Limit
Rules of differentiation
lim[ f ( x) ⋅ g ( x)] = lim f ( x) ⋅ lim g ( x) = L1 ⋅ L2 ;
x→a
x→a
x→a
f ( x) L
f ( x) lim
lim
= x→a
= 1
x→ a g ( x )
lim g ( x) L2
L2 ≠ 0
d u
( )=
dx v
x→a
∞
xr
x2 x3
 x
e = lim1 +  = 1 + x + + +  = ∑
n →∞
2 ! 3!
 n
r =0 r !
n
x
n
 1
lim1 +  = e
n →∞
 n
lim
θ →0
sin θ
θ
ex − 1
=1
= 1 lim
x →0
x
First principal
f ' ( x) = lim
∆x → 0
f (x + ∆x ) − f (x )
∆x
n!
r!(n − r )!
C =C
n
r
Prn = (n − r )!
C +C
n
r
n
n−r
n
r +1
=C
n +1
r +1
( x + y) =
n
x n + C1n x n −1 y + C2n x n − 2 y 2 + ... + Cnn−1 xy n −1 + y n
S1: General term;
S2, Solve for r
Trigonometry
sin (A ± B) = sin A cos B ± cos A sin B
cos (A ± B) = cos A cos B  sin A sin B
2 sin A cos B = sin (A + B) + sin (A − B)
2 cos A cos B = cos (A + B) + cos (A − B)
2 sin A sin B = cos (A − B) − cos (A + B)
tan (A ± B) =
sin 2 x =
tan A ± tan B
1  tan A tan B
1 − cos 2 x
1 + cos 2 x
cos 2 x =
2
2
tan 2 x + 1 = sec 2 x
cot 2 x + 1 = csc 2 x
v
S1: Set f’(x)=0; => x=1,5,or 7
First D Test
x
x<1
x=1
1<x<5
x=5
5<x<7
x=7
7<x
dy/dx
+ve
0
-ve
0
+ve
0
+ve
Maximum value: f(1) Maximum point: (1,f(1))
Minimum value: f(5) Maximum point: (5,f(5))
Stationary point:(7,f(7))
du
dv
−u
dx
dx
v2
Chain Rule
dy dy du
=
⋅
dx du dx
-----Differentiation --->
Normal : y − y1 = −
1
( x − x1 )
f ′( x1 )
d n
( x ) = nx n −1
dx
Tangent : y − y1 = f ' ( x)( x − x1 )
d
(sin x) = cos x
dx
Second D Test
f”(xo)<0
f”(xo)>0
d
(sec x) = sec x tan x
dx
d
(cos x) = − sin x
dx
d
(cot x) = − csc 2 x
dx
d
(csc x) = − csc x cot x
dx
d x
(e ) = e x
dx
d
(ln sec x + tan x ) = sec x
dx
<--- Integration -----
dv d 2 s
ds
a=
=
dt dt 2
dt
S1: formulate equation i.e.
∫ udv = uv − ∫ vdu
v=
dA d (4πr 2 ) dr
=
dt
dr
dt
S3: Substitutes and solve.
i.e.
Application of Integration
b
;
Area=
a
dv
du
∫ u dx dx = uv − ∫ v dx dx
b
xe − > d (e );
x cos x − > d (sin x)
x
∫ (high − low)dx
x
∫ (2πR) Hdy
Shell method:
a
x2
);
2
e x cos x − > d (e x );
x ln x − > d (
b
b
a
a
Vol = ∫ 2πxydx or ∫ 2πxydy
*Use by part X2 or Recurrent
(vertical->dx; horizontal ->dy)
Integration Trigo-Sub
S1’ set f’(x)=0 => x= xo
S2’check f”(xo)
If f”(xo)<0 , (xo, f”(xo)) is a max. pt.
If f”(xo)>0 , (xo, f”(xo)) is a min. pt.
a2 − x2
Point of Inflexion
∫
∫
∫
f”(xo) changes signs
Assymptotes
e.g. y = 2 x − 3 +
2
[x + 5]
Vertical: x = −5
Horizontal/Oblique: y = 2 x − 3
put x = a sin 
a 2 + x 2 , put x = a tan
∫
d
1
(ln x) =
dx
x
d
(ln sec x ) = tan x
dx
Integration by parts
Integration
d x
a = a x ln a
dx
1
d
log a x =
dx
x ln a
Rate of Change,
A = 4πr 2
S2: apply chain rule to diff w.r.t t.
d
(tan x) = sec 2 x
dx
The Binomial Theorem
C rn =
d
dv
du
(uv) = u
+v
dx
dx
dx
Application of Differentiation
Tangent and Normal
∫
∫
∫
∫
a
a
f ( x)dx = ∫ f (t )dt ,
a
x 2 − a 2 , put x = a sec
a
a
a
f ( x)dx = − ∫ f ( x)dx ,
a
a
b
kf ( x)dx = k ∫ f ( x)dx ,
a
b
a
[ f ( x) ± g ( x)]dx
b
a
b
a
a
−a
∫ πR
2
dh
a
b
b
b
Disc method:
f ( x)dx = 0 ,
a
b

b
f ( x)dx ± ∫ g ( x)dx
a
c
b
a
c
f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx
2 a f ( x ) dx even
f ( x )dx =  ∫ 0
0
odd
M.I.
When n=1,
…
It is true for n=1.
Assume the statement is true for
n=k
i.e. …
When n=k+1
…
It is true for n=k+1
By the principle of mathematical
induction, ….
b
Vol = ∫ πx dy
b
2
a
or
Vol = ∫ πy 2 dx
a
Matrix
Properties of Matrices Operation
Properties of Determinants
det AT = det A.
A0 = 0A = 0
det AB = (det A)(det B).
a11 a12
a21 a22
a13
a23 = a 11
A(BC) = (AB)C
a31
a32
a33
A(B ± C) = AB ± AC
a21
a22
a31
a32
(αA)(βB) = A(αβ)B = (αβ)AB
(A + B)T = AT + BT
(AB)T = BTAT .
 A11

