_1_ √(1 − x²)** ## Footnote *Equivalent statements:* y = sin⁻¹ (x) ⇔ sin(y) = x x = sin(y) *Derivative of Both Sides w/ respect to x:* dx/dx = d/dx [sin(y)] 1 = d/dy [sin(y)] • dy/dx 1 = cos(y) • dy/dx ``` _dy_ = _1_ dx cos(y) ``` *Pythagorean trig identity:* _dy_ = _1_ dx √(1 − sin²(y)) *Substitution:* _dy_ = _1_ dx √(1 − x²)

Question 45

**_d_ cos⁻¹ (x) = \_\_\_\_\_ dx**

Answer 45

**− _1_ √(1 − x²)** ## Footnote *Equivalent statements:* y = cos⁻¹ (x) ⇔ cos(y) = x x = cos(y) *Derivative of Both Sides w/ respect to x:* dx/dx = d/dx [cos(y)] 1 = d/dy [cos(y)] • dy/dx 1 = −sin(y) • dy/dx _dy_ = _1_ dx − sin(y) *Pythagorean trig identity:* _dy_ = − _1_ dx √(1 − cos²(y)) *Substitution:* _dy_ = − _1_ dx √(1 − x²)

Question 46

**_d_ tan⁻¹ (x) = \_\_\_\_\_ dx**

Answer 46

**_1_ 1 + x²** ## Footnote *Equivalent statements:* y = tan⁻¹ (x) ⇔ tan(y) = x x = tan(y) *Derivative of Both Sides w/ respect to x:* dx/dx = d/dx [tan(y)] 1 = d/dy [tan(y)] • dy/dx 1 = _1_ • dy/dx cos²(y) _dy_ = cos²(y) dx *Pythagorean trig identity:* = _cos²(y)_ cos²(y) + sin²(y) *Multiply numerator and denominator by same amount:* = _cos²(y)_ • _1 /_ _cos²(y)_ cos²(y) + sin²(y) 1 / cos²(y) = _1_ 1 + ( sin(y) / cos(y) )² *Def of tangent:* = _1_ 1 + tan²(y) *Substitution:* = _1_ 1 + tan²(y)

Question 47

Functions _*f* and *g*_ are _inverses_. In **terms of its inverse**, **f'(x)** = \_\_\_\_\_.

Answer 47

**_1_ g'(f(x))** *The _derivative_ of a function is* *equal to the _reciprocal_ of the _derivative_ of its _inverse_.* * Definitions f(x) = g⁻¹(x) ⇔ g(x) = f⁻¹(x) f(g(x)) = x = g(f(x)) * Now, the work . . . * g(f(x)) = x * _d_ [g(f(x))] = _dx_ dx dx * g'(f(x)) • f'(x) = 1 * f'(x) = _1_ g'(f(x))

Question 48

Given the _table_ below, and that _functions *g* and *h*_ are _inverses_, how could you **determine** **g'(0)**?

Answer 48

g'(0) = **_1_ h'(g(0))** ## Footnote g'(x) = _1_ h'(g(x)) = _1_ h'(g(0)) = _1_ h'(−2) = _1_ −1 = −1

Question 49

What's the best way to **evaluate** * *_d_ [(x + 5)(x − 3)]**? * *dx**

Answer 49

**Expand**, then the **product rule**. ## Footnote *Expressions can be rewritten to make differentiation easier!*

Question 50

What's the * *key** to * *evaluating** * *_d_ [(x² sin(x))³]**? * *dx**

Answer 50

**Label, label, label**. *When applying multiple rules, label them all.* * _Rewriting_: (x² sin(x))³ = x⁶ sin³(x) = f(x) • g(h(x)) * _Labeling_: f(x) = x⁶ g(x) = x³ h(x) = sin(x) * _Differentiating_: f'(x) = 6x⁵ g'(x) = 3x² h'(x) = cos(x) * _d_ [x⁶ sin³(x)] dx * _Product Rule_: = _d_ [f(x)] • g(h(x)) + f(x) • _d_ [g(h(x))] dx dx * _Chain Rule_: = f'(x) • g(h(x)) + f(x) • g'(h(x)) • h'(x) * _Substitution_: = 6x⁵ • sin³(x) + x⁶ • 3 sin²(x) • cos(x) * = 6x⁵ sin³(x) + 3x⁶ sin²(x) cos(x)

