Calculus · Units 1–8 · Quick reference

Calculus 1

Grades 11–12+6 pages

Limits and continuity, every derivative rule, rates and approximation, the Mean Value Theorem, curve sketching, optimization, Riemann sums, the Fundamental Theorem, substitution, area, volume and differential equations.

1Limits2Continuity3Derivative4Rules5Chain & implicit6In context7MVT & tests8Sketching9Optimization10Antiderivatives11Riemann sums12FTC & u-sub13Applications14Diff. equations

1Limits in Depthfirst look (tables, graphs, factoring): Pre-Calc §12

Limit laws

If lim f = L and lim g = M (x → c), the limit of a sum, difference, product, constant multiple, power or root is that combination of L and M; for a quotient, L/M needs M ≠ 0.

Polynomials, and rational functions where defined: just substitute.

Read the form
Substitution gives	Meaning	Next step
a number	that is the limit	done
00	indeterminate	algebra, squeeze, L’Hôpital
k0, k ≠ 0	infinite limit (VA)	check the sign on each side
∞∞	indeterminate	divide by top power

Plug, Peek, Pick

Plug in c. Peek at the form you get. Pick the tool that form calls for: algebra, squeeze, top power or L’Hôpital.

Infinite limits Worked

lim_x→2⁺ x + 1x − 2 = +∞top 3, bottom 0⁺

lim_x→2⁻ x + 1x − 2 = −∞bottom 0⁻: x = 2 is a VA

Squeeze theorem

The bounds only need to hold near c.

g ≤ f ≤ h and lim g = lim h = L ⇒ lim f = L

−x² ≤ x² cos(1/x) ≤ x², so the middle graph is squeezed to 0 at x = 0.

Squeeze Worked

−1 ≤ cos(1/x) ≤ 1 → −x² ≤ x² cos(1/x) ≤ x²

both bounds → 0, so lim_x→0 x² cos(1/x) = 0

Trig limits (x in radians)

lim_x→0 sin xx = 1, lim_x→0 1 − cos xx = 0

EXlim_x→0 sin 5x3x = 53 · lim sin 5x5x = 53

At −∞ with a root Worked

limx→−∞ √(4x² + 1)x − 3
√(x²) = |x| = −x for x < 0watch the sign
= lim −√(4 + 1/x²)1 − 3/x = −2

∞ − ∞: use the conjugate Worked

lim_x→∞ (√(x² + 6x) − x) = lim 6x√(x² + 6x) + x

= lim 6√(1 + 6/x) + 1 = 3

Precise definition

lim_x→c f(x) = L means: for every ε > 0 there is a δ > 0 with |f(x) − L| < ε whenever 0 < |x − c| < δ.

EXlim_x→2 (3x − 1) = 5: |3x − 6| = 3|x − 2| < ε when δ = ε/3

2Continuity & the Intermediate Value Theoremno breaks, no jumps, no holes

Continuous at x = c

f(c) is defined, lim_x→c f(x) exists, and they are equal (Pre-Calc §12: D-E-E).

Polynomials, eˣ, sin x, cos x: continuous everywhere.
Rational, root, log, tan: continuous on their domains.
Sums, products, quotients (denominator ≠ 0) and compositions of continuous functions are continuous.

Fix a removable break Worked

g(x) = x² − 1x − 1 = x + 1 for x ≠ 1

define g(1) = 2the limit

Draw it in one stroke

If you can trace the graph over an interval without lifting the pencil, f is continuous there. Each break type above is a lift.

Intermediate Value Theorem

f continuous on [a, b] and N between f(a) and f(b) ⇒

f(c) = N for some c in (a, b)

Extreme Value Theorem

f continuous on a closed interval [a, b] has an absolute maximum and minimum there (found in §7).

Choose k for continuity Worked

f(x) = kx² for x ≤ 2; 3x + k for x > 2
match at 2: 4k = 6 + kleft value = right limit
k = 2

A root of cos x = x Worked

h(x) = cos x − x: h(0) = 1, h(1) ≈ −0.4597continuous, signs differ

a root lies in (0, 1); it is ≈ 0.7391

On a closed interval

Continuous on [a, b]: on (a, b), from the right at a, from the left at b.

