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mathcity.org Merging man and maths
Exercise 2.5 (Solutions) Calculus and Analytic Geometry, MATHEMATICS 12
Available online @ http://www.mathcity.org, Version: 1.0.2
Some Important Derivative Formulas
0d cdx
= where c is constant 1n nd x nxdx- =
sin cos
cos sin
d x xdxd x xdx
=
=-
2
2
tan sec
cot csc
d x xdxd x xdx
=
=-
csc csc cot
sec sec tan
d x x xdxd x x xdx
=-
=
1
2
1
2
11
11
d Sin xdx xd Cos xdx x
-
-
= - - = -
1
2
12
11
11
d Tan xdx xd Cot xdx x
-
-
=+-=+
1
2
1
2
11
11
d Sec xdx x xd Csc xdx x x
-
-
=-
- =
-
Question # 1(i) Suppose sin 2y x=
sin 2( )y y x xd d + = + sin 2( )y x x yd d = + -
sin 2( ) sin 2x x xd= + - Dividing both sides by xd
sin(2 2 ) sin 2y x x xx x
d dd d
+ -=
2 2 2 2 2 22cos sin2 2x x x x x x
x
d d
d
+ + + -
=
( ) ( )2cos 2 sinx x xx
d dd
+=
Taking limit as 0xd
( ) ( )0 0
2cos 2 sinlim limx x
x x xyx xd d
d ddd d
+=
( ) ( )0sin
2 lim cos 2x
xdy x xdx xd
dd
d= +
( ) ( )0 0sin
2 lim cos 2 limx x
xx x
xd dd
dd
= +
( )2 cos 2 0 (1)x= + 0
sinlim 1q
qq
=Q
2cos2dy xdx
=
Question # 1(ii) Let tan3y x= ( )tan3y y x xd d + = + ( )tan 3 3 tan3y x x xd d = + -
( )( )
sin 3 3 sin3cos 3 3 cos3
x x xx x x
dd
+= -
+ ( ) ( )
( )sin 3 3 cos3 cos 3 3 sin3
cos 3 3 cos3x x x x x x
x x xd d
d+ - +
=+
( )( )
sin 3 3 3cos 3 3 cos3
x x xx x x
dd
+ -=
+ ( )
( )sin 3
cos 3 3 cos3x
x x xdd
=+
Dividing by xd
-
FSc-II / Ex- 2.5 - 2
( )( )
sin 31cos 3 3 cos3
xyx x x x x
ddd d d
= +
Taking limit as 0xd
( )( )0 0sin 3
lim limcos 3 3 cos3x x
xyx x x x xd d
ddd d d
=+
( )( )0
sin 3 1 3limcos 3 3 cos3 3x
xdydx x x x xd
dd d
= +
ing and ing 3 on R.H.S
( )( )0 0
sin 3 13 lim lim3 cos 3 3 cos3x x
xx x x xd dd
d d =
+
( )
13(1)cos 3 3(0) cos3x x
= +
3cos3 cos3x x
= 23
cos 3x=
23sec 3dy xdx
=
Question # 1(iii) Let sin 2 cos2y x x= + ( ) ( )sin 2 cos2y y x x x xd d d + = + + + ( ) ( )sin 2 cos2y x x x x yd d d = + + + -
( ) ( )sin 2 cos2 sin 2 cos2x x x x x xd d= + + + - - ( ) ( )sin 2 2 sin 2 cos 2 2 cos 2x x x x x xd d= + - + + -
2 2 2 2 2 22cos sin2 2
x x x x x xd d + + + - =
2 2 2 2 2 22sin sin2 2
x x x x x xd d + - + - + -
( ) ( ) ( ) ( )2cos 2 sin 2sin 2 sinx x x x x xd d d d= + - + Dividing by xd
( ) ( ) ( ) ( )1 2cos 2 sin 2sin 2 siny x x x x x xx x
dd d d d
d d= + - +