Question 51

What's the * *key** to * *evaluating** * *_d_ [sin (ln (x²))]**? * *dx**

Answer 51

**Label, label, label**. *When applying multiple rules, label them all.* * sin(ln(x²)) = f (g (h(x))) * f(x) = sin(x) g(x) = ln(x) h(x) = x² * f'(x) = cos(x) g'(x) = 1/x h'(x) = 2x * _d_ [sin(ln(x²))] dx * = f' (g (h(x))) • g' (h(x)) • h'(x) * = cos (ln (x²)) • 1/(x²) • 2x * = _2 cos (ln (x²))_ x

Question 52

f(x) = x³ + 2x² f'(x) = 3x² + 4x **f''(x) = \_\_\_\_\_**

Answer 52

**6x + 4** *This is the _second derivative_, or the derivative of the first derivative* *Could also be written as:* _d²y_ [x³ + 2x²] dx²

Question 53

How would you **evaluate** ## Footnote **lim_h→0 _5 log (2 + h) − 5 log (2)_ h**

Answer 53

**Use the derivative** ## Footnote lim_h→0 _5 log (2 + h) − 5 log (2)_ h = 5 lim_h→0 _log (2 + h) − log (2)_ h * The above, by definition, is f'(2)* * So...* * f(x) = log(x)* *f'(x) = _1_* ln(10) x f'(2) *= _1_* ln(10) 2 *Substitution:* 5 • f'(2) = 5 lim_h→0 _log (2 + h) − log (2)_ h = _5_ 2 • ln(10)

Question 54

What is a **related rates problem**?

Answer 54

Applied problems where you find the **rate** at which **one quantity is changing** by relating it to **other quantities** whose **rates are known**. ## Footnote *e.g.:* The radius *r(t)* of a circle is increasing at a rate of 3 centimeters per second. At a certain instant *t*₀, the radius is 8 centimeters. What is the rate of change of the area *A(t)* of the circle at that instant?

Question 55

What are the * *keys** to a * *related rates problem**?

Answer 55

1. **Draw** a diagram 2. **Label**, label, label Label what you know and what you're trying to figure out 3. **Divide and conquer** Use information you know to figure out information you need — in several steps if necessary.

Question 56

Given **√(2) = 4**, how might you **approximate** **√(4.36)**?

Answer 56

**Local linearity** ## Footnote *You can use the line tangent to at a known point to approximate the value of a value that's close, but unknown.*

Question 57

In _words_, What can **L'Hôpital's rule** do?

Answer 57

Use * *derivatives** to find * *limits** with * *undetermined forms** ## Footnote *This is in contrast to what we normally do, which is to use limits to find derivatives (in fact, the definition of derivatives is based on limits).*

Question 58

_Mathematically_, what is **L'Hôpital's rule**?

Answer 58

_Zero Case_: * If: **lim_x→c f(x)** = **0** & **lim_x→c g(x) = 0** & **lim_x→c _f'(x)_ = L g'(x)** * Then: **lim_x→c _f(x)_ = L g(x)** **​**_Infinity Case_: * If: **lim_x→c f(x)** = **±∞** & **lim_x→c g(x)** = **±∞** & **lim_x→c _f'(x)_ = L g'(x)** * Then: **lim_x→c _f(x)_ = L g(x)**

Question 59

How would you **evalulate** * *lim_x→0 _sin(x)_**? * *2x**

Answer 59

**L'Hôpital's rule** ## Footnote Direct substitution yields: lim_x→0 _sin(x)_ = _0_ 2x 0 0/0 is undetermined form. The first two conditions of the rule are met, so we should try to take the derivatives to see if the third condition is also met. = lim_x→0 _cos(x)_ = _1_ 2 2 Therefore . . . lim_x→0 _sin(x)_ = _1_ 2x 2

Question 60

How would you **evalulate** * *lim_x→∞ _4x² − 5x_**? * *1 − 3x²**

Answer 60

**L'Hôpital's rule** | (is one way, although factoring and canceling would also work) ## Footnote Direct substitution yields: lim_x→∞ _4x² − 5x_ = _∞_ 1 − 3x² −∞ This is undetermined form. The first two conditions of the rule are met, so we should try to take the derivatives to see if the third condition is also met. = lim_x→∞ _8x − 5_ = _∞_ −6x −∞ This is still undetermined, so we should try L'Hôpital's rule again: lim_x→∞ _8_ = _−4_ −6 3

Question 61

_Mathematically_, describe the **mean value theorem**?