Calculus 1Calculus · Units 1–8 · continued

3The Derivativeslope of the tangent, rate of change right now (intro: Pre-Calc §13)

Definition

f′(x) = lim_h→0 f(x + h) − f(x)h

f′(a) = lim_x→a f(x) − f(a)x − a

Notation: f′(x), y′, dydx, ddx f(x)
Second: f″(x), d²ydx²: the rate of change of f′
Units: output units per input unit (m/s, $ per item)

f(x) = √x Worked

√(x + h) − √xh · √(x + h) + √x√(x + h) + √x = 1√(x + h) + √xconjugate

h → 0: f′(x) = 12√x

f(x) = 1/x Worked

1/(x + h) − 1/xh = −1x(x + h)combine

f′(x) = −1x²

Tangent & normal lines

y − f(a) = f′(a)(x − a). The normal line has slope −1/f′(a).

EXf(x) = √x at a = 4: f′(4) = ¼ → y = ¼x + 1

f(x) = x³ − 3x (solid) and f′(x) = 3x² − 3 (dashed): f′ = 0 at f's flat points, f′ > 0 where f rises.

Differentiable vs. continuous

Differentiable at c ⇒ continuous at c, but not the reverse. No derivative at:

a corner: |x| at 0
a cusp: x^2/3 at 0
a vertical tangent: ∛x at 0
any break in the graph

Make a piecewise function differentiable Worked

f(x) = x² for x ≤ 1; ax + b for x > 1
continuous: 1 = a + b; equal slopes: 2 = avalue, then slope
a = 2, b = −1

Estimate from a table Worked

t: 0, 2, 4, 6 · T: 70, 74, 80, 88

T′(4) ≈ 88 − 746 − 2 = 3.5 per unit of tsymmetric difference

Meaning with units Worked

C(q) = cost in dollars of q items; C′(100) = 4.5

the 101st item costs about $4.50 moredollars per item

Three names, one number

Slope of the tangent, instantaneous rate, derivative: three names for the same number f′(a).

4Derivative Rulesc, n and a are constants

Core rules
Rule	Derivative
constant	(c)′ = 0
power	(xⁿ)′ = nxⁿ⁻¹ (any real n)
multiple, sum	(cf)′ = cf′; (f ± g)′ = f′ ± g′
product	(fg)′ = f′g + fg′
quotient	(fg)′ = f′g − fg′g²
chain	(f(g(x)))′ = f′(g(x)) · g′(x)

Low d-high minus high d-low (traditional)

For highlow: low times d(high) minus high times d(low), all over low squared. The minus makes order matter.

Power rule Worked

ddx(4x³ − 2x + 5√x)write as powers

= 12x² + 2x² + 52√x

Product Worked

ddx(x² sin x) = 2x sin x + x² cos x

Trigonometric
f	f′	f	f′
sin x	cos x	cos x	−sin x
tan x	sec² x	cot x	−csc² x
sec x	sec x tan x	csc x	−csc x cot x

Co- means minus

The derivatives of the co-functions (cos, cot, csc) all start with a minus sign.

Exponential & log
f	f′
eˣ	eˣ
aˣ	aˣ ln a
ln x	1x (ln\|x\| too)
log_a x	1x ln a

Inverse trig
f	f′
sin⁻¹ x	1√(1 − x²)
cos⁻¹ x	−1√(1 − x²)
tan⁻¹ x	11 + x²
sec⁻¹ x	1\|x\|√(x² − 1)

Quotient Worked

ddx x + 1x² + 1 = (x² + 1) − (x + 1)(2x)(x² + 1)²

= 1 − 2x − x²(x² + 1)²

Exponential, log, inverse trig Worked

ddx(eˣ ln x) = eˣ ln x + eˣxproduct
ddx 3ˣ = 3ˣ ln 3
ddx tan⁻¹(2x) = 21 + 4x²chain

Rules together Worked

ddx x³e^2x = 3x²e^2x + 2x³e^2x = x²e^2x(3 + 2x)product + chain

ddx √(x² + 9) = x√(x² + 9); at x = 4: 45

Higher derivatives

Differentiate again: f(x) = x⁴ → f″ = 12x², f‴ = 24x.

sin x → cos x → −sin x → −cos x → sin x: every 4th derivative repeats.