Taking limit as 0xd
( ) ( ) ( ) ( )0 0
1lim lim 2cos 2 sin 2sin 2 sinx x
y x x x x x xx xd d
dd d d d
d d = + - +
( ) ( ) ( ) ( )0 0 0 0
sin sin2 lim cos 2 lim 2 lim sin 2 lim
x x x x
x xdy x x x xdx x xd d d d
d dd d
d d = + - +
( ) ( )2cos 2 0 (1) 2sin 2 0 (1)x x= + - + Since 0
sinlim 1q
qq
=
2cos2 2sin 2dy x xdx
= -
Question # 1(iv) Let 2cosy x=
2cos( )y y x xd d + = + 2 2cos( ) cosy x y xd d = + -
2 2 2 2( ) ( )2sin sin
2 2x x x x x xd d + + + -
= -
-
FSc-II / Ex- 2.5 - 3 2 2 2 2 2 22 22sin sin
2 2x x x x x x x x x xd d d d + + + + + -
= -
2 2 22 2 22sin sin2 2
x x x x x x xd d d d + + += -
222sin sin
2 2x xx x x x xd dd d
= - + + +
Dividing by xd
2
21 2sin sin2 2
y x xx x x x xx x
d d dd dd d
= - + + +
ing and ing 2xx d +
on R.H.S
2
22 2sin sin2 2
2
xxy x xx x x x x
xx x x
dd d dd d
dd d
+ = - + + + +
22
sin22sin
2 22
xx xx xx x x x
xx x
d dd dd
d d
+ = - + + + +
Taking limit as 0xd
2
2
0 0
sin2lim lim 2sin
2 22
x x
xx xy x xx x x x
xx x xd d
d dd d dd
dd d
+ = - + + + +
2
2
0 0 0
sin22 lim sin lim lim
2 22
x x x
xx xdy x xx x x x
xdx x xd d d
d dd dd
d d
+ = - + + + +
( ) ( )22sin (0) (0) (1) (0)x x= - + + + 22 sindy x x
dx = -
Question # 1(v) Let 2tany x= ( )2tany y x xd d + = + ( )2 2tan tany x x xd d = + - ( )( ) ( )( )tan tan tan tanx x x x x xd d= + + + -
( )( ) ( )( )sin sintan tancos cos
x x xx x xx x x
dd
d
+= + + -
+
( )( ) ( ) ( )( )sin cos sin cos
tan tancos cos
x x x x x xx x x
x x xd d
dd
+ - += + +
+
( )( ) ( )( )sin
tan tancos cos
x x xx x x
x x xd
dd
+ -= + +
+
-
FSc-II / Ex- 2.5 - 4
( )( ) ( )sintan tan
cos cosxx x x
x x xd
dd
= + +
+
Dividing by xd
( )( ) ( )1 sintan tan
cos cosy xx x xx x x x x
d dd
d d d
= + +
+
Taking limit when 0xd
( )( ) ( )0 01 sinlim lim tan tan
cos cosx xy xx x xx x x x xd d
d dd
d d d
= + + +
( )( )0 0
tan tan sinlim limcos cosx x
x x xdy xdx x x x xd d
d dd d
+ + = +
( )( ) ( )
tan 0 tan1
cos 0 cosx xx x
+ +=
+ tan tan
cos cosx xx x+
=
22 tancos
xx
=
22 tan secdy x xdx
=
Question # 1 (vi) Let tany x=
( )tany y x xd d + = + ( )tan tany x x xd d = + -
( )( ) ( )( )tan tan
tan tantan tan
x x xx x x
x x x
dd
d
+ + = + - + +
( )( )
tan tantan tan
x x xx x x
d
d
+ -=
+ +
( )
( )( )
sin1 sincos costan tan
x x xx x xx x x
ddd
+= -
++ +
Now do yourself as above.
