Haroldβs Calculus Notes
Cheat Sheet
26 April 2016
AP Calculus
Limits
Definition of Limit
Let f be a function defined on an open
interval containing c and let L be a real
number. The statement:
lim π(π₯) = πΏ
π₯βπ
means that for each π > 0 there exists a
πΏ > 0 such that
if 0 < |π₯ β π| < πΏ, then |π(π₯) β πΏ| < π
Tip :
Direct substitution: Plug in π(π) and see if
it provides a legal answer. If so then L =
π(π).
The Existence of a Limit
The limit of π(π₯) as π₯ approaches a is L if
and only if:
lim π(π₯) = πΏ
π₯βπ β
lim π(π₯) = πΏ
π₯βπ +
π
Prove that π(π) = π β π is a continuous function.
Definition of Continuity
A function f is continuous at c if for
every π > 0 there exists a πΏ > 0 such that
|π₯ β π| < πΏ and |π(π₯) β π(π)| < π.
Tip: Rearrange |π(π₯) β π(π)| to have
|(π₯ β π)| as a factor. Since |π₯ β π| < πΏ we
can find an equation that relates both πΏ
and π together.
|π(π₯) β π(π)|
= |(π₯ 2 β 1) β (π 2 β 1)|
= |π₯ 2 β 1 β π 2 + 1|
= |π₯ 2 β π 2 |
= |(π₯ + π)(π₯ β π)|
= |(π₯ + π)| |(π₯ β π)|
Since |(π₯ + π)| β€ |2π|
|π(π₯) β π(π)| β€ |2π||(π₯ β π)| < π
π
So given π > 0, we can choose πΉ = |ππ| πΊ > π in the
Definition of Continuity. So substituting the chosen πΏ
for |(π₯ β π)| we get:
1
|π(π₯) β π(π)| β€ |2π| (| | π) = π
2π
Since both conditions are met, π(π₯) is continuous.
π ππ π₯
=1
π₯β0 π₯
πππ
Two Special Trig Limits
1 β πππ π₯
=0
π₯β0
π₯
πππ
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
1
Derivatives
Definition of a Derivative of a Function
Slope Function
Notation for Derivatives
0. The Chain Rule
1. The Constant Multiple Rule
2. The Sum and Difference Rule
3. The Product Rule
4. The Quotient Rule
5. The Constant Rule
6a. The Power Rule
6b. The General Power Rule
7. The Power Rule for x
8. Absolute Value
9. Natural Logorithm
10. Natural Exponential
11. Logorithm
12. Exponential
13. Sine
14. Cosine
15. Tangent
16. Cotangent
17. Secant
(See Larsonβs 1-pager of common derivatives)
π(π₯ + β) β π(π₯)
π β² (π₯) = lim
ββ0
β
π(π₯)
β
π(π)
π β² (π) = lim
π₯βπ
π₯βπ
ππ¦
π
π β² (π₯), π (π) (π₯), , π¦ β² , [π(π₯)], π·π₯ [π¦]
ππ₯
ππ₯
π
[π(π(π₯))] = π β² (π(π₯))πβ² (π₯)
ππ₯
ππ¦ ππ¦ ππ’
=
·
ππ₯ ππ’ ππ₯
π
[ππ(π₯)] = ππ β² (π₯)
ππ₯
π
[π(π₯) ± π(π₯)] = π β² (π₯) ± πβ² (π₯)
ππ₯
π
[ππ] = ππβ² + π π β²
ππ₯
π π
ππ β² β ππβ²
[ ]=
ππ₯ π
π2
π
[π] = 0
ππ₯
π π
[π₯ ] = ππ₯ πβ1
ππ₯
π π
[π’ ] = ππ’πβ1 π’β² π€βπππ π’ = π’(π₯)
ππ₯
π
[π₯] = 1 (π‘βπππ π₯ = π₯ 1 πππ π₯ 0 = 1)
ππ₯
π
π₯
[|π₯|] =
|π₯|
ππ₯
π
1
[ln π₯] =
ππ₯
π₯
π π₯
[π ] = π π₯
ππ₯
π
1
[log π π₯] =
ππ₯
(ln π) π₯
π π₯
[π ] = (ln π) π π₯
ππ₯
π
[π ππ(π₯)] = cos(π₯)
ππ₯
π
[πππ (π₯)] = βπ ππ(π₯)
ππ₯
π
[π‘ππ(π₯)] = π ππ 2(π₯)
ππ₯
π
[πππ‘(π₯)] = βππ π 2 (π₯)
ππ₯
π
[π ππ(π₯)] = π ππ(π₯) π‘ππ(π₯)
ππ₯
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
2
Derivatives
18. Cosecant
19. Arcsine
20. Arccosine
21. Arctangent
22. Arccotangent
23. Arcsecant
24. Arccosecant
25. Hyperbolic Sine
26. Hyperbolic Cosine
27. Hyperbolic Tangent
28. Hyperbolic Cotangent
29. Hyperbolic Secant
30. Hyperbolic Cosecant
