Ex-2-5-FSc-part2-ver-2-2-6
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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: 2.2.6 Some Important Derivative Formulas 0 d c dx • = where c is constant 1 n n d x nx dx - • = sin cos cos sin d x x dx d x x dx • • = =- 2 2 tan sec cot csc d x x dx d x x dx • • = =- csc csc cot sec sec tan d x x x dx d x x x dx • • =- = 1 2 1 2 1 1 1 1 d Sin x dx x d Cos x dx x - - • = - - • = - 1 2 1 2 1 1 1 1 d Tan x dx x d Cot x dx x - - • • = + - = + 1 2 1 2 1 1 1 1 d Sec x dx x x d Csc x dx x x - - • = - - • = - Question # 1 Solution 1(i) Suppose sin 2 y x = sin 2( ) y y x x δ δ ⇒ + = + sin 2( ) y x x y δ δ ⇒ = + - sin 2( ) sin 2 x x x δ = + - Dividing both sides by x δ sin(2 2 ) sin 2 y x x x x x δ δ δ δ + - = 2 2 2 2 2 2 2cos sin 2 2 x x x x x x x δ δ δ + + + - = ( ) ( ) 2cos 2 sin x x x x δ δ δ + = Taking limit as 0 x δ → ( ) ( ) 0 0 2cos 2 sin lim lim x x x x x y x x δ δ δ δ δ δ δ → → + = ( ) ( ) 0 sin 2 lim cos 2 x x dy x x dx x δ δ δ δ → = + ⋅ ( ) ( ) 0 0 sin 2 lim cos 2 lim x x x x x x δ δ δ δ δ → → = + ⋅
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Transcript of Ex-2-5-FSc-part2-ver-2-2-6
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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: 2.2.6
Some Important Derivative Formulas
0d cdx = where c is constant 1n nd x nxdx
=
sin cos
cos sin
dx x
dxd
x xdx
=
=
2
2
tan sec
cot csc
dx xdx
dx xdx
=
=
csc csc cot
sec sec tan
dx x xdx
dx x xdx
=
=
1
2
1
2
11
11
d Sin xdx xd Cos xdx x
=
=
12
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
Solution 1(i) Suppose sin2y x=
sin2( )y y x x + = +
sin2( )y x x y = +
sin2( ) sin2x x x= +
Dividing both sides by x
sin(2 2 ) sin 2y x x xx x
+ =
2 2 2 2 2 22cos sin2 2x x x x x x
x
+ + +
=
( ) ( )2cos 2 sinx x xx
+=
Taking limit as 0x
( ) ( )0 0
2cos 2 sinlim limx x
x x xyx x
+=
( ) ( )0sin
2 lim cos 2x
xdyx xdx x
= +
( ) ( )0 0sin
2 lim cos 2 limx x
xx x
x = +
-
FSc-II / Ex- 2.5 - 2
( )2 cos 2 0 (1)x= + 0
sinlim 1
=
2cos2dy xdx
=
Solution 1(ii) Let tan3y x= ( )tan3y y x x + = + ( )tan 3 3 tan3y x x x = +
( )( )
sin 3 3 sin3cos 3 3 cos3
x x x
x x x
+=
+
( ) ( )( )
sin 3 3 cos3 cos 3 3 sin3cos 3 3 cos3
x x x x x x
x x x
+ +=
+
( )( )
sin 3 3 3cos 3 3 cos3
x x x
x x x
+ =
+
( )( )
sin 3cos 3 3 cos3
x
x x x
= +
Dividing by x ( )
( )sin 31
cos 3 3 cos3xy
x x x x x
= +
Taking limit as 0x
( )( )0 0sin 3
lim limcos 3 3 cos3x x
xyx x x x x
= +
( )( )0
sin 3 1 3limcos 3 3 cos3 3x
xdydx x x x x
= + ing and ing 3 on R.H.S
( )( )0 0
