2018-G12-Math-E

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabNow=dd xy 1 ln 1 + dx  eLearn.Punjab dx x version: 1.1 = 1 .dxx ln 1 + dx = 1 ln 1 + dx x x x x x d xThus lim dd=xy lim  1 ln 1 + dx dxx= 1 lim ln 1 + dx x  x x x x  d x→0 d x→0 d x→0 d x=dy +1  1 dx  x  dx x x dx . ln dlxim→0 x dx → 0 when d x → 0  x = 1 ln e  lim (1 + z ) 1 =e x → z z 0 ( )= 1=.1 1 =  log e 1 xx eNow we ind derivative of the general logarithmic function.Let y = logax then y + d=y loga ( x + d x) and dy = loga ( x + d x) - loga=x log  x +d x = loga 1 + dx  x x =dd xy 1 loga 1 + d=xx  1 . x loga 1 + dx  dx x dx x= 1 log a 1 + dx x x x d xThus =dy lim  1 log a 1 + dx =dxx  1 lim  1 + dx x  dx x x x loga x d x→0 d x→0 d x= +1 lim 1 dx  x  x x dx loga dx x →0 57

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab = 1 loga x  lim (1 + )1 =e  eLearn.Punjab x version: 1.1 z→0 zzor d loga=x  1. 1 =lo=gae 1 1  dx x ln a ln a log a e ( )Find dy=if y dxExample 1: log10 ax2 + bx + cSolution: Let u = ax2 + bx + c Then y= log10u ⇒ dy= 1 1 du u In 10and du = d (ax + bx + c)= a(2x) + b(1)= 2ax + b dx dxTh=us dy d=y . du  1 . 1  du dx du dx u ln 10 dx( )= + ax2 + 1 c ( 2ax b) bx + ln 10( )ord 2ax + b dx log10 ax2 + bx + c  = 2 + bx + c) ln 10 (axExample 2: ( )Differentiate ln x2 + 2x w.r.t. ' x '.( )Solution: Let =y ln x2 + 2x , then( ) ( )d=yd ln 1 .d (Using chain rule)( )dx dx x2 + 2x dx x2 + 2x= x2 + 2x= x2 1 2x .(=2x + 2) 2( x +1) + x2 + 2x  =2x(2x++21x)( )Thus d ln dx x2 + 2x 58

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.12 LOGARITHMIC DIFFERENTIATION Algebraic expressions consisting of product, quotient and powers can be oftensimpliied before diferentiation by taking logarithm.Example 1: Differentiate y = e f (x) w.r.t.' x '.Solution: Here y = e f (x) (i)Taking logarithm of both sides of (i), we haveIn y = f ( x). In e (In e = 1) = f (x)Diferentiating w.r.t x , we get1 . dy = f '(x)y dxSo dy×=y f '=( x) e×f (x) f ' ( x) dx( )or d e f=(x) e f (x) × f ' ( x) dxExample 2: Find derivative of x x2 + 3 x2 +1 = x x2 + 3 ......(i) x2 +1( )Solution:Let yTaking logarithm of both sides, we have( ) ( )l=n y  x x 2++1=3  ln x2 ln x x2 + 3 - ln x2 + 1( ) ( )or ln y= ln x + 1 ln x2 + 3 - ln x2 +1 ......(ii) 2Diferentiating both sides of (ii) w.r.t ‘ x ‘, version: 1.1 59

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab( ) ( )d [In=y] d  In 1 x2 + 1  eLearn.Punjab dx 2 version: 1.1dx x + In x2 + 3 - In1 dy =1 + 1 . 1 3 × 2x - x 1 1 × 2xy dx x 2 x2 + 2+ +=1 x2-x+ 3 2x x x2 +1 = ( x2 + 3) ( x2 + 1) +x (. x x2 + 1) - 2x . x ( x2 + 3) + 3)(x2 + 1) x ( x2 x4 + 4x2 + 3 + x4 + x2 - 2x4 - 6x2 3 - x2 x x2 + 3 x2 +1( )( )= x( x2 + 3)( x2 +1) y(3- x2) x( x2 +1)( x2 +1) x2 + 3 3 - x2 x2 +1 x2 + 3 x2 +1==Thus dy ( )( )x . x dx 3 - x2 x2 + 3 . x2 +1 2 ( )=Example 3: Differentiate (ln x)x w.r.t. ' x '.Solution: Let y = (ln x)x (i)Taking logarithm of both sides of (i) , we have==In y In (In x)x  x In (In x)Diferentiating w.r.t x , =1 dy 1 . In (In x) +x .1 . d ( In x) y dx In x dx =In ( In x ) + x . 1 x . 1 =In ( In x) + 1 x In x In =dy y In ( In x) + 1 x= ( In x)x In ( In x ) + 1 x  dx In In  60

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.13 DERIVATIVE OF HYPERBOLIC FUNCTIONSThe functions deined by: sinh x = ex - e-x , x ∈ R ; cosh x = ex + e-x ; x ∈ R 22 ta=nh x s=inh x ex - e-x ;x∈R cosh x ex + e-xare called hyperbolic functions. The reciprocals of these three functions are deined as: cos ech=x 1= ex 2 ,x∈R - {0}; sinh x - e-x s=ec h x =1 ex 2 ,x∈R cosh x + e-x c=oth 1= ex + e-x , x ∈ R - {0} tanh x ex - e-xDerivatives of sin h x, cos h x and tan h x are found as explained below:d d  1  = 1 dx 2 2( ) ( )dx( sinh x ) = ex - e-x e x - e-x( -1) = 1 ex + e-x = cosh x 2d d  1  = 1 1 dx 2 2 2( ) ( )dx(cosh x) = ex + e-x e x + e-x .( -1) = ex - e-x = sinh x==d [tanh x] d  ex - e-x  ( ) ( ) ( ) ( )ex + e-x ex + e-x - ex - e-x ex - e-x dx ex + e-x ( )ex + e-x 2 dx( ( ) )== e2x + e-2 x +2 - e2x + e-2 x - 2 4 ex + e-x 2 ( )ex + e-x 2 = =ex +2e-x 2 sec h2 x.The following results can easily be proved. version: 1.1 61

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjabd (cos eh x) =coth x cos ech x ; d (sec h x-) tanh x sec h x eLearn.Punjab =-dx dx d (coth x) = - cos ech2 x. dxExample 1: Find dy if y = sinh 2x dxSolution: Let u = 2x, then=y sinh u =⇒ dy cosh u duan=d du dd=x (2x) 2. dxT=hus dy d=y . du cosh=u . du cosh (=2x) .2 2 cosh 2x dx du dx dxor d [sinh 2x] = 2 cosh 2x . dxExample 2: ( )Find dy if y = tanh x2 dxSolution: Let =u x2 ,then =y tanh u ⇒ d=y sec h2 u duan=d du dd=x ( x ) 2x dx( )Thu=s dy dx d=y . du sec h2=u . du sec h×2 x2  2x du dx dxor d tanh x2  = 2x sec h2 x2 dx version: 1.1 62

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab2.14 DERIVATIVES OF THE INVERSE eLearn.Punjab HYPERBOLIC FUNCTIONS: version: 1.1The inverse hyperbolic functions are deined by:=1. =y sinh'-1 x∈ if and if x sinh y ; x, y R =2. y cos=h-1 x if and only if x cos∈h y ∞ ; x∈ 1∞, ), y [0, ]=3. y t=anh-1 x if and∈on-ly if x ∈tanh y ; x ( 1,1) , y R=4. y =coth-1 x if and∈o-nly if x ∈coth- y ; x [ 1,1]' , y R {0} 5. y sec∈h-1 x if a∈nd on∞ly if x= sec h y ; x (0,1`], y [0 , ) =6. y cos ech=-1x if and only if x co∈s ech- y ; x ∈R -{0}, y R {0}The following two equations can easily be derived:( )(i) sinh-1 x = In x + x2 +1 ( )(ii) cosh-1 x= In x+ x2- 1 Proof of (i).=Let y ∈sinh-1 x for x, y R,then =x sinh y ⇒=x ey - e- y 2 ⇒ 2xey =e2 y -1 or e2 y - 2xey - 1 =0 Solving the above equation for ey , we haveey = 2x ± 4x2 + 4 x2 +1 2 =2x ± 2 x2 + 1 =x ± 2As ey is positivefor y ∈ R,so we discard x - x2 +1 )x2 +1(Thus ey =x + x2 +1 ⇒ y =In x + 63