31. Hyperbolic Arcsine
32. Hyperbolic Arccosine
33. Hyperbolic Arctangent
34. Hyperbolic Arccotangent
35. Hyperbolic Arcsecant
36. Hyperbolic Arccosecant
Position Function
Velocity Function
Acceleration Function
Jerk Function
(See Larsonβs 1-pager of common derivatives)
π
[ππ π(π₯)] = β ππ π(π₯) πππ‘(π₯)
ππ₯
π
1
[sinβ1 (π₯)] =
ππ₯
β1 β π₯ 2
π
β1
[cos β1(π₯)] =
ππ₯
β1 β π₯ 2
π
1
[tanβ1(π₯)] =
ππ₯
1 + π₯2
π
β1
[cot β1 (π₯)] =
ππ₯
1 + π₯2
π
1
[sec β1(π₯)] =
ππ₯
|π₯| βπ₯ 2 β 1
π
β1
[csc β1 (π₯)] =
ππ₯
|π₯| βπ₯ 2 β 1
π
[π ππβ(π₯)] = cosh(π₯)
ππ₯
π
[πππ β(π₯)] = π ππβ(π₯)
ππ₯
π
[π‘ππβ(π₯)] = π ππβ2(π₯)
ππ₯
π
[πππ‘β(π₯)] = βππ πβ2 (π₯)
ππ₯
π
[π ππβ(π₯)] = β π ππβ(π₯) π‘ππβ(π₯)
ππ₯
π
[ππ πβ(π₯)] = β ππ πβ(π₯) πππ‘β(π₯)
ππ₯
π
1
[sinhβ1(π₯)] =
ππ₯
βπ₯ 2 + 1
π
1
[coshβ1(π₯)] =
2
ππ₯
βπ₯ β 1
π
1
[tanhβ1 (π₯)] =
ππ₯
1 β π₯2
π
1
[cothβ1(π₯)] =
ππ₯
1 β π₯2
π
β1
[sechβ1(π₯)] =
ππ₯
π₯ β1 β π₯ 2
π
β1
[cschβ1(π₯)] =
ππ₯
|π₯| β1 + π₯ 2
1
π (π‘) = ππ‘ 2 + π£0 π‘ + π 0
2
π£(π‘) = π β² (π‘) = ππ‘ + π£0
π(π‘) = π£ β² (π‘) = π β²β² (π‘)
π(π‘) = πβ² (π‘) = π£ β²β² (π‘) = π (3) (π‘)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
3
Applications of Differentiation
Rolleβs Theorem
f is continuous on the closed interval [a,b],
and
f is differentiable on the open interval (a,b).
If f(a) = f(b), then there exists at least one number c in
(a,b) such that fβ(c) = 0.
π(π) β π(π)
πβπ
π(π) = π(π) + (π β π)πβ²(π)
Find βcβ.
π(π₯)
πΌπ lim π(π₯) = lim
=
π₯βπ
π₯βπ π(π₯)
Mean Value Theorem
If f meets the conditions of Rolleβs
Theorem, then
LβHôpitalβs Rule
π β² (π) =
0 β
{ , , 0 β’ β, 1β , 00 , β0 , β β β} , ππ’π‘ πππ‘ {0β },
0 β
π(π₯)
πβ² (π₯)
πβ²β² (π₯)
π‘βππ lim
= lim β²
= lim β²β²
=β―
π₯βπ π(π₯)
π₯βπ π (π₯)
π₯βπ π (π₯)
Graphing with Derivatives
Test for Increasing and Decreasing
Functions
The First Derivative Test
The Second Deriviative Test
Let f β(c)=0, and f β(x) exists, then
Test for Concavity
Points of Inflection
Change in concavity
Analyzing the Graph of a
Function
x-Intercepts (Zeros or Roots)
y-Intercept
Domain
Range
Continuity
Vertical Asymptotes (VA)
Horizontal Asymptotes (HA)
Infinite Limits at Infinity
Differentiability
Relative Extrema
Concavity
Points of Inflection
1. If f β(x) > 0, then f is increasing (slope up) β
2. If f β(x) < 0, then f is decreasing (slope down) β
3. If f β(x) = 0, then f is constant (zero slope) β
1. If f β(x) changes from β to + at c, then f has a relative
minimum at (c, f(c))
2. If f β(x) changes from + to - at c, then f has a relative
maximum at (c, f(c))
3. If f β(x), is + c + or - c -, then f(c) is neither
1. If f β(x) > 0, then f has a relative minimum at (c, f(c))
2. If f β(x) < 0, then f has a relative maximum at (c, f(c))
3. If f β(x) = 0, then the test fails (See 1π π‘ derivative test)
1. If f β(x) > 0 for all x, then the graph is concave up β
2. If f β(x) < 0 for all x, then the graph is concave down β
If (c, f(c)) is a point of inflection of f, then either
1. f β(c) = 0 or
2. f β does not exist at x = c.
