UNIT-IVDIFFERENTIATION

BASIC CONCEPTS OF DIFFERTIATIONConsider a function y=f(x) of a variable x. Suppose x changes from an initial value x0 to a final value x1. Then the increment in x defined to be the amount of change in x. It is denoted by ∆x. The increment in y namely ∆y depends on the values of x0 and ∆x.If the increment ∆y is divided by ∆x the quotient ∆ y∆x is called the average rate of change of y with respect to x, as x changes from

x0 to x0+∆x. The quotient is given by ∆ y∆x = f (x0+∆ x )−f (x0 )∆xThis fraction is also called a difference quotient.Differentiation using standard formulae

Define differentiation. Solution: The rate of change of one variable quantity with respect to another variable quantity is called Differentiation.(i.e)If y is a function of x, then the rate of change of y with respect to x is called the differential co-efficient of y. It is denoted by dy /dx (or) d ( f (x ))/dx (¿) f ’ ( x )(¿)Df (x )

Find dydx if y=3 sinx+4 cosx−ex. Solution:dydx

=3cosx−4 sinx−ex

Find dydx if y=ex+3 tanx+ log x6. Solution: Given y=ex+3 tanx+6 logx 170

dydx

=ex+3 sec2 x+ 6x

If y=x3−6 x2+7 x+6+cos x+¿ 1x√ x , find dydx . Solution:Given y=x3−6 x2+7 x+6+cos x+x−3

2 dydx

=3 x2−12x+7−sinx+(−32 ) x−32 −1

=2 x−sinx+ −3

2 x52

If y= 3√x +3x4−13√x , find dydx . Solution: Given y= 3

√x+3x4−1

3√xdydx

=3(−12 ) x−12 +12 x3−(−13 )x

−43

¿ −3

2 x32

+12 x3+ 1

3 x43

Find the derivative of y= e7x+sin 3 x+e5 x+3. Solution:dydx

= ddx

(e7x+sin 3 x+e5 x+3 . )=7e7x+3cos3 x+5 e5 x+3

Find dydx if y= log7 x . Solution: we know that ,if y=loga x then dydx = 1x log aedydx

=1xlog7e

Chain Rule (or) Differential coefficient of a function of function If u=f(x), y=f(u) then dydx=dydu . dudx Find dydx , if y=x+1x +( x+ 1x )

3. (L1) Solution:dydx

=1+(−1 ) x−2+3( x+ 1x )2 ddx (x+ 1x )

¿1− 1x2

+3(x+ 1x )2

(1− 1x2 )

171

Find dydx ,if y=loge(2x+3). S olution: dyd x

= 1(2x+3 )

dd x

(2 x+3)

¿ 1(2x+3 )

(2 )= 2(2 x+3)

Differentiate y=cos(x+ y ). Solution:Given, y=cos(x+ y ) Differentiatingboth sidesw . r .¿ xdydx

=−sin ( x+ y )(1+ dydx

)

¿−sin ( x+ y )−dydx

(sin ( x+ y )) dydx (1+sin ( x+ y ) )=−sin (x+ y ) ⟹ dy

dx=

−sin ( x+ y )(1+sin ( x+ y ) )

Differentiate y= log ¿ s∈¿2 x¿ . Solution: Given y= log ¿ sin2 x¿Let u=sin2x⟹ du

dx=2 sinx cosx

∴ y=logu⟹ dydu

=1u

dydx=dydu . dudx=1u .2 sinx cosx= 1sin 2 x

.2 sinx cosx=2 cosxsinx

=2cotx

Find dydx if Y=sec(ax+b). Solution:172

Let y=sec(ax+b)dydx

=a sec(ax+b) tan(ax+b)

Differentiate y=√2x+3+e√ tanx. Solution: y=e( tanx)1/2 dydx

=12

(2x+3 )−1 /2 (2 )+e (tanx )1 /2 ddx

( tanx )1 /2

¿ 1√2 x+3

+e√ tanx 1

2( tanx )−1 /2 d

dx(tanx)