Calculus 1Calculus · Units 1–8 · continued

5Chain Rule, Implicit & Inverse Derivativesfunctions inside functions, and curves that are not functions

Chain rule

For y = f(u) with u = g(x).

dydx = dydu · dudx

DOT

Derivative of the Outside (inside left alone), Times the derivative of the inside. Repeat for each layer.

Chain rule Worked

ddx(3x² + 1)⁵ = 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴
ddx sin³(2x) = 6 sin²(2x) cos(2x)three layers
ddx e−x² = −2x e−x²

Chain rule from a table Worked

g(1) = 2, g′(1) = 3, f′(2) = −4; h = f ∘ g

h′(1) = f′(g(1)) · g′(1) = −4 · 3 = −12

Implicit differentiation

Differentiate both sides with respect to x.
Each y-term gets a factor dydx (chain rule).
Collect the dydx terms and solve.

Circle x² + y² = 25 Worked

2x + 2y y′ = 0 → y′ = −xy

at (3, 4): slope −34

x³ + y³ = 6xy at (3, 3) Worked

3x² + 3y²y′ = 6y + 6xy′product rule on xy
y′ = 2y − x²y² − 2x = 6 − 99 − 6 = −1
tangent line: y = −x + 6

Logarithmic differentiation Worked

y = xˣ → ln y = x ln xtake ln first

y′y = ln x + 1 → y′ = xˣ(ln x + 1)

Derivative of an inverse

where f(a) = b: flip the slope at the matching point.

(f⁻¹)′(b) = 1f′(a)

EXf(x) = x³ + x, f(1) = 2: (f⁻¹)′(2) = 1f′(1) = 14

Inverse sine, implicitly Worked

y = sin⁻¹ x → sin y = x

cos y · y′ = 1 → y′ = 1cos y = 1√(1 − x²)cos y = √(1 − sin² y) ≥ 0

Second derivative, implicitly Worked

x² + y² = 25: y′ = −x/y
y″ = −y − x y′y² = −x² + y²y³ = −25y³substitute y′
at (3, 4): y″ = −2564

6Derivatives in Contextmotion, related rates, approximation and L’Hôpital

Motion on a line
Quantity	Meaning
s(t)	position
v(t) = s′(t)	velocity (signed); speed = \|v\|
a(t) = v′(t) = s″(t)	acceleration
v = 0	at rest; may turn around

s(t) = t³ − 6t² + 9t Worked

v = 3t² − 12t + 9 = 3(t − 1)(t − 3)
at rest at t = 1, 3; a = 6t − 12
t = 0.5: v = 3.75, a = −9 → slowing downsigns differ

Same Signs Speed up

If v and a have the same sign, the object speeds up; opposite signs, it slows down.

Related rates

Draw and label; mark what changes.
Write an equation linking the quantities.
Differentiate with respect to t.
Only then substitute the known values; solve.

Dogs Eat Dinner, Sleep Soundly

Draw, Equation, Differentiate, Substitute, Solve, in that order. Substituting before differentiating freezes the rates at 0.

x² + y² = 10² at every instant.

Sliding ladder Worked

10 ft ladder; foot slides out at 2 ft/s; foot 6 ft out
x² + y² = 100 → y = 8; 2x x′ + 2y y′ = 0
y′ = −6 · 28 = −1.5 ft/s (top slides down)

Filling a cone Worked

cone tip down, radius 3 ft, height 6 ft: r = h/2
V = π12h³ → V′ = π4h² h′
V′ = 2 ft³/min, h = 4: h′ = 12π ≈ 0.159 ft/min

Linear approximation

Near a, f(x) ≈ L(x). Differential: dy = f′(x) dx.