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab ( )⇒ sinh-1 x = In x + x2 +1 eLearn.Punjab version: 1.1 Proof of (ii)=Let y cosh-1 x for x ∈[1, ∞ ), y ∈[0 ,∈), then=x cosh y ⇒=x ey + e- y ⇒ e2y - 2ey=+ 1 0 ......(I) 2Solivng (I) gives, ey =2x ± 4x2 - 4 =2x ± 2 x2 -1 =x ± x2 -1 . 22 ( )ey =x - x2 - 1 can be written as y =In x -x2 1-( )If x = 1, then y = ln 1 - 1 - 1= ln (1=) 0 but ( )ln x - x2 - 1 is negative for all x > 1, that is for each x ∈(1,∞) , y ∉(0,∞),so we discard this value of ey(Thus ey =x + x2 +1 which give y =In x + )x2 - 1 , that is ( )cosh-1 x = In x + x2 -1 .Derivative of sinh-1 x:=Let y ∈sinh-1 x ; x , y RThen x = sinh y=dx cosh y =⇒ dy =1  dy 1  dy dx cosh y dx dx dyor =dy =1 1> (cosh y 0)dx cosh y 1 + sinh2 y( )==dy d ∈sinh-1 x 1 (x R) dx dx 1+ x2 64

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabDerivative of cosh-1 x: eLearn.Punjab version: 1.1=Let y co∈sh-1∞x ; ∈ x ∞[1 ), y [0, ) Then x = cosh y an=d dx sinh y ⇒=dy =1  dy 1  dy dx sinh y dx dx dy or dy = 1 = 1 ( sinh y > 0, as y > 0 ) dx sinh y cosh2 y -1 ( )==Thus dy d c>osh-1 x 1 ( x 1) dx dx x2 -1 ( )As cosh-1 x = In x + x2 -1 ,sod cosh-1 x= 1 1 + 2x 1= 1. x2=- 1 + x 1dx x2 x2 x + x2 -1 x2 - 1 x2 -1 x+ -1 2 -Derivative of tanh-1 x : x ∈(-1, 1) , y ∈ R Let y = tanh-1 x ;   Then x = tanh y and dx = sec h2 ⇒ dy = 1 = dy 1 dy dx sec h2 y dx  dx dy dy = 1 = 1 ( )sec h=2 y- 1 tanh2 y dx 1 - tanh2 y 1- x2 ( )=Thus d tanh-1 x < 1 2 ; -1<x<1or x 1 dx 1 -x The following diferentiation formulae can be easily proved. ( )d coth-1 x = 1 or - 1 ; x >1 1- x2 x2 -1 dx 65

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab ( )d sec h-1 x =- 1 ; 0 <x <1 eLearn.Punjab version: 1.1 dx x 1 - x2 ( )d co-sech-1 x => 1 ; x 0 dx x 1 + x2 ( )or d cos ech-1 x =- 1 ; x ∈R -{0} dx x 1 + x2Example 1: Find=dy if y sinh-1 (ax + b ) dxSolution: Let =u ax + b , then =y sinh-1 u =⇒ dy 1 =dy d=y . du dx 1 + u2 dx du dx 1 . du 1 + u2 dxThus d sinh-1 (ax +=b) 1 .a = du ddx+(=ax b) a  dx dx 1 + (ax + b)2Example 2: =Find dy if y co≤sh (-≤1 sec x) 0 x p /2 dxSolution: Let u = sec x, then =y cosh-1 u ⇒=dy 1 dx u2 -1=and du d=( sec x) sec x tan x dx dx Th=us dy d=y . du 1 . du dx du dx u2 -1 dx===1 (sec x tan x ) 1 ( sec x tan x) sec x tan x sec x or d cosh-1 ( sec x ) = sec x dx 66

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab EXERCISE 2.61. Find f ' ( x) if (ii) =f ( x) x3 1 ( x ≠ 0) (iii) f ( x) = ex (I + ln x) (i) f ( x) = e x-1 ex(iv) f (x) = ex ( )(v) ln ex + e-x (vi) fx = eax - e-ax e-x + 1 eax + e-ax ( )(viii) f (x) = ln e2x + e-2x( )(vii)=f (x) ln e2x + e-2x2. Find dy if (ii) y = x ln x (iii) y = x dx ln x (v) y = ln x2 -1 (i) y = x2 ln x x2 +1 ( )(vi) y = ln x + x2 + 1 (viii) (iv) y = x2 ln 1 (xi) y = e-2x sin 2x ( )(ix) y= e-x x3 + 2x2 + 1 x y = 5e3x-4 (xii) =y ( x +1)x (vii) =y ln (9 - x2 ) (x) y = x esin x (xiii) y = (ln )x ln x y= x2 - 1 ( x + 1)3. Find dy if x3 +1 3/ 2 dx ( )(xiv) (i) y = cosh 2x p xp (ii) y = sinh 3x 22 =(iii) y ta-nh-1<( sin<x) ( )(iv) y = sinh-1 x3 (v) y = ln(tanh x) (vi) y = sinh-1  x  2 version: 1.1 67

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.15 SUCCESSIVE DIFFERENTIATION (OR HIGHER DERIVATIVES): Sometimes it is useful to ind the diferential coeicient of a derived function. If wedenote f ’ as the irst derivative of f, then (f ’)’ is the derivative of f ’ and is called the secondderivative of f .For convenience we write it as f”. Similarly (f ”)’. the derivative of f ”, is called the third derivative of f and is written as f ’”. In general, for n ≥ 4 , the nth derivative of f is written as f (n) . Here we state diferent notations used for derivatives of higher orders.. 1st derivative 2nd derivative 3rd derivative nth derivative y ’ y ’’ y ’’’ y (n) dy d 2 y d 3 y d n y dx dx2 dx3 dxn y1 y2 y3 yn Dy D2 D3 Dn yyy df d 2 f d 3 f d n f dx dx2 dx3 dxnExample 1: Find higher derivatives of the polynomial f ( x )= 1 x4 - 1 x3 + 1 x2 + 2 x + 7 12 6 4Solution: ( ) ( )f ' ( x)= 1 4x3 - 1 3x2 + 1 (2x) + 2 + 0= 1 x3 - 1 x2 + 1 x + 2 12 6 4 322 f '' ( x) = 1 (3x2 ) - 1 ( 2 x ) + 1 (1) + 0 = x2 - x + 1 3 2 2 2 f \"' ( x=) 2x -1 f iv ( x) = 2All other higher derivatives are zero. version: 1.1 68

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab ( )Findd3y eLearn.PunjabExample 2: dx3 if y = ln x+ x2 + a2 =-( )Solution: Give that y = ln x + x2 + a2 (i) version: 1.1Diferentiating both sides of (i) w.r.t. ‘ x ‘ , we have (dx x + x2 + a2 dx )x2 a2=dy +1 + d x= + + 1  2 1× 2x  x x2 + a2 . 1 x2 + a2 = × 1  x2 + a2 + x  x x2 + a2 2 x2 + a2 +That is, dy = 1 (ii) dx x2 + a2Diferentiating (ii) w.r.t. ‘ x ’, we have( ) ( )d 2 ydx2 =ddx  x2 + a2- -1/ 2  =+ 1 x2× a2 -3/ 2 2x 2 d2y =- x (iii) dx2 x2 + a2 3/2( )orDiferentiating (iii) w.r.t. ‘ x ’ , we getd3y ( ) ( )1 . x2 + a2 3/ 2 - x. 3 x2 + a2 1/ 2dx3 = 2 x2 + a2 3/ 2  .2x ( )- ( ) ( ( ) )=x2 + a2 1/ 2x2 +xa2 2+ 3a2 - 3x2- a2 - 2x2 ( )x2 + a2 5/ 2 2x2 - a2 x2 + a2 5/2( )d 3 y =dx3 69