¿ 1√2 x+3

+ e√ tanx

2√ tanxsec2 x

Find dydx , given y= 3√x3+x+1 . Solution: Given y=3√x3+x+1 dydx

=13

[ x3+x+1 ]13−1. d

[ x3+x+1 ]dx

dydx

=13

[ x3+x+1 ]13−1 . (3 x2+1 )

¿ 3 x2+1

3 [x3+x+1 ]23

Differentiate y=log (cos5 (3 x4 ))withrespect ¿ x ' ¿. Solution: Given y=log (cos5 (3 x4 ))

173

Differentiate both sides with respect to x , dydx = ddx {log (cos5 (3 x4 )) }

¿ 1(cos5 (3 x4 ))

. ddx {(cos5 (3 x4 ))}¿ 1

(cos5 (3 x4 ))ddx {(cos (3 x4 ))5 }¿ 1

(cos5 (3 x4 ))5 ¿¿¿ 5cos

4(3x4)(cos5 (3 x4 ) )

¿

¿ 5cos3 x4

¿¿ (−60 x3 ) sin (3 x4)

cos (3 x4) ¿−60 x3 tan3 x4 Differentiate y=(4 x+x−5 )

13 ,withrespect ¿ x ' .Hence show that

dydx= 4 x6−5

3 x83 (4 X6+1)

23 .

Solution: Given y=(4 x+x−5 )

13

∴ dydx

=( 13 )(4 x+x−5 )13−1 ddx

(4 x+ x−5 )¿( 13 )(4 x+x−5 )−23 (4−5 x−6)¿( 13 )(4 x+ 1x5 )

−23 (4− 5

x6 )¿( 13 )( 4 x6+1x5 )

−23 ( 4 x6−5x6 )¿( 13 )( x5

4 x6+1 )23 ( 4 x6−5x6 )¿( 13 ) x

103

(4 x6+1 )23( 4 x

6−5x6 )

¿x103 −6

3 (4 x6+1 )23

(4 x6−5 )¿ x−83 (4 x6−5 )

3 (4 x6+1 )23

¿ 4 x6−5

3x83 (4 X6+1)

23

Thus proved. Find dydx if y=¿ log [ 1+sinx1−sinx ]Solution:

Given y= log [ 1+sinx1−sinx ]

174

dydx

= 11+sinx1−sinx

{ ddx [ 1+sinx1−sinx ]}

¿ [1−sinx1+sinx ]{ (1−sinx ) (cosx )−(1+sinx )(−cosx)(1−sinx)2 }¿ {cosx−sinxcosx−(−cosx−sinx cosx )

(1+sinx )(1−sinx ) }¿ 2cosx1−sin2 x

=2 cosxcos2x

=2 secx

Find dydx if y=√1−sin2 x . Solution: Given y= √1−sin2 x y=(1−sin¿¿2x )¿1/2 dydx

=12(1−sin¿¿2x )

12−1 ddx

(1−sin¿¿2x )¿¿

¿ 12(1−sin¿¿2 x)

−12 (0−2 sinx cosx)¿

¿ −sin 2x2√1−sin 2 x

Find dydx if y=¿ log [ (1+√ x )(1−√x ) ]. Solution:

y= log(1+√ x) – log(1−√ x)

dydx

= 1(1+√x )

ddx

(1+√x )− 1(1−√ x )

ddx

(1−√x )¿ 1(1+√x )

12√x

− 1(1−√ x )

( −12√x

)

