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Flashcards · AP® Calculus AB · Unit 2

Unit 2: Differentiation: Definition and Fundamental Properties flashcards

60 cards on topics 2.1 to 2.10. Pick a topic or study the whole unit.

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2.1 Defining Average and Instantaneous Rates of Change at a Point

What is the limit definition of the derivative of f at x = a?
f′(a) = lim (h → 0) (f(a + h) − f(a))/h, or equivalently f′(a) = lim (x → a) (f(x) − f(a))/(x − a), provided the limit exists. It is the slope of the tangent line at a.
What is the difference between f(a) and f′(a)?
f(a) is a value (the height of the graph). f′(a) is a rate: the instantaneous rate of change at a, with units of output per input.
How do you recognise lim (h → 0) ((3 + h)³ − 27)/h as a derivative?
Find f from the f(a + h) part and check the subtracted number is f(a): it is f′(3) for f(x) = x³. The quotient simplifies to 27 + 9h + h², so the limit is 27.
Use the definition to find q′(1) for q(x) = 5x − x².
3. (q(1 + h) − q(1))/h = (5 + 5h − 1 − 2h − h² − 4)/h = 3 − h for h ≠ 0, which tends to 3.
When does f′(a) fail to exist from the definition?
When the limit of the difference quotient is not a single finite value from both sides, for example at a corner where the left and right limits differ.
What mistakes are common with the limit definition of the derivative?
Writing f(a) + h for f(a + h); subtracting a or f(h) instead of f(a); substituting h = 0 before simplifying (always 0/0); stopping at the quotient without taking the limit; leaving out units.

2.2 Defining the Derivative of a Function and Using Derivative Notation

How is the derivative function f′(x) defined?
f′(x) = lim (h → 0) (f(x + h) − f(x))/h, provided the limit exists. Inside the limit x is fixed and only h changes; cancel h (h ≠ 0) before letting h → 0.
What notations mean the derivative of y = f(x)?
f′(x), dy/dx, y′ and d/dx (f(x)) all mean the same thing. The value at a point is f′(a), or dy/dx at x = a.
What is the equation of the tangent line to y = f(x) at x = a?
y − f(a) = f′(a)(x − a): the point is (a, f(a)) and the slope is the number f′(a), not the formula f′(x).
Find the tangent line to y = x² + 4x at x = 1.
y′ = 2x + 4, so the slope is 6, and the point is (1, 5). Tangent: y − 5 = 6(x − 1), or y = 6x − 1.
What are the units of a derivative?
Units of f divided by units of x. If P(t) is a population and t is in years, dP/dt is in individuals per year.
Why is "f(2) = 9, so f′(2) = 0" wrong?
f(2) is a constant, so differentiating it gives 0, but that is not f′(2). Find f′(x) first, then substitute x = 2.

2.3 Estimating Derivatives of a Function at a Point

How do you estimate f′(a) from a table?
Use the closest data points on an interval that contains a and compute (f(q) − f(p))/(q − p). Divide by the change in x, not the number of rows, and write ≈.
f(1.0) = 3, f(1.2) = 3.8 and f(1.4) = 5.1. Estimate f′(1.2).
Use the entries either side: f′(1.2) ≈ (5.1 − 3)/(1.4 − 1.0) = 2.1/0.4 = 5.25.
What are the forward, backward and symmetric difference estimates of f′(a)?
Forward (f(a + h) − f(a))/h; backward (f(a) − f(a − h))/h; symmetric (f(a + h) − f(a − h))/(2h). The symmetric one is usually most accurate for the same h.
How do you estimate f′(a) from a graph with a tangent line drawn?
Read two grid points on the tangent line (not the curve) and compute rise over run. Through (0, 5) and (4, −1) the slope is −6/4 = −1.5.
How do you interpret h′(7) ≈ 2.4 if h(t) is a balloon's height in metres after t seconds?
At t = 7 seconds the balloon is rising at about 2.4 metres per second. State a rate with units, not a total change.
Why can a calculator's numerical derivative mislead you?
At a corner it averages the two sides and can give a number where no derivative exists. Also use radian mode and give the setup, such as f′(3), before the value.