sin 3 13 lim lim3 cos 3 3 cos3x x
x
x x x x
= +
( )13(1)
cos 3 3(0) cos3x x= +
3cos3 cos3x x
= 23
cos 3x=
23sec 3dy xdx
=
Solution 1(iii) Let sin2 cos2y x x= + ( ) ( )sin 2 cos2y y x x x x + = + + + ( ) ( )sin 2 cos2y x x x x y = + + +
( ) ( )sin 2 cos2 sin 2 cos2x x x x x x = + + + ( ) ( )sin 2 2 sin 2 cos 2 2 cos2x x x x x x = + + +
2 2 2 2 2 22cos sin2 2
x x x x x x + + + =
2 2 2 2 2 22sin sin2 2
x x x x x x + + + +
-
FSc-II / Ex- 2.5 - 3
( ) ( ) ( ) ( )2cos 2 sin 2sin 2 sinx x x x x x = + + Dividing by x
( ) ( ) ( ) ( )1 2cos 2 sin 2sin 2 siny x x x x x xx x
= + + Taking limit as 0x ( ) ( ) ( ) ( )
0 0
1lim lim 2cos 2 sin 2sin 2 sinx x
yx x x x x x
x x
= + +
( ) ( ) ( ) ( )0 0 0 0
sin sin2 lim cos 2 lim 2 lim sin 2 lim
x x x x
x xdyx x x x
dx x x = + +
( ) ( )2cos 2 0 (1) 2sin 2 0 (1)x x= + + Since 0
sinlim 1
=
2cos2 2sin 2dy x xdx
=
Solution 1(iv)
Let 2cosy x= 2cos( )y y x x + = +
2 2cos( ) cosy x y x = +
2 2 2 2( ) ( )2sin sin2 2
x x x x x x + + + =
2 2 2 2 2 22 22sin sin2 2
x x x x x x x x x x + + + + + =
2 2 22 2 22sin sin2 2
x x x x x x x + + +=
222sin sin
2 2x x
x x x x x = + + +
Dividing by x
221 2sin sin
2 2y x x
x x x x xx x
= + + +
ing and ing 2x
x
+
on R.H.S
222 2sin sin
2 22
xx
y x xx x x x x
xx xx
+ = + + + +
-
FSc-II / Ex- 2.5 - 4
22
sin22sin
2 22
xx x
x xx x x x
xx x
+ = + + +
+
Taking limit as 0x
22
0 0
sin2lim lim 2sin
2 22
x x
xx x
y x xx x x x
xxx x
+ = + + +
+
22
0 0 0
sin22 lim sin lim lim
2 22
x x x
xx xdy x x
x x x xxdx
x x
+ = + + + +
( ) ( )22sin (0) (0) (1) (0)x x= + + +
22 sindy x xdx
=
Solution 1(v)
Let 2tany x= ( )2tany y x x + = + ( )2 2tan tany x x x = + ( )( ) ( )( )tan tan tan tanx x x x x x = + + +
( )( ) ( )( )sin sin
tan tancos cos
x x xx x x
x x x
+
= + + +
( )( ) ( ) ( )( )sin cos sin cos
tan tancos cos
x x x x x xx x x
x x x
+ +
= + + +
( )( ) ( )( )sin
tan tancos cos
x x xx x x
x x x
+
= + + +
( )( ) ( )sin
tan tancos cos
xx x x
x x x
= + + +
Dividing by x
( )( ) ( )1 sin
tan tancos cos
y xx x x
x x x x x
= + + +
Taking limit when 0x
( )( ) ( )0 01 sinlim lim tan tan
cos cosx x
y xx x x
x x x x x
= + + +
-
FSc-II / Ex- 2.5 - 5
( )( )0 0
tan tan sinlim limcos cosx x
x x xdy xdx x x x x
+ + = +
( )( ) ( )
tan 0 tan1
cos 0 cosx x
x x
+ +=
+
tan tancos cos
x x
x x
+=
22tancos
x
x=
22tan secdy x xdx
=
Solution 1(vi)
Let tany x=
( )tany y x x + = + ( )tan tany x x x = +
( )( ) ( )( )tan tan
tan tantan tan
x x xx x x
x x x
+ + = + + +
( )( )
tan tan
tan tan
x x x
x x x
+ =
+ +
( )( )( )
sin1 sincos costan tan
x x x
x x xx x x
+=
++ +
Now do yourself as above.