¿ 12√ x [ 1

(1+√x )+ 1

(1−√x ) ]¿ 22√ x¿¿

175

Differentiation using product ruleLet u & v be differentiable functions of x. Then the product of a functionY = u(x).v(x) is differentiable.d(uv)=udv+vdu If y=x2 sinx,finddydx . Solution: By product rule,

dydx

=x2 ddx

(sinx )+sinx ddx

(x2)dydx

=x2cosx+2 xsinx

¿ x (xcosx+2 sinx). Find dydx if y=ex tanx . S olution: Let y=ex tanx By product rule, dydx=¿ ex ddx (tanx )+tanx d

dx(ex )

dydx

=ex sec2x+ex tanx

¿ex (sec2 x+tanx) If y=3x4 ex ,find dydx . Solution: By product rule,dydx=3 (ex ddx (x4 )+x4 d

dx(ex))

dydx=3(x4 ex+4 x3 ex ) =3x3 ex (x+4 ) If y=cosx ex , find dydx . S olution:

176

By product rule,dydx=¿ ex ddx (cosx )+cosx ddx

¿

dydx

=ex (−sinx)+ex cosx=e x(cosx−sinx) If y= x loge x , find dydx . Solution:By product rule,dydx=¿ x ddx ( loge x )+log e x

ddx

(x )

dydx=x 1x + loge x =1+ log e x Differentiate Y=(x2−2¿(3 x+1). (L4) Solution: Let y= (x2−2¿(3 x+1).dydx

=(x2−2 )3+(3 x+1 )(2x )

¿3 (x2−2 ¿ +2 x (3 x+1 ) ¿3 x2−6 + 6x2+2 x ¿9 x2+2x−6 If y=cosecx cotx , find dydx . Solution:By product rule,dydx=¿ cosecx ddx (cotx )+cotx d

dx(cosecx)

dydx

=cosecx ¿

¿−cosecx ¿

If y= (x2+7 x+2¿(ex−logx), finddydx . Solution:By product rule,dydx

=(x2+7 x+2) ddx

(e x−logx)+(ex−logx) ddx

¿ = (x2+7 x+2)(ex−1x ) +(e x−logx )(2 x+7), 177

If y=(6 sinx log10 x ) ,find dydx . Solution:By product rule,dydx=6 {sinx ddx ( log10 x )+log10 xddx

(sinx)}

= 6 {sinx ( 1x ) log10e+ log10 x (cosx)} If y=(ex logxcotx) ,find dydx . Solution:By product rule,

dydx

=ex logx ddx

(cotx )+ex cotx ddx

( logx )+ logxcotx ddx

(ex )

= ex logx(−cosec2 x)+ex cotx( 1x ) logx cotx ex. Differentiatesin2 xcos3 x. Solution: Let y=sin2 x cos3 xdydx

=sin2 x ddx

(cos3 x )+cos3 x ddx

(sin2 x )

¿sin2 x (−sin 3 x ) .3+cos3 x [2 sinx . ddx

( sinx ) ]

¿−3sin2 x sin 3 x+cos 3x [2 sinx cosx ] ¿ sinx[−3 sinx sin 3 x+2cosx cos3x ] Differentiate e4 xsin 4 x. Solution: Let y=¿ e4 xsin 4 xdydx

=e4 x ddx

(sin 4 x )+sin 4 x ddx

(e4 x)¿e4 x [cos 4 x ] ddx

(4 x )+sin 4 x [e¿¿ 4 x . ddx(4 x )]¿

¿e4 x [cos 4 x ](4)+sin 4 x [e¿¿4 x .4 ]¿

178

¿4 e4x [cos 4 x+sin 4 x ]Quotient Rule for DifferentiationLet u & v be differentiable functions of x, then uv is also differentiable d(uv ) = v u,−uv ,

v2

Find y’ , if y= x2

1+ x2 . S olution:

By quotient rule, dydx

=(1+x2 ) d

dx(x2)−x2 d

dx(x2+1 )

(1+x2 )2 dydx= y’= (1+x2 ) (2x )−x2(2 x)(1+x2)2

= 1(1+x2)2

Find y’ , if y= x2−11+x2

. Solution:By quotient rule, dydx

=(1+x2 ) d

dx(x2−1 )−(x2−1) d

dx(x2+1 )

(1+x2 )2 dydx

= y ’=(1+x2 ) (2x )−(x2−1)(2 x )