Question # 1 (vii) Let cosy x= cosy y x xd d + = +
cos cosy x x xd d = + -
2sin sin2 2
x x x x x xd d + + + -= -
Dividing by xd
2sin sin
2 2x x x x x x
yx x
d ddd d
+ + + - = -
Taking limit as 0xd
0 0
sin sin2 2
lim 2 limx x
x x x x x xyx xd d
d ddd d
+ + + - = -
As ( )( )x x x x x x xd d d= + + + - , putting in above
-
FSc-II / Ex- 2.5 - 5
( )( )0
sin sin2 2
2 limx
x x x x x xdydx x x x x x xd
d d
d d
+ + + - = -
+ + + -
( )0 0
sin sin2 2
lim lim
2
x x
x x x x x x
x x xx x xd d
d d
dd
+ + + - = -
+ -+ +
( )
0sin2
(1)0
x x
x x
+ + = -
+ +
( )sin2
xdydx x
= -
Question # 2(i) Assume 2 sec4y x x= Differentiating w.r.t x
2 sec4dy d x xdx dx
=
2 2sec4 sec4d dx x x xdx dx
= +
2 sec4 tan 4 (4 ) sec4 (2 )dx x x x x xdx
= +
2 sec4 tan 4 (4) 2 sec4x x x x x= + ( )2 sec4 2 tan 4 1x x x x= +
Question # 2(ii) Let 3 2tan secy q q= Diff. w.r.t q
3 2tan secdy dd d
q qq q
=
3 2 2 3tan sec sec tand dd d
q q q qq q
= +
3 2 2tan 2sec sec sec 3tan tand dd d
q q q q q qq q
= +
( ) ( )3 2 2 2tan 2sec sec tan sec 3tan secq q q q q q q= + ( )2 2 2 2sec tan 2 tan 3secq q q q= +
Question # 2(iii) Let ( )2sin 2 cos3y q q= - Diff. w.r.t q
( )2sin 2 cos3dy dd d
q qq q
= -
( ) ( )2 sin 2 cos3 sin 2 cos3dd
q q q qq
= - -
( )2 sin 2 cos3 cos 2 (2 ) sin3 (3 )d dd d
q q q q q qq q
= - +
( )( )2 sin 2 cos3 cos2 (2) sin3 (3)q q q q= - + ( )( )2 sin 2 cos3 2cos2 3sin3q q q q= - +
-
FSc-II / Ex- 2.5 - 6 Question # 2(iv) Let cos siny x x= +
( )1 12 2cos( ) sinx x= +
Diff. w.r.t x
( )1 12 2cos( ) sindy d x x
dx dx = +
( ) ( )1 1 12 2 21sin( ) sin sin
2d dx x x xdx dx
-= - +
( ) ( )1 1 12 2 21 1sin( ) sin cos
2 2x x x x- - = - +
1 cos sin2 sin
x xx x
= -
Question # 3(i) Since cosy x y=
cosdy d x ydx dx
=
cos cosd dxx y ydx dx
= +
sin cos (1)dyx y ydx
= +
sin cosdy dyx y ydx dx
- = ( )1 sin cosdyx y ydx
- =
cos1 sin
dy ydx x y
=-
Question # 3(ii) Do yourself as above
Question # 4(i)
Since 1cos1 2
xyx
+=
+
Diff. w.r.t x
1cos1 2
dy d xdx dx x
+=
+
1 1sin1 2 1 2
x d xx dx x
+ += -
+ +
121 1sin
1 2 1 2x d xx dx x
+ + = - + +
121 1 1 1sin
1 2 2 1 2 1 2x x d xx x dx x
-+ + + = - + + +
( ) ( ) ( ) ( )
( )212 1 2 1 1 1 21 1 1 2sin
1 2 2 1 1 2
d dx x x xx x dx dxx x x
+ + - + + + + = - + + +
( )( )
( )( ) ( )( )( )2
12
12
1 2 1 2 1 1 21sin1 2 1 22 1
x x xxx xx
+ + - ++= - + ++
-
FSc-II / Ex- 2.5 - 7
( )( ) ( )
2
12
12
1 21 1 2 2 2sin1 2 1 22 1
xx x xx xx
++ + - -= - + ++
( )( ) ( )
2
12
12
1 21 1sin1 2 1 22 1
xxx xx
++ -= - + ++
( )( ) ( ) 2
12
1 12 2
1 21 1sin2 1 2 2 1 1 2
xxx x x -
++=
+ + +
( )
32
1 1sin1 22 1 1 2
dy xdx xx x
+ =
++ +
Question # 4(ii) Do yourself as above.