L(x) = f(a) + f′(a)(x − a)

Approximate Worked

√4.1 ≈ 2 + ¼(0.1) = 2.025true value 2.02485

sphere r = 10 cm ± 0.1: dV = 4πr² dr ≈ ±125.7 cm³

L’Hôpital’s rule

Only for 00 or ∞∞ (rewrite 0 · ∞, ∞ − ∞ or 1^∞ first); differentiate top and bottom separately.

lim fg = lim f′g′

L’Hôpital Worked

limx→0 eˣ − 1 − xx² = lim eˣ − 12x = lim eˣ2 = 12twice
limx→0⁺ x ln x = lim ln x1/x = lim (−x) = 00 · ∞ → ∞/∞
limx→∞ (1 + 3x)ˣ = e³take ln: 1^∞ → 0/0

Calculus 1Calculus · Units 1–8 · continued

7Mean Value Theorem, Extrema & Derivative Testswhat f′ and f″ say about f

Mean Value Theorem

f continuous on [a, b] and differentiable on (a, b) ⇒ for some c in (a, b):

f′(c) = f(b) − f(a)b − a

Rolle's theorem

Same conditions plus f(a) = f(b): then f′(c) = 0 for some c in (a, b).

EXx³ − 4x on [−2, 2]: 3c² − 4 = 0 → c = ±2√3 ≈ ±1.155

f(x) = x² on [1, 4]: secant slope 5; the tangent at c = 2.5 is parallel.

The speedometer moment

Drive 120 miles in 2 hours and at some instant the speedometer reads exactly 60 mph. That instant is the c of the MVT.

Critical number: c in the domain with f′(c) = 0 or f′(c) undefined
Local extremum: highest or lowest nearby (only at critical numbers)
Absolute extremum: highest or lowest on the whole interval

Closed-interval method Worked

f(x) = x³ − 3x² + 1 on [−1, 4]: f′ = 3x(x − 2)
f(−1) = −3, f(0) = 1, f(2) = −3, f(4) = 17critical numbers and ends
abs. max 17 at x = 4; abs. min −3 at x = −1 and 2

First derivative test

f′ changes + to −: local max
f′ changes − to +: local min
no sign change: neither

Concavity & inflection

f″ > 0: concave up (holds water)
f″ < 0: concave down
Inflection point: concavity changes (f″ changes sign).

Second derivative test

At f′(c) = 0: f″(c) > 0 → local min; f″(c) < 0 → local max; f″(c) = 0 → no conclusion (use the first test).

Both tests on f(x) = x³ − 12x Worked

f′ = 3(x² − 4): critical ±2; rises on (−∞, −2) and (2, ∞)
f″ = 6x: f″(−2) < 0 → local max (−2, 16)
f″(2) > 0 → local min (2, −16); inflection (0, 0)

Plus smiles, minus frowns

f″ positive bends up like a smile, so a flat spot there is a minimum; f″ negative frowns, giving a maximum.

8Curve Sketchingput every clue on one graph

Do I See All Features, Sir?

Domain, Intercepts, Symmetry, Asymptotes, First derivative (rise, fall, extrema), Second derivative (concavity, inflection). Then plot.

f(x) = x⁴ − 4x³ Worked

domain all reals; x-intercepts 0, 4
f′ = 4x²(x − 3): critical 0, 3
f″ = 12x(x − 2): f″ = 0 at 0, 2
local min (3, −27); inflection (0, 0), (2, −16)x = 0 is not an extremum

Sign chart
Interval	f′	f″	Graph
(−∞, 0)	−	+	falls, cup up
(0, 2)	−	−	falls, cup down
(2, 3)	−	+	falls, cup up
(3, ∞)	+	+	rises, cup up

f(x) = x⁴ − 4x³: flat but not turning at 0; minimum at 3.

Reading the graph of f′

f′ above the axis ↔ f rising; below ↔ f falling
f′ crosses zero ↔ f has a local max or min
f′ has a peak or valley ↔ f has an inflection point

With asymptotes: f(x) = x²x² − 1 Worked

VA x = ±1; HA y = 1; even function
f′ = −2x(x² − 1)²: rises for x < 0, falls for x > 0 → local max (0, 0)
f″ = 6x² + 2(x² − 1)³: cup up for |x| > 1, down for |x| < 1; no inflection

f(x) = x²/(x² − 1): the dashed lines are the asymptotes.