12.. DQiufaedrernattiiactEioqnuations eLearn.PunjabExample 3: Find d2y if y3 + 3ax2 + x3 =0 eLearn.Punjab dx2 version: 1.1Solution: Given that y3 + 3ax2 + x3 =0 (i)Diferentiating both sides of (i) w.r.t. ‘ x ‘ , gives d  y3 + 3ax2 + x3= d (0)= 0 dx dx ( )3y2 dy + 3a (2x) + 3x2 =0 ⇒ -y2 dy =+ 2ax x2 dx dx ⇒ dy =- 2ax + x2 (ii) dx y2 Diferentiating both sides of (ii) w.r.t. ‘ x ‘ , gives d2y =( 1) ddx  2axy+2 x2- (( ) )(2a + 2x) y2 =- y22ax2 + x2  2 y dd-yx  dx2 ( )2(a + x) y2 - × - 2ax + x2  y2 =- 2ax + x2 . 2 y y4 ( )( )= - 2 (a + x) y2 + 2ax + x2 2ax + x2  y  y4 ( )= ( a x) + 2  - 2 + y3 + 2ax x2 y4 .y ( ) (- =2 (a + x) -3ax2 - x3 + x2 (2a + x)2  -y3 =-3ax2 )x3 y5 ( )=  - ( a + x ) ( 3a + x) +  - 2x2 4a2 + x2 + 4ax y5 ( )= - + 4a2 + x2 + 4ax 2x2 3a2 + 4ax + x2 - y5- ==2x2y5a2  -2a2 x2 y5 70

21.. DQifuaerdernattiiactEioqnuations eLearn.PunjabExample 1: If x =-a(q sinq=)+, y a(1 cosq ). Then eLearn.Punjab version: 1.1 show that y2 d2y + a =0 dx2Solution: Given that=x a(q + sinq ) (i) and =y a(1 + cosq ) (ii) Differentiating (i) and (ii) w.r.t 'q ' , we get (iii) ddqx= a(1 + cosq ) (iv) and ddqy= a (-sinq ) Using=dy dd=qy . ddqx dy we have dx dq dx dq== a -a sinq ) -sin q 1 + cos q (1 + cos q That is, dy = - 1 sinq (v) dx + cosq Diferentiating (v) w.r.t. ‘ x ’ ddx2-y2 =-ddx  1=+sicnoqs q × d  sin q  dq dq 1+cos q dx = - cos q (1+ cos q ) - sin q (-sin q ) . dq (1 + cos q )2 dx d2y = - cos q +cos2 q +sin2q . dq dx2 dx (1 + cos q )2 =×(11++ccoossqq)2- 1  dd=qx a(1 + cosq )  a(1+ cos q ) 71

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab =1a .(1 + c1os q )2- 1 . 1 =1 + cos-q y  eLearn.Punjab a a version: 1.1  y 2 = a =- 1 × a2 =- a a y2 y2or y2 d2y = -a ⇒ y2 d2y +a =0 dx2 dx2Example 5: Find the irst four derivatives of cos (ax + b).Solution: L=et y cos(ax + b) , then y1 =ddx cos (ax +- b) =+sin (ax b ) . +d ( ax b) dx =- sin (ax + b) × (a + 0) =- a sin (ax + b) y2 =- a d sin ( ax + b) =( -a ) cos ( ax + b) × (a + 0) dx- =a+2 cos (ax b) ( )y3 d ( ax b) - sin (ax + b) × (a + 0) =- a2 dx cos + =-a2= a3 sin (ax + b) y4 = a3 d sin ( ax+ b)= a3× cos (ax+ b)× a= a4 cos (ax+ b) dxExa=mple 6: =If y e-ax +, then show that d3y a3 y 0 dx3( )Solution: As y= e-ax , so dy= d e-ax = e-ax . d (-ax)= e-ax . (-a) dx dx dxThat is dy = -ay ( )e-ax = y dx 72

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNow dy  dy  =d [-ay] ⇒ dd-x2 y2 ==a -dy -( a)( a-y) = ddyx ay  dx dx dx dxor d2y = a2 y (i) dx2Diferentiating (i) w.r.t. ‘ x ‘ we get d  d2y  =d  a 2 y  ⇒ d3y =a2 dy-=a2 (-ay) =a3 y dx dx2 dx dx3 dx Thus d3y + a3 y =0 dx3 ( )I=f y -3Example 7: Sin-1 x , then sho=w that y2 -x a2 a x2 2Solution: y = sin-1 x , so a =y1 d=y =ddx  Sin-1 x  ×1 2 d  x  dx a dx a -  x 1 a ( )= 1= . 1 =a .-1 a2 x2 -1/2 a2 - x2 a a2 - x2 a a2 ( ) ( ) ( )( )y2=d  -1/ 2  =- 1 ×-3/2 -3/ 2 dx a2 - x2 2 a2 - x2 -2x =x a2 - x2 version: 1.1 73

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab EXERCISE 2.71. Find y2 if (i) y= 2x5 - 3x4 + 4x3 + x - 2 (ii)=y (2x + 5)3/2 (iii) =y x + 1 x2. Find y2 if (ii) y = ln  2x + 3  (i) y = x2. e-x 3x + 23. Find y2 if (ii) x3 - y3 =a (=iii) x a=cosq , y a sinq (i) x2 + y2 =a2 (iv)=x a=t2, y bt4 (v) x2 + y2 + 2gx + 2 fy + c =04. Find y4 if (ii) y = cos3 x ( )(iii)=y ln x2 - 9 (i) y = sin 3x( )5. I=f x Sin q=, y Sin mq , Sh-ow that -1 x+2 =y2 xy1 m2 y 0=6. If y =ex si-n x, sho+w that d2y 2 dy 2y 0 dx2 dx eax s-i=n bx, sh+ow th+at ddx2 y2( )=7. If y 2a dy a2 b2 y 0 dx( ) ( )=8. If y=C-os-1 x 2-,prov-e that 1 x2 y2 xy1 2 09. If y = a cos (ln x) + b sin (ln x), prove that x2 d2y + x dy + y =0 . dx2 dx2.16 SERIES EXPANSIONS OF FUNCTIONS A series of the form a0 + a1x + a2x2 + a3x3 + a4x4 + ...... + anxn + ..... is called a power seriesexpansion of a function f ( x) where a0 ,a1,a2 , ... an , ... are constants and x is a variable. We determine the coeicient a0 , a1, a2 , ..., an , ... to specify power series by indingsuccessive derivatives of the power series and evaluating them at x = 0 . That is, version: 1.1 74

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjabf ( x) =a0 + a1x + a2x2 + a3x3 + a4x4 + a5x5 + ...... + an xn + ..... f (0) =a0f ' ( x) = a1 + 2a2x + 3a3x2 + 4a4x3 + 5a5x4 + ...... + na xn 1 + ..... f ' (0) = a1f '' ( x) = 2a2 + 6a3x + 12a4x2 + 20a5x3 + ... + n(n -1) an xn-2 + ... f '' (0) = 2a2f ''' ( x) =6a3 + 24a4x + 60a5x2 + .... f ''' (0) = 6a3f (4=) ( x) 24a4 + 120a5x ........ f (4) (0) = 24a4So we h=ave a0 f=(0) , a1 f=' (0) , a2 f='2' (!0) , a3 f '='3' (!0) , a4 f (4) (0) 4!Following the above pattern, we can write an = f n (0) n!Thus substituting these values in the power series, we havef ( x=) f (0) + f ' (0) x + f '' (0) x2 + f ''' (0) x3 + f (4) (0) x4 + .... + f n (0) xn + .... 2! 3! 4! n!This expansion of f ( x) is called the Maclaurin series expansion.The above expansion is also named as Maclaurin’s Theorem and can be stated as: If f ( x) is expanded in ascending powers of x as an ininite series, thenf ( x=) f (0) + f ' (0) x + f '' (0) x2 + f ''' (0) x3 + f (4) (0) x4 + .... + f n (0) xn + .... 2! 3! 4! n! Note that a function f can be expanded in the Maclaurin series if the function is deinedin the interval containing 0 and its derivatives exist at x = 0 . The expansion is only valid if it is convergent.Example 1: Expand f ( x) = 1 1 x in the Maclaurin series. +Solution: f is d=e=ined at x 0 that is, f (0) 1. Now we ind successive derivatives of f andtheir values at x = 0 . f ' ( x) =(-1)(1 + )x -2 and -f ' (0) =1, f '' ( x) =(-1)(-2) (1 + )x -3 and f '' (0) (=1-)2 2 f ''' ( x) =(-1)(-2) (-3)(1 + )x -4 and f ''' (0) (=1-)3 3 version: 1.1 75