Question # 5(i) Let siny x= and cotu x= Diff. y w.r.t x
sindy d xdx dx
=
cos x= Now diff. u w.r.t x
cotdu d xdx dx
=
2csc x= -
21
cscdxdu x
= -
2sin x= - Now by chain rule
dy dy dxdu dx du
=
( )( )2cos sinx x= - 2sin cosx x= -
Question # 5(ii) Let 2siny x= and 4cosu x= Diff. y w.r.t x
2sindy d xdx dx
=
( )2sin sindx xdx
= 2sin cosx x=
Now diff. u w.r.t x
4cosdu d xdx dx
=
( )34cos cosdx xdx
= ( )34cos sinx x= -
34sin cosx x= -
31
4sin cosdxdu x x
= -
Now by chain rule
-
FSc-II / Ex- 2.5 - 8
dy dy dxdu dx du
=
( ) 312sin cos
4sin cosx x
x x = -
21 sec2
x= -
Question # 6 Since ( )tan 1 tan 1 tany x x+ = -
1 tantan1 tan
xyx
- =
+
1 tan1 1 tan
xx
-=
+ 4
4
tan tan1 tan tan
xx
p
p-
=+
tan4
xp = -
4y xp = -
Diff. w.r.t x
4dy d xdx dx
p = -
0 1= - 1dydx
= -
Question # 7
Since tan tan tan ...y x x x= + + + Taking square on both sides 2 tan tan tan ...y x x x= + + +
tan tan tan tan ...x x x x= + + + + 2 tany x y = + Diff. w.r.t x
( )2 tand dy x ydx dx
= +
22 secdy dyy xdx dx
= + 22 secdy dyy xdx dx
- =
( ) 22 1 secdyy xdx
- =
Question # 8 3cosx a q= , 3siny b q= Diff. x w.r.t q
( )3cosdx d ad d qq q=
( )23cos cosdad
q qq
= ( )23 cos sina q q= -
23 sin cosdx ad
q qq
= - 21
3 sin cosddx aq
q q-
=
Now diff. y w.r.t q
-
FSc-II / Ex- 2.5 - 9
( )3sindy d bd d qq q=
( )23sin sindbd
q qq
= 23 sin cosb q q=
Now by chain rule dy dy ddx d dx
qq
=
2 213 sin cos
3 sin cosb
aq q
q q= -
tanba
q= -
tandya bdx
q = - tan 0dya bdx
q + =
Question # 9 ( )cos sinx a t t= + and ( )sin cosy a t t t= -
Do yourself
Derivative of inverse trigonometric formulas
(i) 12
11
d Sin xdx x
- =-
See proof on book page 76
(ii) 12
11
d Cos xdx x
- -=-
Proof Let 1cosy x-= where [ ]0,x p cos y x = Diff. w.r.t x
cosd dxydx dx
= sin 1dyydx
- =
1sin
dydx y
= -
2
11 cos y
-=
- Since sin y is positive for [ ]0,x p
2
11 x-
=-
(iii) 1 21
1d Tan xdx x
- =+
See proof on book at page 77
(iv) 1 21
1d Cot xdx x
- -=+
Proof Let 1coty x-= cot y x = Diff. w.r.t x
cotd dy xdx dx
= 2csc 1dyydx
- =
21
cscdydx y
- =
-
FSc-II / Ex- 2.5 - 10
21
1 cot y-
=+
2 21 cot cscy y+ =Q
21
1dydx x
= -+
(v) 12
11
d Sec xdx x x
- =-
Proof Let 1secy x-= sec y x = Diff. w.r.t x
secd dy xdx dx
= sec tan 1dyy ydx
=
1sec tan
dydx y y
=
2
1sec sec 1y y
=-
2 21 tan secy y+ =Q
12
11