(1+x2)2= 4 x

(1+ x2)2

Find y’ , if y= 2x−34 x+5 . Solution:By quotient rule ,dydx

=(4 x+5 ) d

dx(2 x−3 )−(2 x−3) d

dx(4 x+5 )

(4 x+5 )2

179

dydx

= y ’= (4 x+5 ) (2 )−(2 x−3 )4(4 x+5)2

¿ 8 x+10−8 x+12(4 x+5)2

¿ 22(4 x+5)2

Find y’ , if y= logxsinx . Solution:By quotient ruledydx

=(sinx ) d

dx( logx )−(logx) d

dx( sinx )

( sinx )2 dydx

= y ’=sinx(1x )−logx(cosx )

(sinx)2

¿ sinx−x cosxlogxx sin2 x

Find y’ , if y= log x2ex

. Solution: y=2 logxex By quotient rule, dydx

=(ex ) d

dx(2 logx )−(2 logx) d

dx(ex )

(ex)2

dydx

= y ’=ex (2 1x )−2 logx(ex )

(ex)2=2 ex−x (2 logx(ex ))

x (ex)2 ¿2e x(1−xlogx)

x e2x

180

Find y’ , if y= x2+e x

(cosx+logx) . Solution:By quotient rule, dydx

=(cosx+logx) d

dx( x2+e x)−(x2+ex ) d

dx(cosx+logx)

(cosx+logx)2 dydx

= y ’= (cosx+logx ) (2x+e x)+( x¿¿2+ex)(−sinx+ 1x )

(cosx+logx)2¿

Find y’ , if y= sinx+cosxsinx−cosx . Solution:By quotient rule, dydx

=(sinx−cosx ) d

dx(sinx+cosx )−(sinx+cosx) d

dx( sinx−cosx )

(sinx−cosx )2

dydx= y ’=

( sinx−cosx ) (cosx−sinx )−(sinx+cosx )(cosx+sinx)(sinx−cosx)2

¿−(sinx−cosx)2−(sin x+cosx)2

(sinx−cosx)2

Differentiate y= (x7−47)(x−4) . Solution:By quotient rule, dydx

=(x−4 ) d

dx(x7−47 )−(x7−47) d

dx( x−4 )

( x−4 )2

dydx

= y ’=( x−4 ) (7 x6 )−(x7−47 )(1)

(x−4)2

181

¿7 x7−28 x6−x7+47¿ ¿

(x−4)2=(6 x7−28 x6+47)

(x−4)2

Differentiate Y= (x+1)(x¿¿2+1)¿ . S olution:

y= (x+1)( x¿¿2+1)¿

d yd x

=(x2+1 )−( x+1 ) 2x

(x¿¿2+1)2¿ ¿ x2+1−2x2−2x

(x¿¿2+1)2=−x2−2 x+1(x¿¿2+1)2¿

¿

Differentiate x2−x+1x2+x+1

Solution:Let y= x

2−x+1x2+x+1

dydx

=(x2+x+1 ) (2 x−1 )−(x2−x+1 )(2 x+1)

(x2+ x+1 )2

¿(2x3−x2+2 x2−x+2 x−1 )−(2x3+x2−2 x2−x+2x+1 )

(x2+x+1 )2¿ 2x

3+x2+x−1−2 x3+x2−x−1(x2+x+1 )2

¿ 2 x2−2(x2+x+1 )2

= 2(x2−1)(x2+x+1 )2

Differentiate secxlogx . Solution:

182

Let y= secxlogx

dydx

=logx ( secx tanx )−secx . 1

x( logx )2

¿ x logx secx tanx−secxx (logx )2

¿s ecx [xtanx logx−1]

x (logx )2

Find the derivative of ex+e− xex−e− x

. Solution: Let y= ex+e−x

e x−e−x

dydx

=(ex−e− x) (ex−e−x )−(ex+e−x)(ex+e−x )