1
A =
adj  A12
A
A
 13
A21
A22
A23
−1
A31 

A32 
A33 
n -1
(b)(AB)-1 = B-1A-1 ,
-1
A)
-1 n
(e) (A ) = (A )
Area of ABCD = AB × AD
1
AB × AD
2
Volume of Parallel pine = (b × c) • a
Area of ABD=
1
(b × c) • a
2
1
Volume of tetrahedron = (b × c ) • a
6
Pyramid with parallelogram base =
1
(b × c) • a
3
a33
2.
b1
b2
b3
c1
a1 b1
c2 = − a3 b3
c3
a2 b2
b1
b2
kb3
c1
c2
kc3
b3
c3
a1 + ka2
b1 + kb2
c1 + kc2
c3 =
a2
b2
c2
c2
a3
b3
c3
c1
a1
a1
a3
b1
b1
b3
c1
x y
c2 + a2 b2
z
c2
c3
c3
and
a3
b3
a3
b3
c1
c1 = 0
c3
(3) a ⋅ b = a ⋅ c ⇒ b = c
Or | OP | = x 2 + y 2 + z 2
| OP |= x 2 + y 2
If OP ≠ 0, sin θ =
cos θ =
4.
x
x2 + y2
y
x + y2
2
,
tan θ =
and
and
then a ⋅
b = x2i + y2j + z2k ,
b = x1x2 + y1y2 + z1z2
cos θ =
y
,
x
The unit vector in the direction of a is
x1 x 2 + y 1 y 2 + z 1z 2
a⋅b
=
2
a b
x1 + y 1 2 + z1 2 ⋅ x 2 2 + y 2 2 + z 2 2
If a ⋅ b = 0 ,a is perpendicular to b.
a
.
Vector Products
a
a × b=|a||b|sin θ n, where θ is the angle between a and b and n
 a11x + a12 y + a13z = b1

a 21x + a 22 y + a 23z = b 2
a x + a y + a z = b
32
33
3
 31
5.
mi + nj = xi + yj
⇒
m=x
6.
mi + nj = 0
⇒
m=n=0
7.
a is a unit vector iff a = 1 .
and
n=y
is the unit vector
(a) a × a = 0
(b) a × b = –b × a
(c) k(a × b) = (ka) × b = a × (kb) .
1 Cramer’s Rule
1.Parallelism
(d) |a × b| = |b × a|
∆ x determinant of A with x column replaced
If the vectors p = mi + nj and q = xi + yj are parallel, then
(e) For non-zero vectors a and b , a × b = 0 if and only if a
x y
x m
=
=
and
m n
y n
and b are parallel.
2. Division of a Line Segment
(h) |a × b|2 = |a|2|b|2 – (a ⋅ b)2
AB such that AP : PB = m : n, then
Vectors in R3
∆y
∆x
∆
, y=
, z= z ;
det A
det A
det A
2 Gaussian Elimination
x=
Application of Vectors:
Volume of triangular prism =
a31
(2) i ⋅ j = j ⋅ k = k ⋅ i = 0.
OP = xi + y j + zk.
+ a 13
System of Linear Equation
-1 -1

A
=
a33
Or
– a 12 a21 a23
3.
a1
det A = a2
a3
a3
(Properties of inverses)
(c) (AT)-1 = (A-1)T , (d) (
a32
a1 + x b1 + y c1 + z
a1 b1
a2
b2
c2 = a2 b2
Non-singular => det A ≠ 0
(a) (A-1)-1 = A,
a23
(1) i ⋅ i = j ⋅ j = k ⋅ k = 1.
If a = x1i + y1j + z1k
a1
k det A = a2
ka3
Inverse of Matrix
a22
Scalar product(Dot Product)
a ⋅ b = | a || b |cos θ, where θ is the angle between a and b.
OP = xi + yj.
1.
AI = IA = A
(A± B)C = AC ± BC
Vectors
1 a b e 


reduce to echelon form.  0 1 d f 
0 0 1 g 


mQB + nQA
m+n
n
m
a+
b
, i.e. p =
n+m
m+n
QP =
3. Use inverse of Matrix
AX = B
If det A ≠ 0, Unique Solution
and
−1
X=A B
*For Homogeneous
AX=0 ,
If det A ≠ 0, then X =0, trivial solution.
If det A = 0, AX = B has ∞ or no solutions (use Gaussian
Elimination to Solve)
(i)
3. Coplanar
Vol=0 => Scalar Triple Product =0
c = x3i + y3 j + z3k
then
(ii) i × j = k
,
k × j = –i
i × k = –j)
,
j×k=i
(iii) If a × b = a × c
y1
y2
y3
,
k × i = j.
(j × i = –k
⇒ b ≠c (e.g. a and b are parallel, and
c is 0).
Scalar Triple Product (b × c) • a
If a = x1i + y1 j + z1k , b = x2i + y2 j + z2k and
x1
a ⋅ (b × c) = x2
x3
i × i = j × j = k × k = 0.
If a = x1i + y1 j + z1k and b = x2 i + y 2 j + z 2k are two
non-zero vectors, and θ is the angle between them, then
z1
z2 .
z3
i
a × b = a b sin θ nˆ = x1
x2
j
y1
y2
k
z1 .
z2
,