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab f (4) ( x) =(-1)(-2) (-3)(-4)(1 + )x -5 and f (4) (0) (=1-)4 4 Following the pattern, we can write f (n) (0) = (-1)n n Now substituting f -(0) =-1, f '=(0) =1, f '' (0) ( 1)2 2 .- f ''' (-0) =( 1-)=3 3, f (4) =(0) ( 1)4 4,.... f (n) (0) ( 1)n n in the formula.f ( x) =f (0) + f '(0) x + f '' (0) x2 + f ''' (0) x3 + f (4)+(0) x4 =... + f (n) (0) xx ,...4 23 n + f (n) (0) xx+,... we have n 1 = 1+ (-1) x + (-1)2 2 x2 + (-1)3 3 x3 + (-1)4 4 x4 + ... + (-1)n n xn + ...1+ x 2 3 4n Thus, the Maclaurin series for 1 is the geometric series with the irst term 1 andcommon ratio -x. 1+ xNote: Applying the formula S = a1 , we have 1- r 1 - x + x2 - x=3 + ... 1 - =(1- x ) 1 1+ xExample 2: Find the Maclaurin series for sin xSolution: Let f=( x) sin x. T=hen f (=0) sin 0 0. 0; f ' ( x) =cos x and f ' (0) -=cos 0 =1 ; f '' (-x)=s=in x and f '' (0) sin 0 f ''' ( x) =- cos x and f ''' (0) =-cos 0 =-1 ; f (4) ( x) =-( -sin x) =-sin x and f =(4) (0) s=in (0) 0.f (5) = ( x) = cos x and f (5) (0)= cos 0= 1, f (6) ( x)= - sin xand=f (6) (0) =0 ; f (7) cos x and f (7) (0-) 1 =- Putting these values in the formula version: 1.1 76

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjabf ( x) =f (0) + f ' (0) x + f '' (0) x2 + f ''' (0) x3 + f (4) (0) x4 + f (5) (0) + ...,we have4 23 5sin x+=0 1+.x 0+x2 -+1 x3 0+x4 1+ x5 0+x+6 -1+x7 ... 2 3 4 56 7=x - x3 + x5 - x7 + ...... 357Example 3: Expand ax in the Maclaurin series. Solution: Let f ( x) = ax ,then ===f ' ( x) ax ln a, f '' ( x) ax (ln a)2 , f ''' ( x) ax (ln a)3==f (4) ( x) ax(ln a)4 , ..., f (n) ( x) ax (ln a)(n) . Putting x = 0 in f ( x) , f ' ( x) , f '' ( x) , f ''' ( x) , f (4) ( x) , ... f (n) ( x) , we get f (0=) a=0 1, f '(0=) a0 ln =a ln a, f '' (0=) (ln a)2 , f ''' (0) (ln a )3==f (4) (0) (ln a)4 , ... , f (n) (0) (ln a)n . Substituting these values in the formulaf ( x=) f (0) + f '(0) x + f '' (0) x2 + f ''' (0) x3 + ... + f (n) (0) xn + ..., we have 23 nax =1 + (ln a ) . x + (ln a)2 x2 + (ln a)3 x3 + ... + (ln a)n xn + ...n 23Note: If we put a = e in the above expansion, we getex =1 + x + x2 + x3 + ... + xn + . .. ( In e = 1) 23 nReplacing x by 1, we havee =1 + 1 + 1 + 1 + ... + 1 23 n version: 1.1 77

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 4: Expand (1 + x)n in the Maclaurin series.Solution: Let f ( x=) (1 + x)n , thenf ' (=x) n(1 + )x n-1 , f '' ( x) = n(n -1) (1 + )x n-2f ''' ( x) =n(n -1)(n - 2)(1 + )x n-3 , f (4) ( x) =n(n -1)(n - 2)(n - 3)(1 + )x n-4Putting x = 0 , we getf (0) =(1 + 0)n =1, f '(0) =n(1 + )0 n-1 =n ,f '' (0) = n(n -1)(1 + )0 n-2 = n(n -1)f ''' (0) =n(n -1) (n - 2)(1 + )0 n-3 =n(n -1)(n - 2) ,f (4) (0) =n(n -1)(n - 2)(n - 3)(1 + )0 n-4 =n(n -1)(n - 3)Substituting these values in the formulaf ( x) =f (+0) f '(0)+. x f ''(0)+x2 f '''(0)+x3 ... , we have 23(1 + x)n +=1 n+. x n(n -1) +x2 n(n -1)(n - 2) x3 + ... 232.17 TAILOR SERIES EXPANSIONS OF FUNCTIONS: If f is deined in the interval containing ' a' and its derivatives of all orders exist atx = a , then we can expand f ( x) as f ( x)= f (a) + f ' (a)( x - a) + f ''(a) ( x - a)2 + f (''' a) ( x - a)3 23 + f (4) (a) ( x - a)4 + ... + f (n) (a) ( x - a)n + ... 4nLet f ( x) = a0 + a1 ( x - a) + a2 ( x - a)2 + a3 ( x - a)3 + a4 ( x - a)4 + ... +an ( x - a)n + ...Obviously f (a) = a.0 (  putting x = a , all other terms vanish ) version: 1.1 78

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab f ' ( x) =+ a1 2a-2 ( x +a) 3a-3 ( x +a)2 4a-4 ( x +a)3 + ... na-n ( x )a+n-1 ... f '' ( x)= 2a2 + 6a3 ( x - a) +12a4 ( x - a)2 + ... + n(n -1) an ( x - )a n-2 + ... f ''' ( x) = 6a3 + 24a4 ( x - a) + ......Putting x = a , we get f ' (a=) a1; f '' (a=) 2a2 ⇒ a2= f '' (a)=; f ''' (a) 6a3 2⇒ a3 =f ''' (a) 3 Following the above pattern , we have f ( )(a) Substituting the values of a0 ,a1,a2 ,a3 ,..., , w e g e t f ( x=) f (a) + f ' (a)( x - a) + f '' (a) ( x - a)2 + f ''' (a) ( x - a)3 + ... 23 + f (n) (a) ( x - a)n + ... n This expansion is the Taylor series for f at x = a . The expansionisonly valid if it isconvergent . If a = 0, then the above expansion becomes f ( x=) f (0) + f ' (0) x + f '' (0) x2 + f '' (0) x3 + ...+ f (n) (0) xn + ...n 23 which is the Maclaurin series for f at x = a . Replacing x by x + h and a by x , the expansion in (A) can be written asf ( x + h=) f ( x) + f ' ( x)h + f '' ( x) h2 + f ''' ( x) h3 + ...+ f (n) ( x) hn + ... (B)n 23 The expansions in (B) is termed as Taylor’s Theorem and can be stated as: If x and hare two independent quantities and f ( x + h) can be expanded in ascending power of h asan ininite series, then version: 1.1 79

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjabf ( x + h=) f ( x) + f ' ( x)h + f '' ( x) h2 + f ''' ( x) h3 + ...+ f (n) ( x) hn + ... n 23Example 1: Find the Taylor series expansion of In (1 + x) at x = 2.Solution: Let f ( x) = ln (1+ x) , then f (2)= ln (1+ 2)= ln 3 Finding he successive derivatives of ln (1+ x) and evaluating them at x = 2 f ' (x) = 1 1 x and f '=(2) 1=+1 2 1 + 3 f '' ( x-) =( 1+) (1 x) and f '' (2) =-(1 + )2 -2 =- 1 9 f ''' ( x) =(-1)(-2)(1 + x)-3 and f ''' (2) = 2 . (1 + 2)-3= 2 27f (4) ( x) =(-1) (-2)(-3)(1 + x)-4 =(-1)3 3(1 + )x -4 and f (4) (2) 3=- 81The Taylor series expansions of f at x = a isf ( x=) f (a) + f ' (a) .( x - a) + f '' (a) ( x - a)2 + f ''' (a) ( x - a)3 + ...... 23Now substituting the relative values, we haveln (1 + x) = ln 3+ 1 (x - 2) + -1 (x - 2)2 + 2 (x - 2)3 + - 3 ( x - 2)4 + .... 9 27 81 32 3 4= ln 3+ x-2 - ( x - 2)2 + ( x - 2)3 - ( x - 2)4 + .... 1.3 2.32 3.33 4.34Example 2: Use the Taylor series expansion to ind the value of sin 310.Solution: We take a = 30° = p 6Let f=( x) sin x, th=en f  =p6  sin p 1 6 2 version: 1.1 80