(ex−e− x)2¿

(ex−e− x )2−(e x+e−x )2

(e x−e−x )2

¿ (e2x+e−2x−2)−(e2x+e−2x+2)(ex−e− x )2

¿ (e2x+e−2x−2−e2x−e−2 x−2)(ex−e− x)2

= −4(ex−e−x )2

Differentiate y= t e2 t

2costwith respect ¿ t '.Solution:

Given y= t e2 t

2cost Let u=t e2 t , v=2cost dudt

=( t ) (2e2 t )+(e2 t ) (1 )=2t e2 t+e2 t , dvdt

=−2 sint

∴ dydt

= v u'−uv 'v2

¿(2cost ) [2 t e2t+e2t ]−(t e2 t )(−2 sint )

(2cost )2¿ 4 t e

2 t cost+2 e2 t cost+2t e2 t sint4 cos2 t

¿2e2 t [2t cost+cost+t sint ]

4 cos2 t(i .e ) dy

dt= e2 t

2cos2 t(2t cost+cost+ t sint )

Differentiation of Parametric functions If x and y are expressed in terms of a third variable t , then the third variable is called the parameter, equation containing a parameter is known as parametric equation.

183

(ie) If x = f(t), y = g(t) then dydx = dydtdxdt

If x=√ t , y=t+ 1t ,then find dydx . Solution: dydx

=dy /dtdx /dt

x=√ t⟹ dxdt

= 12√ t

y=t+ 1t⟹ dy

dt=1−1

t 2 dydx

=(1− 1

t 2)

12√t

=2 (t 2−1 )√ t

t 2=2

(t 2−1 )

t32

If x=a (1+cosθ); y=a(θ+sinθ¿ then, find dydx . Solution: dxdθ

=a (0−Sinθ )=−a sin θ dydθ

=a (1+cosθ)

dydx

=dy /dθdx /dθ

=a(1+cosθ)−a sin θ

¿−cos2 θ

2

sin θ2cos θ

2

¿−cot θ2

Find dydx , if x=a¿. Solution:184

Given x=a ( t−sint ) , y=a (1−cost )dxdt

=a (1−cost ) , dydt

=a(0+sint )

∴ dydx

=

dydtdxdt

=a(sint )

a(1−cost)= sint1−cost

Find dydx , if x=ct , y= ct . Solution:Given x=ct , y= c

tdxdt

=c , dydt

=c (−1)t2

∴ dydx

=

dydtdxdt

=−ct 2× 1c=−1t 2

Find dydx , if x=acost , y=bsint . Solution: Given x=acost , y=bsintdxdt

=−asint , dydt

=bcost∴ dydx

=

dydtdxdt

= bcost−asint

=−bacot t

Find dydx , if x=acos2t , y=b sin2 t. Solution: Given x=acos2t , y=b sin2 t dxdt

=a2cost (−sint ) , dydt

=2bsint cost∴ dydx

=

dydtdxdt

= 2b sint cost−2acost sint

=−ba

Logarithmic Differentiation

185

Take the logarithm of the given function, then differentiate. This method is useful for those functions in which the base and index both are variables. Differentiate Y= sinx x. Solution: y= s inx xTaking log on both sides ,we get,log y= log sinxxlog y=xlogsinx

Differentiate y=x√x . Solution: Let y= x√xTaking log on both sides ,we get, log y=¿ logx√x log y=¿ √ x logxDifferentiating both sides w.r.to x, 1y dydx=√x 1x + logx( 1

2√ x )

¿ 1√x (1+ logx2 )¿x√ x( 1√ x (1+ logx2 )) Differentiate ( x−2 )(x−1)

( x+1 )( x−3) . Solution:Let y=

( x−2 )(x−1)( x+1 )(x−3) Taking log on both sides

logy=log [ ( x−2 )(x−1)( x+1 )(x−3) ]

¿ log [ ( x−2 ) ( x−1 ) ]−log [ ( x+1 )(x−3)]

186

¿ log (x−2)+ log (x−1)−log (x+1)−log (x−3) Differentiate both sides with respect to x ',1ydydx