d Sec xdx x x
- =-
sec y x=Q
(vi) 12
11
d Csc xdx x x
- = --
See on book at page 77
Question # 10(i)
Let 1 xy Cosa
-=
Diff. w.r.t x 1dy d xCos
dx dx a-=
2
1
1
d xdx ax
a
- = -
2
2
1 1
1
d xa dxx
a
-=
-
2 2
2
1 1 (1)aa x
a
-=
-
2 2
1aaa x
-=
-
2 2
1a x
-=
- Ans
Question # 10(ii)
Let 1 xy Cota
-=
Diff w.r.t x 1dy d xCot
dx dx a-=
21
1
d xdx ax
a
= +
( )2 22
1 1 d xa x a dx
a
= +
( )2
2 21 1a
a x a=
+ 2 2
aa x
=+
-
FSc-II / Ex- 2.5 - 11 Question # 10(iii)
Let 11 ay Sina x
-=
Diff. w.r.t x 11dy d aSin
dx a dx x-=
2
1 1
1
d aa dx xa
x
= -
( )12 2
2
1 1 da xa dxa x
x
-= -
( )22 2
x xa x
-= --
22 21xxa x
= - -
2 2
1x a x
= --
Ans
Question # 10(iv)
Let 1 21y Sin x-= - Diff. w.r.t x
1 21dy d Sin xdx dx
-= -
( )2
22
1 11 1
d xdx
x= -
- - ( ) ( )2 2
2
121 1 1 1
21 1dx xdxx
-= - -
- +
( )
( )2 2
12
1 1 1 22 1
xx x
= --
2
11
xx x
= - -
2
11 x
= --
Question # 10(v)
Let 2
12
11
xy Secx
- += -
Diff. w.r.t x 2
12
11
dy d xSecdx dx x
- += -
2
222 2
2 2
1 111 1 1
1 1
d xdx xx x
x x
+= - + +
- - -
( ) ( )
( ) ( ) ( ) ( )( )
2 2 2 2
22 2 22 22
2 2 2
1 1 1 11
11 111 ( 1)
d dx x x xdx dx
xx xxx x
- + - + - =
- + - - + - -
( ) ( )
( )( ) ( )( )( )
2 2
224 2 4 22
2 2
1 2 1 21
12 1 2 111 ( 1)
x x x x
xx x x xxx x
- - + = -+ + - + + + - -
( )
( )( )
( )
22 2 2
22 4 2 4 2 2
1 2 1 1
1 2 1 2 1 1
x x x x
x x x x x x
- - - - = + + + - + - -
( )
( )( )2 2
1 2 21 4
xx x
= -+
( )2
41 2
xx x
-=
+
( )221x
-=
+ Ans
-
FSc-II / Ex- 2.5 - 12 Question # 10(vi)
Do yourself as above.
Question # 10(vii) Do yourself as above.
Question # 11
Since 1y xTanx y
-= 1 xy x Tany
- =
Diff. w.r.t x 1dy d xx Tan
dx dx y- =
( )1 1d x x dx Tan Tan xdx y y dx
- - = +
( )121 1
1
d x xx Tandx y yx
y
-
= + +
12 2 22
(1)1dyy x xdxx Tan
y x y yy
-
- = + +
2 2x dy yy x
y x dx x = - + +
2
2 2 2 2dy xy x dy ydx y x y x dx x
= - ++ +
2
2 2 2 2dy x dy xy ydx y x dx y x x
+ = ++ +
2 2
2 2 2 21 1x dy y x
y x dx x y x
+ = + + +
dy ydx x
= Proved
Question # 12 Since ( )1tany pTan x-= 1 1Tan y pTan x- - = Differentiating w.r.t x
1 1d dTan y p Tan xdx dx
- -=
2 21 1
1 1dy p
y dx x =
+ + ( ) ( )2 21 1dyx p ydx + = +
( ) ( )2 211 1 0x y p y + - + = Since 1dy ydx =
------ The End ------ 01 September 2005
Made by: Atiq ur Rehman ([email protected]), http://www.mathcity.org