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNow taking the successive derivative of sin x and evaluating them at p , we have 6f ' ( x) = cos x and f '= p6  c=os p 3 6 2f '' ( x) = - sin x and f ''-p6  -=s=in p -1 6 2f '''( x) = -cos x and f '''-p6  -=c=os p 3 6 2f (4) ( x) =- (- sin x) =sin x and f (4) = p6  s=in p 1 6 2Thus the Taylor series expansion at a = p is 6sin x =1 + 3  x - p  + -1  x - p 2 + - 3  x - p 3 + ... 2 2 6 2 6 6 2 2 3 =1 + 3  x - p  - 1  x - p 2 - 2 3  x - p 3 + .... 2 2 6 22 6 3 6( )For x =310-, x p = 3-10 300 =≈10 .017455 6sin 310 ≈ 1 + 3 (.017455) - 1 (.017455)2 - 3 (.017455)3 2 2 4 12 ≈ .5 + .015116 - 0.000076 ≈ .5150Example 3: Prove that e x+=h ex 1 + h+ h2 + h3 + .... 2 3Solution: Le=t f ( x=+ h) ex+h , then f ( x) ex ...(i)By successive derivatives of (i) w.r.t ‘x’ we have =f '( x) e=x , f '' ( x) e=x , f '''( x) ex etc .By Taylor’s Theorem we have version: 1.1 81

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab f ( x + h)= f ( x) + h f ' ( x) + h2 f '' ( x) + h3 + f ''' ( x) + ... eLearn.Punjab version: 1.1 23Putting the relative values, we getex+h =ex + h ex + h2 ex + h3 ex + ... 23 = ex 1 + h + h2 + h3 + ... 2 3 EXERCISE 2.81. Apply the Maclaurin series expansion to prove that:(i) ln (1 + x) = x - x2 + x3 - x4 + ...... 222(ii) cos x =1 - x2 + x4 - x6 + ...... 246(iii) 1 + x = 1 + x - x2 + x3 + ...... 2 8 16(iv) ex =1 + x + x2 + x3 + ...... 23(v) e2x =1 + 2x + 4x2 + 8x3 + ...... 232. Show that:cos( x + h)= cos x - h sin x - h2 cos x + h3 sin x + ...... 23and evaluate cos 61°.3. Show that 2x+h =2x{1 + (ln 2) h + (ln 2)2 h2 + (ln 2)3 h3 + ...} 23 82

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.18 GEOMETRICAL INTERPRETATION OF A DERIVATIVE Let AB be the arc of the graph of f deined by the equation y = f ( x).Let P( x, f ( x)) and Q( x + d x. f ( x + d x)) be twoneighbouring points on the arc AB where x ,x +d x∈Df . The line PQ is secant of the curve and it makes∠XSQ with the positive direction of the x -axis. (Seethe igure 2.21.1) Drawing the ordinates PM ,QN andperpendicular PR to NQ , we haveRQ = NQ - NR = NQ - MP = f ( x + d x) - f ( x) and PR = MN = ON - OM = x + d x - x = d x Thus tan m∠XSQ = tan m∠RPQ= R=Q f (x +d x)- f (x) PR dx Revolving the secant line PQ about P towards P, some of its successive positionsPQ1, PQ2, PQ3,... are shown in the igure 2.21.2. Points Qi (i = 1,2,3,...) are getting closer andcloser to the point P and PRi i.e; d xi (i = 1, 2, 3, ...) are approaching to zero. In other words we can say that therevolving secant line approaches the tangentline PT as its limiting position at P while d xapproaches zero, that is,tan m∠ XSQ →tan m∠XTP whend x → 0or f (x +d x)- f ( x) →tan m∠XTP as dx→0 dx (x +d x) - f=( x) so lim f dx tan m∠XTP d x→0 version: 1.1 83

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab or f '=( x) tan m∠XTPThus the slope of the tangent line to the graph of f at ( x, f ( x)) is f '( x) .Example 1: Discuss the tangent line to the graph of the function | x | at x = 0 .Solution: Let f (x) = |x| so f (=0) |=0| 0 and, f (0 + d x) =| 0 + d x |=|d x |, f (0 +d x) - f (0) = |d x|- 0 and f (0 +d x) - f (0) = |d x| dx dx dx dx Thus f ' (0) = lim d x→0 Because d x = d x when d x > 0 and d x = - d x when d x < 0so we consider one-sided limits dL=x→im0+ dd xx dL=x→im0+ dd xx 1 dxand Lim dx = Lim -d x = -1 dx d x→0- d x→0- The right hand and left hand limits are not equal, therefore, the Lim dx does notexist. dx d x→0 This means that f ' (0) ,the derivative of f at x = 0 does not exist and there is no tangentline to the graph of f and x = 0(see the igure 2.21.3). version: 1.1 84

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 2: Find the equations of the tangents to the curve x2 - y2 - 6 y =0 at the pointwhose abscissa is 4.Solution. Given that x2 - y2 - 6 y =0 (i)We irst ind the y-coordinates of the points at which the equations of the tangents are tobe found. Putting x = 4 is (i) gives 16 - y2 - 6 y =0 ⇒ y2 + 6 y -16 =0or y = -6 ±=36 + 64 -=6 ± 100 -6 ± 10 , that is , 2 22 y= -6 + 10= 4= 2 or y = -6 -10 = -16 = -8 22 22Thus the points are (4, 2) and (4, - 8).Diferentiating (i) w.r.t. ‘ x ’ we have2x - 2 y dy - 6 dy =0 ⇒ 2 dy ( y + 3) =2x ⇒ dy =y +x 3 dx dx dx dxThe slope of the tangent to (i) at (4, 2=) = 2=+4 3 4. 5Therefore, the equation of the tangent to (i) at (4, 2) is y-2 = 4 ( x - 4 ) ⇒ 5y -10 = 4x -16 5or 5=y 4x - 6The slope of the tangent to (i) at (4, - 8) = 4 3 = -4 -8 + 5Therefore the equation of the tangent to (i) at (4, - 8) is y - ( -8) =- 4 ( x - 4) 5 5y + 40 =-4x +16 ⇒ 4x + 5y + 24 =0 version: 1.1 85

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab2.19 INCREASING AND DECREASING FUNCTIONS Let f be deined on an interval (a, b) and let x1, x2 ∈(a,b) . Then (i) f is increasing on the interval (a, b) if f(x2) > f(x1) whenever x2 > x1 (ii) f is decreasing on the interval (a, b) if f(x2) < f(x1) whenever x2 > x1 We see that a diferentiable function f is increasing on (a,b) if tangent lines to its graphat all points (x, f(x)) where xd(a, b) have positive slopes, that is, f ’ (x) > 0 for all x such that a < x < b and f is decreasing on (a, b) if tangent lines to its graph at all points ( x, f ( x)) wherex ∈(a,b) , have negative slopes, that is, f '( x) < 0 for all x such that a < x < b Now we state the above observation in the following theorem.Theorem: version: 1.1 Let f be a diferentiable function on the open interval (a,b). Then (i) f is increasing on (a,b) if f ' ( x) > 0 for each x ∈(a,b) (ii) f is decreasing on (a,b) if f ' ( x) < 0 for each x ∈(a,b) Let f ( x) = x2 , then f ( x2 ) - f ( x1 ) = x22 - x12 = ( x2 - x1 )( x2 + x1 ) If x1,x2 ∈(-∞, 0) and x2 > x1 ,, then 86

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab f ( x2 ) - f ( x1 ) < 0 ( x2 - x1 > 0 and x2 + x1 < 0)⇒ f ( x2 ) < f ( x1 )⇒ f is decreasing on the interval (-∞,0)If x1,x2∈(0,∞) and x2 > x1 , thenf ( x2 ) - f ( x1 ) > 0 ( x2 - x1 > 0 and x2 + x1 > 0)⇒ f ( x2 ) > f ( x1 )⇒ f is increasing on the interval (0, ∞) =Here f ' ( x) <2x and f ' (∈x) -∞0 for all x ( ,0) , therefore, f is decreasing on the interval (-∞,0) Also f ' ( x) > 0 for all x ∈(0,∞) , so f is increasing on the interval (0, ∞).From the above theorem we can conclude that1. f ' ( x1 ) < 0 ⇒ f is decreasing at x12. f ' ( x1=) 0 ⇒ f is neither increasing nor decreasing at x13. f ' ( x1 ) > 0 ⇒ f is increasing at x1Now we illustrate the ideas discussed so far considering the function f deined asf ( x=) 4x - x2 (I)To draw the graph of f, we form a table of some ordered pairs which belongs to fx -1 0 1 2 3 4 5y = f ( x) -5 0 3 4 3 0 -5 version: 1.1 87

