1. Module 1
INTRODUCTION TO QUANTITATIVE
TECHNIQUES
Topic:
Continuity and Differentiation
(Basic Mathematics)
2. (1)
ype Using standaYd Form
-a maY-1
a o
Theorem Um
X- a
16 4-2 = 4 2 =
3
m
Um
Examle4 Find Gm M 2 - 2
- 2 - b
Um - b
Examble 2 Evaluate m
b
8b 3 b 8
3- b
m - b
For cAice
Ans: LJi - bJb
Find Gm
- b
T im = 8o am mEN,
nd
-22
a Um
m - a
Find the Value of a 4 m
3/-1-I Ans:Eva luate m
.
- 2
3. Comtmuity at a point
A Functi on () 1S
Scd to be Coimuous at a
puint X =a of its domain ifF
m ( ) = +(a) .
X a
Thus
i s continuous at x=a) 2 um f(«) =
la)
im ) = m F = l a ) LH.L = R.H.L = Value ot the funcion
Xat
Algeba of lsntinunus tunch on Let and q be tuwo eal uncims, Coinuovs at
= a. Let be a real mumber. Then
Conhnuous at = a (i is Cmtimuovs at Xza
(in' + IS Conhnuous at = a (iw is cnthnuus at n= a
1S ntinuvs at x =a provialed fla)E (V) s Coninunus at x a, pTovided g(4) o
(V
Examle 4 Test the Ccrinuity o he unchm(») at oigin,where
= o
Soluhon e ob seve hat
= Um (-h) =
um -h)
LHL 4+ X=o)= m t(x) =
m +o-h) =um f(-h) - -
4. (3)
m= üm
ho
- -1,
and, RHL at x = o) Um ( ) = im +l0+h) = Um h=m
m 1
h
im
ho
Ths we havee Am ) Um +(
Csminu us at the OYi gin
Hence ( is mot
Example 2 D s u s s he Cntinü
he funckn() at ael/2 where
< L
F() Ans: Discaninvn
-
S-4 is Cominus
at x =1.
4x3 < 2
Example 3 Shos hat (x) =
unchom is lomhnuus
Examle 4
D E t e m 1 n e e
the Value of k fr which he toll0LIngtunchon
is Cnhnunns
a t 3
3
3
3
5. (4-)Soluion Smce ( is CntinuUuus at - 3 .
m + ) e+(3
3
L F 3) - k]
=> m k
x)x+3) UmX+3) =K 6=ku m m
3
3 - 3
Exambles. Find the Value ohe Constant s o hat he uncHon aiven belosis
Canhnuuv s at e -1
Ane: A=-4
2a b x>1
Example C. T he funch on F given F(A) s
Sa -2b
COninus at e 1,ind the values a a and b LAns: a e
3, be2
-4 +a < 4
Example 7. Let
x-4 is Cntnuuus at e4- ind
=4-
a b the Values 6 a omd
4+b 7 4
Ans: a21, b -1J
-+
6. (S)Derivative ot a fumctiom
Let unch on, Then he Value y depends uon they ) be a given Canhnubs
value af and it chamges with a
change in he value of . we use theward.increment to
denote a Small choamqe a . imeeage. or clecvease m the value ot and y
Let Ayy be am increment im Co Ye.spomodung to an imcnu m ent Ax n x. Then
y f ) omol y+Ay = ( * )
On
Subtracng re e
Ay tAx)-$()
A (+Aw)-
A A
Ay Um
+ A ) - f(n)
Gm
A
Ax o A o A
the dei vahve oy difeYenha1 Caciuent
The above imi t exists nitely is called
denote o by
o y ( itm Yesheck to and it is cdenoteol by
dc
ditferenhafm.
The rocess P nding he clerivak ve. is knoLn a
The Value st ( ) obtuneol by puhng X= a is called the
Serivative at a pnnt
derivaive ot (») at a and it is dlenoted by'(a) a
Geametrically, he derivative ot a unchm ( ) at a
puint = a nupausemhs the slobe
o the tangent to the Curve x) at Ca,tla))
7. (6)
Example 1 Shous that the unchum is diffexnhable at 3 and dbtain s
derivetive at - 3.
Solu hum nave
3 + h ) - ( 3 )
+h6h -1 hré
whith aplroache 6ah->o
Hene (is ditternhable at x 3 amd it derivahive at x23 is f (3- 6.