(See Haroldβs Illegals and Graphing Rationals Cheat
Sheet)
f(x) = 0
f(0) = y
Valid x values
Valid y values
No division by 0, no negative square roots or logs
x = division by 0 or undefined
limβ π(π₯) β π¦ and lim+ π(π₯) β π¦
π₯ββ
π₯ββ
limβ π(π₯) β β and lim+ π(π₯) β β
π₯ββ
π₯ββ
Limit from both directions arrives at the same slope
Create a table with domains, f(x), f β(x), and f β(x)
If π β(π₯) β +, then cup up β
If π β(π₯) β β, then cup down β
f β(x) = 0 (concavity changes)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
4
Approximating with
Differentials
π(π₯π )
πβ²(π₯π )
π¦ = ππ₯ + π
π¦ = π β² (π)(π₯ β π) + π(π)
Newtonβs Method
Finds zeros of f, or finds c if f(c) = 0.
π₯π+1 = π₯π β
Tangent Line Approximations
Function Approximations with
Differentials
Related Rates
π(π₯ + βπ₯) β π(π₯) + ππ¦ = π(π₯) + π β² (π₯) ππ₯
Steps to solve:
1. Identify the known variables and rates of change.
π
(π₯ = 15 π; π¦ = 20 π; π₯ β² = 2 ; π¦ β² = ? )
π
2. Construct an equation relating these quantities.
(π₯ 2 + π¦ 2 = π 2 )
3. Differentiate both sides of the equation.
(2π₯π₯ β² + 2π¦π¦ β² = 0)
4. Solve for the desired rate of change.
π₯
(π¦ β² = β π₯ β² )
π¦
5. Substitute the known rates of change and quantities
into the equation.
15
3π
(π¦ β² = β β¦ 2 =
)
20
2 π
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
5
Summation Formulas
π
β π = ππ
π=1
π
βπ =
π=1
π
π(π + 1) π2 π
=
+
2
2 2
β π2 =
π=1
π
π(π + 1)(2π + 1) π3 π2 π
=
+ +
6
3
2 6
2
π
3
β π = (β π ) =
π=1
π
Sum of Powers
π=1
π2 (π + 1)2 π4 π3 π2
=
+ +
4
4
2
4
π(π + 1)(2π + 1)(3π2 + 3π β 1) π5 π4 π3 π
β π4 =
=
+ + β
30
5
2
3 30
π=1
π
β π5 =
π=1
π
β π6 =
π=1
π
β π7 =
π=1
π2 (π + 1)2 (2π2 + 2π β 1) π6 π5 5π4 π2
=
+ +
β
12
6
2
12 12
π(π + 1)(2π + 1)(3π4 + 6π3 β 3π + 1)
42
π2 (π + 1)2 (3π4 + 6π3 β π2 β 4π + 2)
24
π
πβ1
(π + 1)π+1
1
π+1
ππ (π) = β π =
β
β(
) ππ (π)
π+1
π+1
π
π
π=1
π
π=0
π
π
2
β π(π + 1) = β π + β π =
π=1
π
Misc. Summation Formulas
π=1
π=1
π(π + 1)(π + 2)
3
1
π
β
=
π(π + 1) π + 1
π=1
π
β
π=1
1
π(π + 3)
=
π(π + 1)(π + 2) 4(π + 1)(π + 2)
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
6
Numerical Methods
π
π
π0 (π₯) = β« π(π₯) ππ₯ = lim β π(π₯πβ ) βπ₯π
βπββ0
π
π=1
where π = π₯0 < π₯1 < π₯2 < β― < π₯π = π
and βπ₯π = π₯π β π₯πβ1
and βπβ = πππ₯{βπ₯π }
Types:
ο· Left Sum (LHS)
ο· Middle Sum (MHS)
ο· Right Sum (RHS)