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab The graph of f is shown in the igure 2.22.1. eLearn.Punjab From the graph of f, it is obvious that y rises from 0 to 4 as x increases from 0 to 2 andy falls from 4 to 0 as x increases from 2 to 4. In other words, we can say that the function f deined as in (I) is increasing in theinterval 0 < x < 2 and is decreasing in the interval 2 < x < 4. The slope of the tangent to the graph of f at any point in the interval 0 < x < 2 , in whichthe function f is increasing is positive because it makes an acute angle with the positivedirection of x-axis. (See the tangent line to the graph of f at (1, 3)). But the slope of the tangent line to the graph of f at any pointin the interval2 < x < 4 in which the function f is decreasing is negative as it makes an obtuse angle with thepositive direction of x-axis. (See the tangent line to the graph of f at (3, 3)). As we know that the slope of the tangent line to the graph of f at ( x, f ( x)) is f ' ( x) , sothe derivative of the function f i.e., f ' ( x) , is positive in the interval in which f is increasing andf ' ( x) , is negative in the interval in which f is decreasing. The function f under consideration is actually increasing at each x for which f ' ( x) > 0 . version: 1.1 88

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjabi.e. 4 - 2x > 0 ⇒ -2x > - 4 ⇒x<2Thus it is increasing in the interval (-∞, 2) . Similarly we can show that it is decreasing,in the interval (2, ∞) .Now we give an analytical approach to the above discussion. Let f be an increasing function in some interval in which it is diferentiable. Let x andx + d x be two, points in that interval such that x + d x > x . As the function f is increasing in the interval, it conveys the fact that f(x + dx) > f(x). Consequently we have, f ( x + d x) - f ( x) > 0 and ( x + d x) - x > 0 , that is, f(x + dx) - f(x) > 0 and dx > 0or f (x +d x)- f (x) > 0 dxThe above diference quotient becomes one-sided limit lim f (x +dx)- f (x) d x→0+ dxAs f is diferentiable, so f ‘ (x) exists and one sided limit must equal to f ‘ (x).Thus f ‘ (x) > 0Example 1: Determine the values of x for which f deined as f ( x) = x2 + 2x - 3 is(i) increasing (ii) decreasing.(iii) ind the point where the function is neither increasing nor decreasing.Solution: The table of some ordered pairs satisfying f ( x) = x2 + 2x - 3 is given below: x -4 -3 -2 -1 0 1 2 y = f(x) 5 0 -3 -4 -3 0 5 version: 1.1 89

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabThe graph of f is shown in the igure2.22.2.f ' ( x=) 2x + 2(i) The condition f ' ( x) > 0 ⇒ 2x + 2 > 0 ⇒ 2x > - 2which gives x > -1 , so the function f deined asf ( x) = x2 + 2x - 3 is increasing in the interval ( -1,∞) .(ii) And the condition f ' ( x) < 0 ⇒ 2x + 2 < 0 ⇒ 2x < -2 which gives x < -1, so the function f underconsideration in the example I is decreasing in theinterval (-∞,-1) .(iii) The function is neither increasing nor decreasing where f ' ( x) = 0 , that is, 2x + 2 =0 ⇒ x =-1.If x =-1 then f (-1) =(-1)2 + 2(-1)- 3 =-4. Thus f is neither increasing nor deceasing atthe point (-1, -4).Note: Any point where f is neither increasing nor decreasing is called a stationarypoint, provided that f ‘ (x) = 0 at that point.Example 2: Determine the intervals in which f is increasing or it is decreasing if f ( x) =x3 - 6x2 +9xSolution. f ' ( x) = 3x2 -12x + 9 = 3( x2 - 4x + 3) =3( x -1)( x - 3) f '(x) > 0 ⇒ x2 - 4x + 3 > 0 ⇒ ( x -1)( x - 3) > 0 version: 1.1 90

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab ( x -1) ( x - 3) > 0 ‘ in the intervals (-∞ ,1) and (3,∞) f ' (x) < 0 ⇒ ( x -1)( x - 3) < 0 ( x -1)( x - 3) < 0 if x > 1 and x < 3 that is 1 < x < 32.20 RELATIVE EXTREMA Let (c - d x, c + d x) ⊆ Df , , (domain of a function f), whered x is small positive number. If f (c) ≥ f ( x) for all x ∈(c - d x, c + d x) then the functionf is said to have a relative maxima at x = c . Similarly if f (c) ≤ f ( x) for all x ∈(c - d x, c + d x) , thenthe function f has relative minima at x = c . Both relative maximum and relative minimum arecalled in general relative extrema. The graph of a function is shown in the adjoining igure.It has relative maxima at x = b and x = d . But at x = a andx = c , it has relative minima. Note that the relative maxima at x = d is not the highest point of the graph.2.21 CRITICAL VALUES AND CRITICAL POINTS If c ∈ Df and f ' (c) =0 or f ' (c) does not exist, then the number c is called a critical valuefor f while the point (c. f(c)) on the graph of f is named as a critical point.Note: There are functions which have extrema (maxima or minima) at the pointswhere their derivatives do not exist. For example, the derivatives of the function f and fdeined as. version: 1.1 91

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab f (x) = xand f(x) = 22 - x x>0 + x x≤0do not exist at (0, 0) and (0, 2) respectively. But f has minima at (0, 0) and f has maxima at(0, 2). See the adjoining igures. Those critical points on the graph of f at whichf ' ( x) = 0 are called stationary points of f. Now we discuss relative maxima and relativeminima of the diferentiable function f deined as: y =f ( x) =x3 - 3x2 + 4 ....(1)Graph of f is drawn with the help of some ordered pairs tabulated as below: X -3/2 -1 -1/2 0 1/2 1 3/2 2 5/2 3 Y -49/8 0 25/8 4 27/8 2 5/8 0 7/8 4Now diferentiating (i) w.r.t. ' x' we getf ' (x) = 3x2 - 6x = 3x (x - 2)f '(x) = 0 ⇒ 3x( x - 2) =0 ⇒=x 0 or=x 2Now we consider an interval (-d x ,d x) in the neighbourhood of x = 0 . Let 0 - e is apoint in the interval (-d x,0) We see thatf ' (0 - e ) = 3(-e ) (-e - 2 ) ( f =' ( x) 3x ( x - 2)) = 3e (e + 2) > 0 (e > 0,e + 2 > 0)That is f ' ( x) is positive for all x ∈(-d x, 0 ) .Let 0 + e1 is a point in the interval (0,d x) , then we have f '(0 +=e1 ) 3(e1 )(e1 - 2) -=3e1-(2 e<1 ) 0 (2 - e1 > 0,e1 > 0) , that is,f ' ( x) is negative for all x ∈(0 ,d x) version: 1.1 92

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab We note that f ' ( x) > 0 before x = 0, f =' ( x) 0=at x <0 and f ' ( x) 0 after x = 0 . The graph of f shows that it has relative maxima at x = 0. Thus we conclude that a function has relative maxim=a at x c if f ' ( x) > 0 , before=x c=f ' (c) 0 and f ' ( x) < 0 after x = c. Considering an interval (2 - dx, 2 + dx) in the neighbourhood of x = 2 we ind the valuesof f ‘ (2-e) and f ‘ (2 + e) when 2 - ed(2 - dx, 2) and 2 + ed(2, 2 + dx) f '(2 - e )= 3(2 - e )(2 - e - 2)  f ='( x) 3x( x - 2) = 3(2 - e )(-e ) (e > 0, 2 - e > 0) =-3e (2 - e ) < 0 (e > 0, 2 + e > 0)and f '(2 + e )= 3(2 + e )(2 + e - 2) = 3e (2 + e ) > 0We see that f '( x) < 0 before x = 2, f '( x) = 0 at x = 2 and f '( x) > 0 after x =2 .It is obvious from the graph that it has relative minima at x = 2. Thus we conclude that a function has relative mini=ma at x c if f '( x) < 0 beforex =c, f ' ( x) ==0 at x c and>f ' ( x) 0 =after x c .First Derivative Rule: Let f be diferentiable in neighbourhood of c where f '(c) = 0.1. If f '( x) changes sign from positive to negative as x increases through c, then f (c) the relative maxima of f.2. If f '( x) changes sign from negative to positive as x increases through c, then f (c) is the relative minima of f. version: 1.1 93