= 1x−2

+ 1x−1

− 1x+1

− 1x−3

dydx

= y [ 1x−2

+ 1x−1

− 1x+1

− 1x−3 ]

¿ [ ( x−2 )(x−1)( x+1 )(x−3) ] [ 1

x−2 +1x−1−

1x+1−

1x−3 ]

Find dydx , given y=√( x−1 ) ( x−2 ) ( x−3 )(x−4) . Solution: Given y=√( x−1 ) ( x−2 ) ( x−3 )(x−4)

By taking log on both sides log y=1

2log [ ( x−1 ) ( x−2 ) ( x−3 )(x−4)]¿ 12 [ log ( x−1 )+log ( x−2 )+ log ( x−3 )+ log (x−4)]

Differentiate both sides with respect to x '1ydydx

=12 [ 1x−1

+ 1x−2

+ 1x−3

+ 1x−4 ]∴ dy

dx= y . 1

2 [ 1x−1

+ 1x−2

+ 1x−3

+ 1x−4 ]

¿ (√( x−1 ) ( x−2 ) ( x−3 )(x−4)) . 12 [ 1x−1

+ 1x−2

+ 1x−3

+ 1x−4 ]

If xm yn=( x+ y )m+n , then show that y ,= yx . Solution: Given xm yn=( x+ y )m+n

Taking log on both sides log (xm yn )=log ( x+ y )m+n

log xm+log yn= (m+n ) log (x+ y) mlogx+n logy=(m+n ) log (x+ y)

187

Differentiate both sides with respect to x , m . 1x +n . 1y . y ,= (m+n ) . 1

x+ y(1+ y ,)

⟹ ny. y ,− (m+n )

x+ y. y ,= (m+n )

x+ y−mx⟹ y ,( ny−m+n

x+ y )= x (m+n )−m(x+ y )x (x+ y)

⟹ y ,( n ( x+ y )− y (m+n)y ( x+ y ) )= x (m+n )−m(x+ y )

x (x+ y )

⟹ y ,[ nx+ny−my−nyy (x+ y ) ]=mx+nx−mx−myx (x+ y)⟹ y ,[ nx−myy (x+ y) ]= nx−my

x (x+ y)⟹ y ,= y

x

Differentiate (tanx)secx. Solution:Let y=(tanx)secx

Taking log on both sides logy=log(tanx)secxlogy=secx log (tanx) Differentiate both sides with respect to x ,

1ydydx

=secx 1tanx

( sec2 x )+secxtanx . log ( tanx)

⟹ dydx

= y { 1cosx

cosxsinx

( sec2 x )+secxtanx . log ( tanx)}¿( tanx)secx {cosecx (sec2 x )+secxtanx . log (tanx)}

If x y= y x , then prove that dydx=y ( y−xlogy)x (x− ylogx) .

Solution: Given x y= y x Taking log on both sides188

y logx=x logy Differentiate both sides with respect to x ' y ( 1x )+logx . dydx=x . 1y dydx +logy .1⟹ y

x+ dydx. logx= x

y. dydx

+ log y dydx ( log x− x

y )=log y− yx

dydx ( y log x−xy )= x logy− y

x∴ dydx

= y (xlogy− y)x ( ylogx−x)

= y ( y−xlogy)x (x− ylogx)

Find dydx , if (cosx ) y=(siny )x. Solution: Given (cosx ) y=(siny )x

By taking log on both sides log [ (cosx ) y ]=log [ ( siny ) x ] ⟹ y log ( cosx )=x log (siny)

Differentiate both sides with respect to x 'y [ 1cosx

. (−sinx )]+ log (cosx ) . dydx

=x . 1siny

(cosy ) dydx

+ log ( siny ) .1

dydx [ log (cosx )−xcoty ]=log ( siny )+ y tanx⟹ dy

dx= log (siny )+ y tanxlog (cosx )−xcoty

If x y=ex− y Provethat y ,= logx(1+logx)2

. Solution: Given x y=e x− y

By taking log on both sides log x y=loge x− y

189

y logx=( x− y ) . loge ⟹ y logx=( x− y ) --------------------------(1) Differentiate both sides with respect to x,by product ruley . ddx