Calculus 1Calculus · Units 1–8 · continued

9Optimizationthe best size, shape or price

SOCKS

Sketch, write the Objective, write the Constraint, Kill a variable with it, Solve f′ = 0 and check ends or f″. Answer with units.

Fence along a river Worked

600 m of fence, three sides: A = x(600 − 2x)
A′ = 600 − 4x = 0 → x = 150A″ = −4 < 0: max
150 m × 300 m, area 45,000 m²

Open box from a 12 in square Worked

cut x-inch corners: V = x(12 − 2x)², 0 < x < 6
V′ = (12 − 2x)(12 − 6x) = 0 → x = 2
V = 2 · 8² = 128 in³

Box volume V(x): the peak is the answer.

Best price Worked

q items sell at p = 40 − 0.5q dollars
R = q(40 − 0.5q); R′ = 40 − q = 0 → q = 40
price $20, revenue $800R″ = −1 < 0

Cheapest can, 500 cm³ Worked

S = 2πr² + 2πrh, h = 500πr² → S = 2πr² + 1000r
S′ = 4πr − 1000r² = 0 → r = ∛(250π) ≈ 4.30 cm
h = 2r ≈ 8.60 cmheight = diameter

Closest point on y = √x to (3, 0) Worked

minimize D = (x − 3)² + xsquare of distance
D′ = 2(x − 3) + 1 = 0 → x = 2.5
point (2.5, √2.5) ≈ (2.5, 1.58)

10Antiderivativesundo the derivative

Indefinite integral

F is an antiderivative of f if F′ = f. Then ∫ f(x) dx = F(x) + C: every antiderivative differs by a constant.

Initial value problem Worked

f′(x) = 3x² + 2, f(1) = 5
f = x³ + 2x + C; 1 + 2 + C = 5 → C = 2
f(x) = x³ + 2x + 2

From acceleration Worked

a = −9.8 m/s², v(0) = 20, s(0) = 2

v = −9.8t + 20, s = −4.9t² + 20t + 2one constant per step

Basic antiderivatives
∫ f(x) dx	= F(x) + C
xⁿ (n ≠ −1)	xⁿ⁺¹n + 1
1x	ln\|x\|
eˣ; aˣ	eˣ; aˣln a
cos x; sin x	sin x; −cos x
sec² x; csc² x	tan x; −cot x
sec x tan x; csc x cot x	sec x; −csc x
11 + x²; 1√(1 − x²)	tan⁻¹ x; sin⁻¹ x

Up one, divide by it, add C

Power rule backwards: raise the power up one, divide by the new power, then + C. (It fails only for x⁻¹, which gives ln|x|.)

Term by term Worked

∫(6x² − 4x + 3eˣ) dx = 2x³ − 4 ln|x| + 3eˣ + C

∫(√x + 1x²) dx = 23x^3/2 − 1x + Cpowers ½ and −2

11Riemann Sums & the Definite Integralarea as a limit of rectangles

Riemann sums

Split [a, b] into n pieces of width Δx = b − an; add f(sample) · Δx. Samples: left ends, right ends or midpoints. Trapezoid = average of left and right.

Right-endpoint rectangles for x² on [0, 2], n = 4: R₄ = 3.75.

x² on [0, 2], n = 4, Δx = 0.5 Worked

L₄ = 0.5(0 + 0.25 + 1 + 2.25) = 1.75
R₄ = 0.5(0.25 + 1 + 2.25 + 4) = 3.75
M₄ = 2.625; T₄ = 2.75; exact 83 ≈ 2.667

Rising: left lags, right runs ahead

For an increasing f, left sums underestimate and right sums overestimate. Falling f: the reverse.

Trapezoid rule

T_n = Δx2[f(x₀) + 2f(x₁) + ⋯ + 2f(x_n−1) + f(x_n)]

EXx² on [0, 2], n = 4: 0.25[0 + 2(0.25 + 1 + 2.25) + 4] = 2.75

Definite integral

Net signed area: area above the x-axis minus area below.