Solution 1(vii)
Let cosy x=
cosy y x x + = +
cos cosy x x x = +
2sin sin2 2
x x x x x x + + + =
Dividing by x
2sin sin2 2
x x x x x x
yx x
+ + +
=
Taking limit as 0x
0 0
sin sin2 2
lim 2 limx x
x x x x x x
yx x
+ + +
=
As ( )( )x x x x x x x = + + + , putting in above
-
FSc-II / Ex- 2.5 - 6
( )( )0sin sin
2 22 lim
x
x x x x x x
dydx x x x x x x
+ + + =
+ + +
( )0 0sin sin
2 2lim lim
2
x x
x x x x x x
x x xx x x
+ + +
=
+ + +
( )
0sin
2 (1)0
x x
x x
+ +
=
+ +
( )sin2
xdydx x
=
Question # 2
Solution # 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= +
Solution # 2(ii)
Let 3 2tan secy = Diff. w.r.t
3 2tan secdy dd d
=
3 2 2 3tan sec sec tand d
d d
= +
3 2 2tan 2sec sec sec 3tan tand dd d
= +
( ) ( )3 2 2 2tan 2sec sec tan sec 3tan sec = + ( )2 2 2 2sec tan 2 tan 3sec = +
-
FSc-II / Ex- 2.5 - 7
Solution # 2(iii)
Let ( )2sin 2 cos3y = Diff. w.r.t
( )2sin 2 cos3dy dd d
=
( ) ( )2 sin 2 cos3 sin 2 cos3dd
=
( )2 sin 2 cos3 cos 2 (2 ) sin 3 (3 )d dd d
= +
( )( )2 sin 2 cos3 cos2 (2) sin3 (3) = + ( )( )2 sin 2 cos3 2cos2 3sin3 = +
Solution # 2(iv)
Let cos siny x x= +
( )1 12 2cos( ) sinx x= + Diff. w.r.t x
( )1 12 2cos( ) sindy d x xdx dx
= +
( ) ( )1 1 12 2 21sin( ) sin sin2
d dx x x x
dx dx
= +
( ) ( )1 1 12 2 21 1sin( ) sin cos2 2
x x x x
= +
1 cos sin2 sin
x x
x x
=
Question # 3
Find dydx
if (i) cosy x y= (ii) sinx y y= Solution 3(i) Since cosy x y=
cosdy d
x ydx dx
=
cos cosd dx
x y ydx dx
= +
( sin ) cos (1)dyx y ydx
= +
sin cosdy dyx y ydx dx
+ = ( )1 sin cosdyx y ydx
+ =
cos
1 sindy ydx x y
=+
-
FSc-II / Ex- 2.5 - 8
Solution 3 (ii) Since sinx y y=
( sin )dx d y ydx dx
=
1 sin sind dyy y ydx dx
= +
cos sindy dyy y ydx dx
= + ( cos sin ) dyy y ydx
= +
1cos sin
dydx y y y
=+
Question # 4 Find the derivative w.r.t. x
(i) 1cos1 2
x
x
+
+ (ii) 1 2sin
1x
x
+
+
Solution 4(i)
Since 1cos1 2
xyx
+=
+
Diff. w.r.t x
1cos
1 2dy d xdx dx x
+=
+
1 1sin
1 2 1 2x d xx dx x
+ +=
+ +
121 1
sin1 2 1 2
x d xx dx x
+ + = + +
121 1 1 1
sin1 2 2 1 2 1 2
x x d xx x dx x
+ + + = + + +
( ) ( ) ( ) ( )( )2
12 1 2 1 1 1 21 1 1 2
sin1 2 2 1 1 2
d dx x x x
x x dx dxx x x
+ + + + + +
= + + +
( )( )
( )( ) ( )( )( )2
12
12
1 2 1 2 1 1 21sin
1 2 1 22 1
x x xx
x xx
+ + ++=
+ ++
( )( ) ( )2
12
12
1 21 1 2 2 2sin
1 2 1 22 1
xx x x
x xx
++ + =
+ ++
( )( ) ( )2
12
12
1 21 1sin
1 2 1 22 1
xx
x xx
++ =
+ ++
-
FSc-II / Ex- 2.5 - 9
( )( ) ( ) 2
12
1 12 2
1 21 1sin
2 1 2 2 1 1 2
xx
x x x
++=
+ + +
( ) 321 1
sin1 22 1 1 2
dy xdx xx x
+ =
++ +
Solution 4(ii) Do yourself as above.
Question # 5
Differential (i) sin x w.r.t cot x (ii) 2sin x w.r.t 4cos x . Solution 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
csc
dxdu x
=
2sin x=
Now by chain rule
dy dy dxdu dx du
=
( )( )2cos sinx x= 2sin cosx x=
Solution 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=
-
FSc-II / Ex- 2.5 - 10
31
4sin cosdxdu x x
=
Now by chain rule
dy dy dxdu dx du
=
( ) 312sin cos 4sin cosx x x x
=
21 sec2
x=
Question # 6
If ( )tan 1 tan 1 tany x x+ = , show that 1dydx
= .
Solution
Since ( )tan 1 tan 1 tany x x+ = 1 tan
tan1 tan
xyx
=+
1 tan1 1 tan
x
x
=
+
4
4
tan tan
1 tan tanx
x
pi
pi
=
+
tan4
xpi
=
4y xpi =
Diff. w.r.t x
4dy d
xdx dx
pi =
0 1= 1dy
dx =
Question # 7
If tan tan tan ...y x x x= + + + , prove that ( ) 22 1 secdyy xdx
= .