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabNote: 1. A stationary point is called a turning point if it is either a maximum point or a minimum point. 2. If f ‘ (x) > 0 before the point x = a, f ‘ (x) = 0 at x = 0 and f ‘ (x) > 0 after x = 0, then f does not has a relative maxima. See the graph of f (x) = x3. In this case, we have f '( x) = 3x2 , that is, f '(0 - e ) =3(-e )2 =3e 2 > 0 and f '(0 + e ) = 3(e )2= 3e 2 > 0 The function f is increasing before x = 0 and also it is increasing after x = 0. Such a point of the function is called the point of inlexion.Second Derivative Test: We have noticed that the irst derivative f '( x) of a function changes its sign frompositive to negative at the point where f has relative maxima, that is, f ‘ is a decreasingfunction in the neighbouring interval containing the point where f has relative maxima. Thus f ''( x) is negative at the point where f has a relative maxima. But f '( x) of a function f changes its sign from negative to positive at the point where fhas relative minima, that is, f ’ is an increasing function in the neighbouring interval containingthe point where f has relative minima. Thus f ''( x) is positive at the point where f has relative minima.Second Derivative Rule Let f be diferential function in a neighbourhood of c where f '(c) = 0 . Then 1. f has relative maxima at c if f ''(c) < 0. 2. f has relative minima at c if f ''(c) > 0 . version: 1.1 94

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 1: Examine the function deined as f ( x) =x3 - 6x2 + 9x for extreme values.Solution: f ' ( x) = 3x2 -12x + 9 = 3( x2 - 4x + 3) = 3( x -1)( x - 3)First Method If x= 1 - e where e is very very small positive number, then ( x -1)( x - 3) =(1- e -1)(1- e - 3) =(-e )(-e - 2) =e (2 + e ) > 0 that is , f ' ( x) > 0 before x=1. For x =1 e , we have + ( x -1)( x - 3) =(1+ e -1)(1+ e - 3) =(e )(-2 + e ) =-e (2 - e ) < 0 That is, f ' ( x) < 0 after x =1 As f ' ( x) > 0 before x =1, f ' ( x) =0=at x 1 and f ' ( x)<0 a=fter x 1 Thus f has relative maxima at x =1 and f (1) =-1- 6 + 9 =4. Let x= 3 - e , then ( x -1)( x - 3) =(3 - e -1)(3 - e - 3) =(2 - e )(-e ) =-e (2 - e ) < 0 That is f ‘(x) < 0 before x = 3. For x = 3 + e (x - 1) (x - 3) = (3 + e - 1)(3 + e - 3)= (2 + e)(e) > 0 That is, f ' ( x) > 0 after x =3. As f ' ( x) < 0 before x = 3 , f ' ( x) at x = 3 and f ' ( x) > 0 after x=3 , Thus f has relative minima at=x 3. and f (=3) 3(3)2 -12(3) +=9 0 Second Method: f '' ( x) = 3(2x - 4) = 6( x - 2) f '' (1) =6(1- 2) =- 6 < 0 , therefore, f has relative maxima at x =1 and f (1) =(1)3 - 6(1)2 + 9(1) =1- 6 + 9 =4 f '' (3) =6(3 - 2) =6 > 0,therefore f has relative minima at x = 3 and f (3) = 27 - 54 + 27 = 0 version: 1.1 95

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.PunjabExample 2: Examine the function deined as f ( x)= 1+ x3 for extreme valuesSolution: Given that f ( x)= 1+ x3Diferentiating w.r.t. ' x' we get f ' ( x) =3x2 f ' (x) = 0 ⇒ 3x2 =0 ⇒ x =0 f '' ( x) = 6x and f ''=(0) 6=(0) 0The second derivative does not help in determining the extreme values. f ' (0 - e ) = 3(0 - e )2 = 3e 2 > 0 f ' (0 + e ) = 3(0 + e )2 = 3e 2 > 0As the irst derivative does not change sign at x = 0 , therefore (0, 0) is a pointof inlexion.Example 3: Discuss the function deined as f=( x) sin x + 1 cos 2x for extreme values inthe interval (0 , 2p ). 22Solution: Given that f=( x) sin x + 1 cos 2x 22 f ' ( x) =cos x + 1 (-2 sin 2x) =cos x - 1 sin 2x 22 2 =cos x 1 (2 sin x -cos x) cos x 2 s=in x cos x - 2( )= cos x 1- 2 sin xNow f ' ( x) = 0 ( )⇒ cos x 1- 2 sin x =0 ⇒ cos x =0 ⇒ x =p ,3p 22or 1 - 2 sin x =0 ⇒ sin x =1 ⇒ x =p ,3p 2 44Diferentiating (i) w.r.t. ‘ x ’ , we have version: 1.1 96

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab f '' (-x) =s-in x 1 (cos×-2x=) 2- sin x 2 cos 2x 2As f ''  p  =- sin p - 2 cosp =-1 - 2 × (-1) = 2 -1 > 0 2 2and f ''  3p  =- sin 3p - 2 cos 3p =-(-1) - 2 (-1) =1+ 2 >0 2 2Thus f ( x) has minimum value=s for x p=and x 3p 2 2As f ''  p - =sin-p 2 c-os=p - 1 - =2<. 0 10 4 4 22 2and f ''  3p - =sin-3p 2 co-s =3p - 1 - =2<. 0 1 0 4 4 22 2Thus f ( x) has minimum value=s for x p=and x 3p 44 EXERCISE 2.91. Determine the intervals in which f is increasing or decreasing for the domainmentioned in each case.(i) f ( x) = sin x ; x ∈(-p ,p )(ii) f ( x) = cos x ; x ∈∈ ( --22p,2,p)2 (iii) f ( x)= 4 - x2 ; x(iv) f ( x) = x2 + 3x + 2 ; x ∈(-4,1)2. Find the extreme values for the following functions deined as:(i) f ( x)= 1- x3 (ii) f ( x) = x2 - x - 2(iii) f ( x) = 5x2 - 6x + 2 (iv) f ( x) = 3x2(v) f ( x) = 3x2 - 4x + 5 (vi) f ( x) = 2x3 - 2x2 - 36x + 3 version: 1.1 97

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab(vii) f ( x=) x4 - 4x2 (viii) f ( x) =( x - 2)2 ( x -1)(ix) f ( x) = 5 + 3x - x33. Find the maximum and minimum values of the function deined by the following equation occurring in the interval [0,2p ] f=( x) sin x + cos x.4. Show that y = ln x has maximum value at x = e . x5. Show that y = xx has a minimum value at x = 1 . eApplication of Maxima and Minima Now we apply the concept of maxima and minima to the practical problems. We irstform the functional relation of the form y = f(x) from the given information and ind themaximum or minimum value of f as required. Here we solve some examplesrelating to maxima and minima problems.Example 1: Find two positive integers whose sum is 9 and the product of one withthe square of the other will be maximum.Solution: Let x and 9 - x be the two required positive integers such that version: 1.1 x(9 - x)2 will be maximum. Let f (=x) x(9 - x)2 . Then f ' (=x) 1 . (9 - x)2 + x . 2(9 - x) × (-1) = (9 - x)[9 - x - 2x] = (9 - x)(9 - 3x) = 3(9 - x)(3 - x) f ' ( x) = 0 ⇒ 3(9 - x)(3 - x) = 0 ⇒ x = 9 or x = 3 In this case x = 9 is not possible because 9 - x = 9 - 9 = 0 which is not positive integer. f '' ( x) = 3(-1)(3 - x) + (9 - x) × (-1) = 3[-3 + x - 9 + x] 98

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab = 3[2x -12] = 6( x - 6)As f '' (3) =6(3 - 6) =6(-3) =-18 which is negative.Thus f ( x) gives the maximum value if x = 3 , so the other positive integer is 6 because9 - 3 = 6.Example 2: What are the dimensions of a box of a square base having largestvolume if the sum of one side of the base and its height is 12 cm.Solution: Let the length of one side of the base (in cm) be x and the height of the box (incm) be h, then V=x2hIt is given that x + h =12 ⇒ h = 12 - xThus V=x2 (12 - x) anddV= 2x(12 - x) + x2 (-1)= 24x - 3x2= 3x(8 - x)dx dV = 0 ⇒ 3x(8 - x) = 0 . In this case x cannot be zero, dxso 8 - x = 0 ⇒ x = 8. ddx2V=2 24 - 6x which is negative for x = 8Thus V is maximum if x = 8(cm) and h = 12 - 8 = 4(cm)Example 3: The perimeter of a triangle is 20 centimetres. If one side is of length 8centimetres, what are lengths of the other two sides for maximum area of the triangle?Solution: Let the length of one unknown side (in cm) be x , then the length of the otherunknown side (in cm) will be 20 - x - 8 = 12 - x . Let y denote the square of the area of the triangle, then we have( )=y 10(10 - 8)(10 - x)(10 -12 + x) =s 2=0 10 and area of the triangle )s(s - a)(s - b)(s - c) 2= 10.2(10 - x)( x - 2) = 20 -x2 +12x - 20 version: 1.1 99

