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# Continuity and differentiation

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# Continuity and differentiation

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### Continuity and differentiation

1. 1. Module 1 INTRODUCTION TO QUANTITATIVE TECHNIQUES Topic: Continuity and Differentiation (Basic Mathematics)
2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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