Solution
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
-
FSc-II / Ex- 2.5 - 11
( )2 tand dy x ydx dx
= +
22 secdy dyy xdx dx
= + 22 secdy dyy xdx dx
=
( ) 22 1 secdyy xdx
=
Question # 8
If 3cosx a = , 3siny b = , show that tan 0dya bdx
+ = .
Solution
3cosx a = , 3siny b = Diff. x w.r.t
( )3cosdx d ad d = ( )23cos cosda
d
=
( )23 cos sina =
23 sin cosdx ad
= 21
3 sin cosddx a
=
Now diff. y w.r.t
( )3sindy d bd d = ( )23sin sindb
d
=
23 sin cosb =
Now by chain rule dy dy ddx d dx
=
22
13 sin cos3 sin cos
ba
=
tanba
=
tandy
a bdx
= tan 0dya bdx
+ =
Question # 9
Do yourself
-
FSc-II / Ex- 2.5 - 12
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 pi
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 pi
2
11 x
=
(iii) 1 21
1d Tan xdx x
=
+
See proof on book at page 77 (iv) 1 2
11
d Cot xdx x
=
+
Proof Let 1coty x= cot y x = Diff. w.r.t x
cotd dy xdx dx
= 2csc 1dyy
dx =
21
csc
dydx y
=
21
1 cot y
=
+
2 21 cot cscy y+ =
21
1dydx x
= +
(v) 12
11
d Sec xdx x x
=
Proof Let 1secy x= sec y x = Diff. w.r.t x
-
FSc-II / Ex- 2.5 - 13
secd dy xdx dx
= sec tan 1dyy ydx
=
1sec tan
dydx y y
=
2
1sec sec 1y y
=
2 21 tan secy y+ =
12
11
d Sec xdx x x
=
sec y x=
(vi) 12
11
d Csc xdx x x
=
See on book at page 77
Question # 10
Solution 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
dx
a dxxa
=
2 2
2
1 1 (1)aa x
a
=
2 2
1aaa x
=
2 2
1a x
=
Ans
Solution 10(ii)
Let 1cot xya
=
Diff w.r.t x 1cot
dy d xdx dx a
=
-
FSc-II / Ex- 2.5 - 14
21
1
d xdx ax
a
=
+
( )2 22
1 1 dx
a x a dxa
=
+
( )2
2 21 1a
a x a
=
+ 2 2
a
a x
=
+.
Solution 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 22
1 1 da x
a dxx ax
=
( )22 2x xx a
=
22 2
1xxx a
=
2 2
1x x a
=
Ans
Solution 10(iv)
Let 1 21y Sin x= Diff. w.r.t x
1 21dy d Sin xdx dx
=
( )2
22
1 11 1
dx
dxx
=
( ) ( )2 22121 1 1 1
21 1d
x xdxx
=
+
( ) ( )2 2 121 1 1 2
2 1x
x x
=
2
11
x
x x=
2
11 x
=
Solution 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 x
x x
x x
+=
+ +
-
FSc-II / Ex- 2.5 - 15
( ) ( )( ) ( ) ( ) ( )
( )2 2 2 2
22 2 22 22
2 2 2
1 1 1 1111 11
1 ( 1)
d dx x x x
dx dxxx xx
x x
+ +
= + +
( ) ( )( )( ) ( )( )
( )2 2
224 2 4 22
2 2
1 2 1 2112 1 2 11
1 ( 1)
x x x x
xx x x xx
x 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 21 2 2
1 4x
x x=
+ ( )2
41 2
x
x x
=
+ ( )2
21x
=
+ Ans
Solution 10(vi) Do yourself as above.
Solution 10(vii) Do yourself as above.
Question # 11
Show that dy ydx x
= if 1y xTanx y
= .
Solution
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 Tan
dx y yxy
= + +
12 2 2
2
(1)1dyy x
xdxx Tany x y y
y
= + +
2 2x dy yy x
y x dx x
= + +
-
FSc-II / Ex- 2.5 - 16
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
If ( )1tan ,y pTan x= show that ( ) ( )2 211 1 0.x y p y+ + = Solution
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 =
Error Analyst
Muzammil Ahsan 2009-11 Govt. Post Graduate Collage Jauharabad Distt. Khushab
Nain Malik 2012-14
Muhammad Usman Saleem (2013-15) DPS & IC Jauharabad Distt. Khushab
Abeer Butt (2013-15) FCC
Book: Exercise 2.5
Calculus and Analytic Geometry Mathematic 12 Punjab Textbook Board, Lahore. Edition: May 2013.
Made by: Atiq ur Rehman ([email protected]) Available online at http://www.MathCity.org in PDF Format (Picture format to view online). Page Setup used: A4. Printed: September 25, 2014.