Riemann Sum
π
π
π0 (π₯) = β« π(π₯) ππ₯ β β π(π₯Μ
π ) βπ₯ =
π
π=1
βπ₯[π(π₯Μ
1 ) + π(π₯Μ
2 ) + π(π₯Μ
3 ) + β― + π(π₯Μ
π )]
πβπ
where βπ₯ =
Midpoint Rule
π
1
and π₯Μ
π = 2 (π₯πβ1 + π₯π ) = ππππππππ‘ ππ [π₯πβ1 , π₯π ]
Error Bounds: |πΈπ | β€
π
πΎ(πβπ)3
24π2
π1 (π₯) = β« π(π₯) ππ₯ β
π
βπ₯
[π(π₯0 ) + 2π(π₯1 ) + 2π(π₯3 ) + β― + 2π(π₯πβ1 )
2
+ π(π₯π )]
πβπ
where βπ₯ =
π
and π₯π = π + πβπ₯
Trapezoidal Rule
Error Bounds: |πΈπ | β€
π
πΎ(πβπ)3
12π2
π2 (π₯) = β« π(π₯)ππ₯ β
π
Simpsonβs Rule
βπ₯
[π(π₯0 ) + 4π(π₯1 ) + 2π(π₯2 ) + 4π(π₯3 ) + β―
3
+ 2π(π₯πβ2 ) + 4π(π₯πβ1 ) + π(π₯π )]
Where n is even
πβπ
and βπ₯ = π
and π₯π = π + πβπ₯
πΎ(πβπ)5
Error Bounds: |πΈπ | β€ 180π4
[MATH] fnInt(f(x),x,a,b), [MATH] [1] [ENTER]
TI-84 Plus
TI-Nspire CAS
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
Example: [MATH] fnInt(x^2,x,0,1)
1
1
β« π₯ 2 ππ₯ =
3
0
[MENU] [4] Calculus [3] Integral
[TAB] [TAB]
[X] [^] [2] [TAB]
[TAB] [X] [ENTER]
7
Integration
(See Haroldβs Fundamental Theorem of Calculus Cheat Sheet)
β« π β² (π₯) ππ₯ = π(π₯) + πΆ
Basic Integration Rules
Integration is the βinverseβ of
differentiation, and vice versa.
π
β« π(π₯) ππ₯ = π(π₯)
ππ₯
π(π₯) = 0
β« 0 ππ₯ = πΆ
π(π₯) = π = ππ₯ 0
β« π ππ₯ = ππ₯ + πΆ
1. The Constant Multiple Rule
2. The Sum and Difference Rule
The Power Rule
π(π₯) = ππ₯ π
β« π π(π₯) ππ₯ = π β« π(π₯) ππ₯
β«[π(π₯) ± π(π₯)] ππ₯ = β« π(π₯) ππ₯ ± β« π(π₯) ππ₯
β« π₯ π ππ₯ =
π₯ π+1
+ πΆ, π€βπππ π β β1
π+1
πΌπ π = β1, π‘βππ β« π₯ β1 ππ₯ = ln|π₯| + πΆ
π
The General Power Rule
If π’ = π(π₯), πππ π’β² = π(π₯) then
ππ₯
π’π+1
π β²
β« π’ π’ ππ₯ =
+ πΆ, π€βπππ π β β1
π+1
π
Reimann Sum
β π(ππ ) βπ₯π ,
π€βπππ π₯πβ1 β€ ππ β€ π₯π
π=1
βββ = βπ₯ =
π
Definition of a Definite Integral
Area under curve
π
lim β π(ππ ) βπ₯π = β« π(π₯) ππ₯
ββββ0
π
Swap Bounds
Additive Interval Property
πβπ
π
π
π=1
π
β« π(π₯) ππ₯ = β β« π(π₯) ππ₯
π
π
π
π
π
β« π(π₯) ππ₯ = β« π(π₯) ππ₯ + β« π(π₯) ππ₯
π
π
π
The Fundamental Theorem of
Calculus
π
β« π(π₯) ππ₯ = πΉ(π) β πΉ(π)
π
π₯
π
β« π(π‘) ππ‘ = π(π₯)
ππ₯
π(π₯)
π
π
β« π(π‘) ππ‘ = π(π(π₯))πβ² (π₯)
ππ₯
The Second Fundamental
Theorem of Calculus
π
β(π₯)
π
β« π(π‘) ππ‘ = π(β(π₯))ββ² (π₯) β π(π(π₯))πβ²(π₯)
ππ₯
π(π₯)
Mean Value Theorem for
Integrals
The Average Value for a
Function
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
π
β« π(π₯) ππ₯ = π(π)(π β π) Find βπβ.