2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist

What is the link between differentiability and continuity?
If f is differentiable at a, then f is continuous at a. So a hole, jump or asymptote at a means f′(a) does not exist. The converse is false.
What are three ways a continuous function can fail to be differentiable?
A corner (finite one-sided slopes that differ), a cusp (slopes go to −∞ on one side and +∞ on the other) and a vertical tangent (slopes unbounded with the same sign on both sides).
Is |2x − 6| differentiable at x = 3?
No. It is continuous there, but the one-sided limits of the difference quotient are −2 and 2, so there is a corner.
How do you check differentiability at the join of a piecewise function?
Check continuity first (a jump rules it out). Then compare the one-sided limits of the difference quotient, using the actual f(a) from the piece that includes a. Matching slopes alone is not enough.
Should you write f′(a) = ∞ for a vertical tangent?
No. Say f′(a) does not exist because the difference quotient is unbounded.
Does a formula containing |x| always have a corner at 0?
No. Absolute values cause corners only where the inside changes sign and the slopes differ. For example x|x| is differentiable at 0.

2.5 Applying the Power Rule

What is the power rule?
d/dx [xⁿ] = n·xⁿ⁻¹ for any constant exponent n: whole, negative, fractional or irrational. It applies only to a single power of x.
How do you differentiate 1/x³?
Rewrite as x⁻³. The derivative is −3x⁻⁴ = −3/x⁴. Lowering a negative exponent makes it more negative: −3 − 1 = −4.
What is the derivative of √x, and where does it exist?
√x = x^(1/2), so the derivative is (1/2)x^(−1/2) = 1/(2√x), which exists only for x > 0.
Let f(x) = x^(5/3). Find f′(8).
f′(x) = (5/3)x^(2/3), and 8^(2/3) = 4, so f′(8) = 20/3.
Is x^(1/3) differentiable at 0?
No. f′(x) = 1/(3x^(2/3)) has a zero denominator at 0: the graph has a vertical tangent there.
When can you not use the power rule directly?
When the variable is in the exponent (2ˣ is not a power function), when the base is not plain x, as in (2x + 1)³ (expand, or use the chain rule later), or on a product of factors without combining them first.

2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple

What are the constant, constant multiple, sum and difference rules?
d/dx [c] = 0; d/dx [k·f] = k·f′; d/dx [f ± g] = f′ ± g′. Together with the power rule they differentiate any polynomial.
Why does a constant term have derivative 0?
It only shifts the graph up or down, which does not change any slope. So y = x³ and y = x³ − 40 have the same derivative.
Let f(x) = 6/x − 2√x. Find f′(1).
f(x) = 6x⁻¹ − 2x^(1/2), so f′(x) = −6x⁻² − x^(−1/2), and f′(1) = −6 − 1 = −7.
How do you differentiate (x + 2)(x − 5) with only these rules?
Expand first: x² − 3x − 10, so the derivative is 2x − 3. These rules do not split products or quotients.
f′(3) = 2 and g′(3) = −5. Find the derivative of 4f − g at x = 3.
4(2) − (−5) = 13, by linearity.
What mistakes are common with the basic rules?
Keeping the constant term (d/dx [x² + 7] = 2x); losing a coefficient (d/dx [5x] = 5); differentiating π² or e as if they contained x; sign slips such as d/dx [−3x⁻²] = +6x⁻³.