Example2 hinol the slope ot the tanqent to he CuTve y = #)e J i at the puint
9, 2)
By geometnical interpstahiun, he slape of the tanqent ts the Curve
y J at 1,) = (1)
Sutm
Um F h) -
fl1) ath-
h
4+h-3 m th-3 h +3m
1+h +3
-+h)-9 m
h o ath +3
+ot3
slakHencetha tangest Line toY Jat 1,2) h slope-
8. Ans. 8here x) +2+1
Example 3. hnd 3 hy deinition
Examble 4. oleiniton, nd'2 and s)when () ax+1x+4 Lns:1,11
Example4.
ExambleS. Fav the funchum given b ) =
6x+8, rove that
S) 3 t(2) =
(8)
Example 6. I the unchon ) = A+1x-4 (5) =
9, find a. LAns:à=9J
Devivatives of Some Shandard +unctims
2) =d
4 log24)a7o d(a =
a l0gea.3.
6. Ce)6
eCloa.) logaS d
Rule2 Cctt) c
tw]
hundamertal Rule tur Ditterentahim
Rule 4
Rule 3
9. Example 1. Dfteeniate the fsllouima umchoms
i 2 e+4-14. CI (J*
Soluiom (1 Let
Cx) + (T) - 2 (e) +4 9
d
--1
+(-1 x e 4
- - +
Let
= (J*+ 2.
2 L
ey (x)d (2) +()
10. Example 2 Fimd the derivatives o
X+1 Cx-2) (b) 3+3+e-2 ee(a
Ans (ay 3
(C S+6) lx-3) (b 3loa3 + e
(C 3x-16 x21
Rule oduct Rule T ( ) amd x) anL two derenhable fun cions, then
( . g ) is also dierenable Such Hhat
F ) g(x»= t . Lat)+ 3) )d
Example 4 DTereniate the olMowing uneions'
73 (b (+3). log C ( Su+6) A+23
(a
Solu Hm Let y =
/3e4)
, e) +e.d (la) e )d
e - e s
2
(b Let y= +Bx). log
+3). og*) +loqx (+3x) = +3x)-L + log 2x+3)
d
+3)t(20+3) log
11. (C Let ye(S1t6)Cx*+2)
' y (sx+t). a (+2) + 2) d ( s 46)d
= (3) (x-sx * 6) +6+2).(2x-5)
3-15x+ 18 +2'-Sx*t 4x -10
ST-2ox+18x+ 4- -10
Example 2 ind the derivatives
(C ( 31 +2) (x+2)2
(a) e (b +3+1).log
3 - 2 - 4
(C
Ans. (a e ( t I ) ( b (2.+3) log + +3 +1 (C 3-2% 4
Rule uotient Rule T (a) and gx) a dtfeeniable funch uns wTt. hen
are also diferenhable Su hat
gx) 4)
-
ttx).Cgt)
dd L0
Example 1 Diferenhate the flouang unchons
(a |4+ log
-
Soluion Let y x -
(a
-
12. dyd
(x-1)a+)-(at 1) (M-)=-1--=
- 1 ) 2
2
(-1)*(X-1)
(b Letyy= og
d
)0gx)- -log *) 4 ) . - log x+1)= (%+1)-2M+)}log
( ) (+
Examlple 2 Hind the cderivatives ot
b) ++Y Ans:(as (4)(b)(b) apx +2sp(a
antb
ax+b)J
Kle 6Derivati o a ncion t a funchon (Choin Rule)
T+y lt) and t= g(x) ther d =a
d
This may be further extended to thne uncthm
T lt), t u) and u h(x hen
dy dlt y d
ddu
d dt
1 Ty =lo9 + J a + )
nd d
d
As y e loq(xta+)
Example
Soluim
- loglog(xrdd*a) a +d
13. (a 4 )
Ja+x
Examble 2. Hnd the derivative ut e
Let y
e-
e d. (eyle+2")-e'+2)4;(C.2)Soluim.
(c
(ee)CeN-e') - (e+)(e'+*)(e"_e)-( e ë . 4
e - ) (e-x)
Examle3 Hind the derivatives of-
Ans( a - /(a
Ans loe x Clog4)
(b og +lb910 loq,+ lg
rove that (1-x) AL +y = o
(C T+ y A
+
(d T y C1-e)JI-Shous hat
d
2.(Ans
(e
2 -2
+e
Ans
e -2