π
π
1
β« π(π₯) ππ₯
πβπ π
8
Integration Methods
1. Memorized
See Larsonβs 1-pager of common integrals
β« π(π(π₯))πβ² (π₯)ππ₯ = πΉ(π(π₯)) + πΆ
Set π’ = π(π₯), then ππ’ = πβ² (π₯) ππ₯
2. U-Substitution
β« π(π’) ππ’ = πΉ(π’) + πΆ
π’ = _____
ππ’ = _____ ππ₯
β« π’ ππ£ = π’π£ β β« π£ ππ’
π’ = _____
ππ’ = _____
3. Integration by Parts
π£ = _____
ππ£ = _____
Pick βπ’β using the LIATED Rule:
L β Logarithmic : ln π₯ , log π π₯ , ππ‘π.
I β Inverse Trig.:
tanβ1 π₯ , sec β1 π₯ , ππ‘π.
A β Algebraic:
π₯ 2 , 3π₯ 60 , ππ‘π.
T β Trigonometric: sin π₯ , tan π₯ , ππ‘π.
E β Exponential:
π π₯ , 19π₯ , ππ‘π.
D β Derivative of: ππ¦βππ₯
π(π₯)
β«
ππ₯
π(π₯)
where π(π₯) πππ π(π₯) are polynomials
4. Partial Fractions
Case 1: If degree of π(π₯) β₯ π(π₯)
then do long division first
Case 2: If degree of π(π₯) < π(π₯)
then do partial fraction expansion
β« βπ2 β π₯ 2 ππ₯
Substutution: π₯ = π sin π
Identity: 1 β π ππ2 π = πππ 2 π
5a. Trig Substitution for βππ β ππ
β« βπ₯ 2 β π2 ππ₯
Substutution: π₯ = π sec π
Identity: π ππ 2 π β 1 = π‘ππ2 π
5b. Trig Substitution for βππ β ππ
β« βπ₯ 2 + π2 ππ₯
5c. Trig Substitution for βππ + ππ
6. Table of Integrals
7. Computer Algebra Systems (CAS)
8. Numerical Methods
9. WolframAlpha
Substutution: π₯ = π tan π
Identity: π‘ππ2 π + 1 = π ππ 2 π
CRC Standard Mathematical Tables book
TI-Nspire CX CAS Graphing Calculator
TI βNspire CAS iPad app
Riemann Sum, Midpoint Rule, Trapezoidal Rule, Simpsonβs
Rule, TI-84
Google of mathematics. Shows steps. Free.
www.wolframalpha.com
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
9
Partial Fractions
Condition
Example Expansion
Typical Solution
(See Haroldβs Partial Fractions Cheat Sheet)
π(π₯)
π(π₯) =
π(π₯)
where π(π₯) πππ π(π₯) are polynomials
and degree of π(π₯) < π(π₯)
If degree of π(π₯) β₯ π(π₯) π‘βππ ππ ππππ πππ£ππ πππ ππππ π‘
π(π₯)
(ππ₯ + π)(ππ₯ + π)2 (ππ₯ 2 + ππ₯ + π)
π΄
π΅
πΆ
π·π₯ + πΈ
=
+
+
+
2
2
(ππ₯ + π) (ππ₯ + π) (ππ₯ + π)
(ππ₯ + ππ₯ + π)
π
β«
ππ₯ = π ππ|π₯ + π| + πΆ
π₯+π
Sequences & Series
(See Haroldβs Series Cheat Sheet)
lim ππ = πΏ (Limit)
πββ
Sequence
Example: (ππ , ππ+1 , ππ+2 , β¦)
π(1 β π π )
π
=
πββ
1βπ
1βπ
π = lim
Geometric Series
Convergence Tests
Series Convergence Tests
Taylor Series
only if |π| < 1
where π is the radius of convergence
and (βπ, π) is the interval of convergence
(See Haroldβs Series Convergence Tests Cheat Sheet)
1. Divergence or ππ‘β Term
6. Ratio
2. Geometric Series
7. Root
3. p-Series
8. Direct Comparison
4. Alternating Series
9. Limit Comparison
5. Integral
10. Telescoping
(See Haroldβs Taylor Series Cheat Sheet)
π(π₯) = ππ (π₯) + π
π (π₯)
+β
Taylor Series
=β
π=0
π (π) (π)
π (π+1) (π₯ β )
(π₯ β π)π +
(π₯ β π)π+1
π!
(π + 1)!
where π₯ β€ π₯ β β€ π (worst case scenario π₯ β )
and lim π
π (π₯) = 0
π₯β+β
Copyright © 2015-2016 by Harold Toomey, WyzAnt Tutor
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