2.7 Derivatives of cos x, sin x, eˣ, and ln x

What are the derivatives of sin x, cos x, eˣ and ln x?
d/dx sin x = cos x; d/dx cos x = −sin x; d/dx eˣ = eˣ; d/dx ln x = 1/x for x > 0. Trig results need radians.
What is the derivative of a constant such as e², ln 7 or cos 1?
0. They contain no x. A common mistake is writing d/dx e² = e² or d/dx ln 7 = 1/7.
Find lim (h → 0) [e^(3 + h) − e³]/h.
e³. It is the difference quotient for f(x) = eˣ at a = 3, so the limit is f′(3) = e³.
What is the derivative of xᵉ + eˣ?
e·x^(e − 1) + eˣ. xᵉ is a power (power rule); eˣ is an exponential (its own derivative).
What is the slope of y = sin x at x = π/3?
cos(π/3) = 1/2.
Why does d/dx ln x = 1/x connect to eˣ?
ln x is the reflection of eˣ in y = x, so its slopes are reciprocals: at height y = ln x, the slope of eˣ at the matching point is x, giving 1/x.

2.8 The Product Rule

What is the product rule?
(f·g)′ = f′·g + f·g′: differentiate one factor at a time and add the results. The derivative of a product is not the product of the derivatives.
How can you check that (fg)′ ≠ f′g′?
x²·x³ = x⁵ has derivative 5x⁴, but 2x·3x² = 6x³. Only f′g + fg′ = 2x·x³ + x²·3x² = 5x⁴ is correct.
Differentiate x³eˣ.
3x²eˣ + x³eˣ = x²(x + 3)eˣ. Factoring eˣ helps when solving y′ = 0.
f(1) = 4, f′(1) = 2, g(1) = −3 and g′(1) = 1. Find (f·g)′(1).
f′(1)g(1) + f(1)g′(1) = 2(−3) + 4(1) = −2.
Differentiate sin x cos x.
cos x·cos x + sin x·(−sin x) = cos²x − sin²x.
Do you need the product rule for 5 ln x?
No. A constant factor needs only the constant multiple rule: the derivative is 5/x.

2.9 The Quotient Rule

What is the quotient rule?
If q = f/g with g ≠ 0, q′ = (g·f′ − f·g′)/g²: bottom times derivative of top, minus top times derivative of bottom, over bottom squared.
Differentiate x/(x + 1).
((x + 1)(1) − x(1))/(x + 1)² = 1/(x + 1)².
When is it faster not to use the quotient rule?
When one part is a constant or the fraction simplifies. For example 3/x² = 3x⁻², with derivative −6x⁻³ = −6/x³.
f(1) = 2, f′(1) = 3, g(1) = 1 and g′(1) = 4. Find (f/g)′(1).
(g(1)f′(1) − f(1)g′(1))/g(1)² = (3 − 8)/1 = −5.
What mistakes are common with the quotient rule?
Wrong order in the numerator (flips the sign); differentiating the denominator instead of squaring it; dropping brackets so the minus misses part of f·g′; using f′/g′.
Where does the derivative of a quotient not exist?
Where the denominator g(x) is 0, because the quotient itself is undefined there.

2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

What are the derivatives of tan x, cot x, sec x and csc x?
tan x → sec²x; cot x → −csc²x; sec x → sec x tan x; csc x → −csc x cot x. The "co" functions get a minus sign.
How do you derive d/dx tan x = sec²x?
Write tan x = sin x/cos x and use the quotient rule: (cos²x + sin²x)/cos²x = 1/cos²x = sec²x.
Find the slope of y = tan x at x = π/3.
sec²(π/3) = 2² = 4.
Differentiate sec x + cot x and evaluate at π/4.
sec x tan x − csc²x. At π/4: √2 × 1 − (√2)² = √2 − 2.
Why is tan x increasing on every interval of its domain?
Its derivative sec²x ≥ 1 is always positive, so the slope is at least 1. Similarly −csc²x ≤ −1, so cot x is always decreasing.
What mistakes are common with these trig derivatives?
Writing d/dx tan x = sec x (missing square); d/dx sec x = tan x (missing the sec x factor); dropping the minus on cot or csc; confusing sec x = 1/cos x with csc x = 1/sin x.

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