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjabdy =20( -2 x + 12) =-40 ( x - 6)dxdy = 0 ⇒ x =6dxAs d2y is -ve,so x = 6 gives the maximum area of the triangle. dx2The length of other unknown side = 12 - 6 = 6 (cm)Thus the lengths of the other two sides are 6 cm and 6 cm.Example 4: An open box of rectangular base is to be made from 24 cm by 45cmcardboard by cutting out square sheets of equal size from each corner and bending thesides. Find the dimensions of corner squares to obtain a box having largest possiblevolume.Solution: Let x (in cm) be the length of a side of each square sheet to be cut of from eachcomer of the cardboard. Then the length and breadth of the resulting box (in cm) will be45 - 2x and 24 - 2x respectively. Obviously the height of the box (in cm) will be x . Thus thevolume V of the box (in cubic cm) will be given by V = x(24 - 2x)(45 - 2x) = 2x(12 - x)(45 - 2x) ( )= 2x 540 - 69x + 2x2d=V( )anddx 2 1. 2x2 - 69x + 540 + x(4x - 69)( )= 2 6x2 -138x + 540= 12 x2 - 23x + 90 = 12( x - 5)( x -18)dV = 0 ⇒ 12( x - 5)( x -18) =0 =⇒ x 5 o=r x 18dx ⇒ x =5 [if x =18, then 12-x =12 -18 =-6, that is ,V is negative which is not possible]=ddx2 y2 12(2x - 23) version: 1.1 100

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjabd 2V is negative for x = 5 because 12(2 × 5 - 23)= 12 ( -13)dx2Thus V will be maximum if the length of a side of the corner square to be cut of is 5 cm.Example 5: Find the point on the graph of the curve y = 4 - x2 which is closest tothe point (3, 4).Solution: Let l be distance between a point ( x, y) on the curve y= 4 - x2 and the point (3 ,4). Thenl = ( x - 3)2 + ( y - 4)2( )= ( x - 3)2 + 4 - x2 - 4 2 ( )( x, y)is on the curve y= 4 - x2= ( x - 3)2 + x4Now we ind x for which l is minimum.( )=dl dx 2. -( x -+13)2 + x4 . 2( x 3) 4x3 ( )= 1 .2 2x3 + x - 3 2l version: 1.1 101

12.. DQiufaedrernattiiactEioqnuations eLearn.Punjab eLearn.Punjab= 1 ( 2 x3 + x - 3) l = 1 ( x - 1) ( 2 x 2 + x - 3) l( )dl = 0 ⇒ 1( x -1) 2x2 + 2x + 3 =0 =⇒ x -1 0 or 2x2 +=2x + 3 0dx l( )⇒ x =1 2x2 + 2x + 3 =0l is positive for 1- e and 1+e where e is very very small positive real number.Also 2x2 + 2x + 3 =2  x2 + x + 1  + 5 = 2  x + 1 2 + 5 is positive,for x =1 -e 4 2 2 2and x= 1+ eThe sign of dl depends on the factor x -1. dxx - 1 is negative for x = 1 - e because x - 1 = 1 - e -1=-e ..... (i)x - 1 is positive for x = 1 + e because x - 1 = 1 + e - 1 = e ..... (ii)From (i) and (ii), we conclude that dl changes sign from -ve to +ve at x = 1. dxThus l has a minimum value at x = 1.Putting x= 1 in y= 4 - x2 , we get the y-coordinate of the required point whichis 4 - (1)2 =3Hence the required point on the curve is (1, 3). EXERCISE 2.101. Find two positive integers whose sum is 30 and their product will be maximum.2. Divide 20 into two parts so that the sum of their squares will be minimum.3. Find two positive integers whose sum is 12 and the product of one with the square of the other will be maximum.4. The perimeter of a triangle is 16 centimetres. If one side is of length 6 cm, what are length of the other sides for maximum area of the triangle?5. Find the dimensions of a rectangle of largest area having perimeter 120 centimetres. version: 1.1 102

21.. DQifuaerdernattiiactEioqnuations eLearn.Punjab eLearn.Punjab6. Find the lengths of the sides of a variable rectangle having area 36 cm2 when its perimeter is minimum.7. A box with a square base and open top is to have a volume of 4 cubic dm. Find the dimensions of the box which will require the least material.8. Find the dimensions of a rectangular garden having perimeter 80 metres if its area is to be maximum.9. An open tank of square base of side x and vertical sides is to be constructed to contain a given quantity of water. Find the depth in terms of x if the expense of lining the inside of the tank with lead will be least.10. Find the dimensions of the rectangle of maximum area which its inside the semi-circle of radius 8 cm as shown in the igure.11. Find the point on the curve y = x2 - 1that is closest to the point (3, -1).12. Find the point on the curve y = x2 + 1 that is closest to the point (18, 1). version: 1.1 103

version: 1.1CHAPTER Integration3 Animation 3.1: Integration Source and credit: eLearn.Punjab

13.. IQntueagdrraatitoicn Equations eLearn.Punjab eLearn.Punjab3.1 INTRODUCTION When the derived function (or diferential coeicient) of a function is known, thenthe aim to ind the function itself can be achieved. The technique or method to ind sucha function whose derivative is given involves the inverse process of diferentiation, calledanti-derivation or integration. We use diferentials of variables while applying methodof substitution in integrating process. Before the further study of anti-derivation, we irstdiscuss the diferentials of variables.3.1.1 Differentials of Variables Let f be a diferentiable function in the interval a < x < b, deined as y = f(x), then =d y f (x + d x) - f (x) ==and dlxi→m0 dd xy lim f (x+ d x) - f (x) f ′(x), that is dx d x→0 dy = f ′( x ) dx We know that before the limit is reached, dy difers from f ‘ (x) by a very small real dxnumber e.Let d y = f ′ ( x) + e where e is very small d xor d y = f ′( x)d x + e d x (i) The term f ' ( x)d x being more important than the term e dx, is called the diferential ofthe dependent variable y and is denoted by dy (or df) Thus dy = f ' ( x)d x (ii) A=s dx (=x)' d x (1)d x, so the diferential of x is denoted by dx and is deined by the relation dx = dx. The equation (ii) becomes dy = f ’ (x) dx (iii)Note. Instead of dy, we can write df, that is, df = f ‘ (x) dx where f ‘ (x) being coeicient ofdiferential is called diferential coeicient. version: 1.1 2

31.. IQntueagdraratitoicn Equations eLearn.Punjab eLearn.Punjab3.1.2 Distinguishing Between dy and dy. The tangent line is drawn tothe graph of y = f(x) at P(x, f(x) andMP is the ordinate of P, that is,MP = f(x). (see Fig. 3.1) Let dx be small number, then thepoint N is located at x + dx’on the x-axis.Let the vertical line through N cut thetangent line at T and the graph of f at Q.Then the point Q is (x + dx, f(x + dx)), so dx = dx = PR and dy = RQ = RT + TQ = tan jdx + TQ  tanj = RT  PRwhere j is the angle which the tangent PT makes with the positive direction of the x-axis. or dy = f ‘ (x)dx + TQ (∴ tan jdx = f ‘ (x)) ⇒ dy = dy + TQ We see that dy is the rise of f for a change dx in x at x where as dy is the rise of thetangent line at P corresponding to same change dx in x. The importance of the diferential is obvious from the igure 3.1. As dx approaches 0,the value of dy gets closer and closer to that of dy, so for small values of dx, dy = dy or dy = f ‘ (x)dx [a dy = f ‘ (x)dx] (iv) We know that dy = f(x + dx) - f(x) f(x + dx) = f(x) + dy (v) But dy c dy, so (vi) f(x + dx) c f(x) + dy f(x + dx) c f(x) + f ‘ (x)dx version: 1.1 3