( logx )+logx . dydx

=(1−dydx )⟹ y . 1x+logx dy

dx=1−dy

dx ⟹ logx . dy

dx+ dydx

=1− yx ⟹ dy

dx(1+ logx )=1− y

x−−−−−−−−−−(2)

From (1) , we get, y logx+ y=x ⟹ y (1+logx )=x

⟹ y= x1+ logx∴Equation (2 )⟹ dy

dx(1+logx )=1−

( x1+logx )x

¿1− 11+logx

¿ 1+logx−11+logx

¿ logx1+logx

∴ dydx

= logx(1+ logx )2

Differentiate x3 x+2with respect ¿ x. Solution: Let y=x3 x+2 By taking log on both sides log y=log x3 x+2 (i .e ) logy=(3 x+2 )logx

Differentiate both sides with respect to x , ∴ 1ydyd x

=(3 x+2 )( 1x )+ (logx ) (3 )Hence , dydx

= y [3 x+2x +3 logx ]¿ x3x+2[ 3x+2x +3 logx ]190

Differentiation of Implicit functions If the relation between x and y is given by an equation of the form f(x,y) = 0, then the function is called implicit function Find dydx if x2a2+ y2

b2 = 1. Solution:Given, x2a2

+ y2

b2 = 1, Differentiatingboth sidesw . r .¿ x

1a2 (2 x )+ 1b2(2 y¿ dydx =0

2 yb2dydx

=−2 xa2

dydx

=−2xa2

x b2

2 y=−( ba )

2

( xy ) Differentiate xy2 ¿k . (L4) Solution: Given , xy2 ¿k , Differentiatingboth sidesw . r .¿ xx (2 y dydx )+ y2 .1=0 2 xy dy

dx=− y2

dydx

=− y2 x

Find dydx , given x2+ y2+x+ y+λ=0. (L4) Solution:

191

Given x2+ y2+x+ y+λ=0 Differentiate both sides with respect to x , 2 x+2 y dy

dx+1+ dy

dx+0=0

⟹ dydx

(1+2 y )=−(1+2x )∴ dydx

=−(1+2x )1+2 y

Maxima (or) MinimaStationary point (or) Turning point The point at which the function changes its nature is called the turning point. Stationary points on a graph where the gradient is zero.Maxima (or) Minima: At a point where the function changes from an increasing function to a decreasing function , the function attains its maximum value (ie) the value of the function that point is greater than all other values in the neighbourhood on either side of the turning point.Working rule to find maxima or minima of a given function; Find dydx and equate it to zero Find the roots of dydx = 0. Let it be a1 , a2… ..an. These points x=

a1 , x=a2…. x=anare called turning points Find d2 y

dx2

Find( d2 ydx2

¿¿at x=a1, ( d2 ydx2 ¿¿at x=a2,…… ( d2 ydx2

¿¿at x=an

If ( d2 ydx2

¿¿at x=a1= _ ve, then we have a max value at x=a1 = +ve, then we have a min value at x=a1 If ( d2 y

dx2¿¿at x=a1= 0 then

192

find ( d3 ydx3

¿¿at x=a1= _ ve, then we have a max value at x=a1 = +ve, then we have a min value at x=a1

Define stationary points. Solution: Stationary points are points on a graph where the gradient is zero. Find the stationary points on the graph of y=2 x2+4 x3. (L1)Solution:Given y=2x2+4 x3 dydx

=4 x+12 x2 At stationary points , dy

dx=0

⟹4 x+12 x2=0 ⟹4 x (1+3 x)=0 ⟹4 x=0(¿)(1+3 x)=0

⟹ x=0(¿) x=−13

d2 ydx2

=4+24 xWhen x=0 , d2 ydx2

=4>0 is positive.