∫_a^b f(x) dx = lim_n→∞ Σ f(xᵢ*) Δx

Properties

∫_a^a f = 0; ∫_b^a f = −∫_a^b f
∫_a^b f + ∫_b^c f = ∫_a^c f
∫(cf ± g) = c∫f ± ∫g

By geometry Worked

∫₋₃³ √(9 − x²) dx = half a circle of radius 3 = 9π2

∫₀⁴ (x − 2) dx = 0equal triangles above and below

Calculus 1Calculus · Units 1–8 · continued

12The Fundamental Theorem & Substitutionderivatives and integrals undo each other

FTC, part 1

With a function as the upper limit: ddx ∫_a^g(x) f(t) dt = f(g(x)) g′(x).

ddx ∫_a^x f(t) dt = f(x)

EXddx ∫₀^x² cos t dt = 2x cos(x²)

FTC, part 2

F is any antiderivative of f; the + C cancels.

∫_a^b f(x) dx = F(b) − F(a)

Evaluate Worked

∫13 (2x + 1) dx = [x² + x]13 = 12 − 2 = 10
∫0π sin x dx = [−cos x]0π = 2
∫1e 1x dx = 1

Net change

∫_a^b F′(x) dx = F(b) − F(a): integrating a rate gives the total change.

u-substitution

Let u = the inside function; du = u′(x) dx.
Rewrite everything in u (change the limits too).
Integrate, then return to x if indefinite.

Inside and its derivative

Look for a function inside another with its derivative (up to a constant) standing outside. That inside is u.

Substitution Worked

∫ 2x(x² + 1)⁴ dx = (x² + 1)⁵5 + Cu = x² + 1
∫ x ex² dx = ½ex² + Cdu = 2x dx
∫01 x√(1 − x²) dx = ½∫01 √u du = 13u = 1 − x², limits 1 → 0

13Applications of the Integralaverage value, motion, area and volume

Average value

Some c in [a, b] has f(c) = f_avg (MVT for integrals).

f_avg = 1b − a ∫_a^b f(x) dx

EXx² on [0, 3]: 13 · 9 = 3, reached at c = √3

Displacement vs distance Worked

v(t) = t² − 4t + 3 on [0, 3] (v < 0 on (1, 3))
displacement ∫03 v dt = 0
distance ∫03 |v| dt = 43 + 43 = 83split where v = 0

Area between curves

Or ∫ [right − left] dy. Find a and b where the curves meet.

A = ∫_a^b [top − bottom] dx

Region between y = 4 − x² and y = x + 2.

That region Worked

meet: 4 − x² = x + 2 → x = −2, 1

∫₋₂¹ (2 − x − x²) dx = 92top − bottom

Top minus bottom

Area: top curve minus bottom curve, left to right. Volume: square first, then subtract, πR² − πr², never π(R − r)².

Disks & washers (about the x-axis)

r = 0 gives a disk. R = outer radius, r = inner radius.

V = π∫_a^b [R(x)² − r(x)²] dx

y = √x on [0, 4] spun about the x-axis; each slice is a disk of radius R = √x.

Volumes Worked

disk: π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = 8π ≈ 25.13

washer, y = x over y = x² on [0, 1]: π∫₀¹ (x² − x⁴) dx = 2π15 ≈ 0.419

14Differential Equationsequations for an unknown function

Differential equation: relates y, its derivatives and x
General solution: a family, with + C
Particular solution: the one member fitting an initial condition

Separation of variables

Put every y on the dy side and every x on the dx side.
Integrate both sides (one + C is enough).
Solve for y; use the initial condition.

Slope field for y′ = x − y; the curve through (0, 1) is y = x − 1 + 2e^−x.

SIP

Separate, Integrate, Plug in the initial value. Only then is the answer one curve instead of a family.

Separate Worked

dydx = xy, y(0) = 3 → dyy = x dx

ln|y| = x²2 + C → y = 3e^x²/2y(0) = 3