Answer 61

Given that function *f* is: * **differentiable** over the * *open interval (*a*, *b*)** and * **continuous** over the * *closed interval [*a*, *b*]**, There is at least * *one number *c*** on the * *interval** from *a* to *b* such that **f'(c) = _f(b) − f(a)_ b − a**

Question 62

_Graphically_, describe the **mean value theorem**?

Answer 62

An _arc_ between _two endpoints_ has a **point** at which the **tangent** to the arc is **parallel** to the **secant** through its endpoints

Question 63

Does the * *mean value theorem** * *apply** to this function?

Answer 63

**No**. For the MVT to apply, the **function must be differentiable** over (a, b). *The function has only two possible tangent lines, neither of which is parallel to the secant between x = a and x = b.*

Question 64

Does the * *mean value theorem** * *apply** to this function?

Answer 64

**No**. For the MVT to apply, the **function must be continuous** over [a, b]. *All possible tangent lines are necessarily decreasing, while the secant line is increasing.*

Question 65

Does the **mean value theorem** apply to this function over the interval **[−6, −2]**? ## Footnote *There is a vertical tangent at x = −5*

Answer 65

**No**, because of the _vertical tangent_. The function _must be differentiable_ over (−6, −2) for the MVT to apply.

Question 66

Does the **mean value theorem** apply to this function over the interval **[−4, −1]**? ## Footnote *There is a vertical tangent at x = −5*

Answer 66

**Yes**. ``` The function is continuous over [−4, −1] and differentiable over (−4, −1). ```

Question 67

Does the **mean value theorem** apply to this function over the interval **[−1, 2]**? ## Footnote *There is a vertical tangent at x = −5*

Answer 67

**No**. The function _must be continuous_ over (−1, 2) for the MVT to apply.

Question 68

Does the **mean value theorem** apply to this function over the interval **[0, 4]**? ## Footnote *There is a vertical tangent at x = −5*

Answer 68

**Yes**. ``` The function is continuous over [0, 4] and differentiable over (0, 4). ```

Question 69

Does the **mean value theorem** apply to this function over the interval **[0, 5]**? ## Footnote *There is a vertical tangent at x = −5*

Answer 69

**No**, because of the _"sharp" turn_. The function _must be differentiable_ over (0, 5) for the MVT to apply.

Question 70

_Mathematically_, describe the **extreme value theorem**?

Answer 70

Given that function *f* is: * *continuous** over the * *closed interval [*a*, *b*]**, There are **values *c* and *d*** in the interval [*a*, *b*] such that **f(c) _\<_ f(x) _\<_ f(d)**.

Question 71

_Graphically_, describe the **extreme value theorem**?

Answer 71

Where function *f* is **continuous over [a, b]**, there is an **absolute maximum value of f** over the interval and an **absolute minimum value of f** over the interval.

Question 72

Visualize why an **interval must be continuous** for the **extreme value theorem to apply**.

Answer 72

Question 73

Visualize why an **interval must be closed** for the **extreme value theorem to apply**.

Answer 73

Question 74

What is a **critical point**?

Answer 74

An x-value: * **within the domain of function *f*** (graphed below) * where * **f'(x) = 0** or * **f'(x) is undefined** *Green is local max — f'(x) is _undefined_ Red is local min — f'(x) is _undefined_ Orange is global max — f'(x) is _zero_*

Question 75

How do you find an **increasing or decreasing interval** of a function?

Answer 75

**Differentiate** and evaluate **around critical points**. *f'(x) will be _positive where f(x) is increasing_ and _negative where f(x) is decreasing_.*

Question 76

How do you find **global/absolute extrema**? ## Footnote *(aka the first derivative test)*

Answer 76

**Same as local/relative extrema,** ***except*** you must **check the edges** in both directionsbecause endpoint can be global/absolute extrema. * **Differentiate** * **Find** * **critical points** and * **where f(x) is undefined** * **Analyze intervals using first derivative** * **Find extremum points**, which will be where * **f(x) is defined** and * **f'(x) changes signs** * **Check edges**

Question 77

How do you find **local/relative extrema**? ## Footnote *(aka the first derivative test)*

Answer 77

* **Differentiate** * **Find** * **critical points** and * **where f(x) is undefined** * **Analyze intervals using first derivative** * **Find extremum points**, which will be where * **f(x) is defined** and * **f'(x) changes signs**