⟹ x=0 is a point of minimum value.When x=−1

3, d

2 ydx2

=−4<0is negative .

⟹ x=−13is a point of maximumvalue .

At x=0⟹ y=0.

193

At x=−13⟹ y=2( 19 )+4(−127 )= 2

27

∴Maximum point is(−13 , 227 ) Minimum point is (0,0) Determine the stationary points on the graph of y=x3-3x+1. State their nature. (L6)Solution: Given y=¿ x3-3x+1dydx

=3 x2−3

At stationary points , dydx

=0

⟹3x2−3=0⟹3 (x2−1 )=0⟹ x2=1

⟹ x=±1d2 ydx2

=6 x .

When x=1 , d2 ydx2

=6>0 is positive. ⟹ x=1is point of minimum value .

When x=−1 , d2 ydx2

=−6<0is negative .

⟹ x=−1 isa point of maximumvalue .

At x=1 ; y=−1 At x=−1 , y=3.

∴ Maximum point is ¿,3). Minimum point is (1 ,−1).Leibnitz theorem If u and v be any two functions of x,then the nth derivative of the function of y=uv is

194

Dn(uv )=¿Dn(u ) v+n c1Dn−1 (u )D (v )+nc2D

n−2 (u )D2 ( v )±−−−−−−−∓UDn(v) It is useful for finding the nth differential coefficient of a product State Leibnitz theorem. (L1) Solution:If u and v be any two functions of x,then the nth derivative of the function of y=uv is Dn(uv )=¿Dn(u ) v+n c1D

n−1 (u )D (v )+nc2Dn−2 (u )D2 ( v )±−−−−−−−∓UDn

(v) Find the nth differential coefficient of y=xm. (L1) Solution: Let y=xm ,Then y1=mxm−1 , y2=m (m−1 ) xm−2 ,

y3=m (m−1 )(m−2) xm−3 In general, yn=m (m−1 ) (m−2 )…(m−n+1)xm−n

Determine the nth differential coefficient of x2eaxusing Leibnit z ' s rule . (L6)Solution:Leibnit z ' s Formula is Dn(uv )=¿Dn(u ) v+n c1D

n−1 (u ) D (v )+nc2Dn−2 (u )D2 ( v )±−−−−−−−∓uDn(v)

Let y=x2eax , u= eax, v=x2D(u)=aeax ,D (v )=2 x D2 (u)=¿a2eax, D2 (v )=2 D3 (u )=a3 eax D3 (v )=0

195

. . . . . . . . . . . . . . . . . . Dn (u )=an eax, Dn(uv )=¿Dn(u ) v+n c1D

n−1 (u )D (v )+nc2Dn−2 (u )D2 ( v )±−−−−−−−∓uDn(v) =aneax x2+¿ nc1an−1 eax 2x+ nc2an−2 eax 2+0 =an x2 eax+¿ nc12x an−1 eax+¿ nc22 an−2eax .

Determine the nth differential coefficient of x3 loge x ,using Leibnitz’s rule. (L6)Solution: Let y=x3 loge x u=logx v=¿ x3D (u )=1

xD(v)=3 x2D2 (u )=−1

x2D2 (v )=6 xD3 (u )= 2

x3D3 ( v )=6

D4 (u )=−6x4D4 ( v )=0

D5 (u )=24x5

D6 (u )=−120x6

Dn (u )=(−1)n−1 (n−1 ) !xn

,Dn−1 (u )=¿¿….. Leibnitz’s rule is Dn(uv )=¿Dn(u ) v+n c1D

n−1 (u )D (v )+nc2Dn−2 (u )D2 ( v )±−−−−−−−∓uDn( v )

Dn(uv )=¿Dn(u ) v+n c1D

n−1 (u ) D (v )+nc2Dn−2 (u )D2 ( v )+nc3D

n−3 (u )D3 ( v )+nc4Dn−4 (u )D4 (v )

¿(−1)n−1 (n−1 )!

xnx3+nc1¿¿

196

197