NM – MI THI B AI PAS T P AP ER QUESTI ONS
STU DENT S AR E RE QU E STED T O T AK E A P RI NT OF THI S FI L E & T HEN PR ACTI SE
01.
= (2x 3 – 7) 5 . log (tan x )
y
02.
= si n (4x 2 – 3)
y
(x 2 – 2) 4
STEP 1 :
d log(tan x)
STEP 1 :
dx
d si n (4x 2 – 3 )
=
1
.
d tan x
tan x
dx
dx
= cos (4x 2 – 3) . d (4x 2 – 3)
=
1
sec 2 x
.
dx
tan x
= cos(4x 2 – 3) . 8x
=
1
.
1
cos 2 x
si n x
STEP 2
cos x
=
d (x 2 – 2) 4
1
dx
si n x . cos x
=
= 4 (x 2 – 2) 3 .
2
d (x 2 – 2)
dx
2 sin x . cos x
=
= 8x . cos(4x 2 – 3)
= 4 (x 2 – 2) 3 . 2x
2
si n 2x
= 8x (x 2 – 2) 3
= 2 . cosec 2x
STEP 3
STEP 2 :
y
d (2x 3 – 7) 5
= si n (4x 2 – 3)
(x 2 – 2) 4
dx
dy =
= 5 (2x 3 – 7) 4 . d (2x 3 – 7)
dx
dx
=
5 (2x 3
–
7) 4
.
dx
6x 2
=
=
= ( 2 x 3 –7 ) 5 . 2 cose c 2
8 x . (x 2 4 ) 4 . co s(4 x 2 – 3 )  . 8 x (x 2 – 2 ) 3 si n( 4 x 2 –3 )
dx
=
( 2 x 3 –7 ) 5 . 2 co se c2 x + l og(ta nx ).3 0 x 2 .( 2x 3 – 7 ) 4
= ( 2 x 3 –7 ) 4
( x 2  4 ) 4 .8 x . co s(4 x 2 – 3 )  sin (4 x 2 – 3 ) . 8 x( x 2 – 2 ) 3
(x 2 – 2) 8
( 2 x 3 – 7 ) 5 d l og( tan x ) + l og ( ta nx ) d (2 x 3 – 7 ) 5
dx
=
2
A R R A N G I N G T H E T E RM S
= (2x 3 – 7) 5 . log (tan x )
dx
–
2) 4
(x 2 – 2) 8
STEP 3 :
dy =
dx
(x 2
= 30x 2 .(2x 3 – 7) 4
y
( x 2  4 ) 4 d si n(4x 2 –3)  sin (4x 2 –3) d ( x 2  4 ) 3
+
8 x (x 2 – 2 ) 3 ( x 2 4 ). co s( 4x 2 – 3 )  si n(4 x 2 – 3 )
(x 2 – 2) 8
30 x 2 .( 2 x 3 – 7 ) 4 .l og( tan x )
=
2 (2 x 3 – 7 )co sec2 x + 3 0x 2 ..l o g( ta nx )
8 x (x 2 4 ). co s(4 x 2 – 3 )  sin (4 x 2 – 3 )
(x 2 – 2) 5
1
03.
y
04.
= x.cos 2 x
y
= log (cos 5x)
x 2 + 3x – 1
(1 + x) 3
STEP 1 :
STEP 1 :
d x.cos 2 x
d log (cos 5x)
dx
= x
dx
d cos 2 x + cos 2 x . d x
dx
=
dx
1
d cos 5x
cos 5x
= x.2 cos x d cos x + cos 2 x . 1
=
dx
1
dx
. ( sin 5x) .
cos 5x
= x. 2 cos x ( sin x) + cos 2 x
=
1
d 5x
dx
. ( sin 5x) . 5
cos 5x
=  x . 2 sin x cos x + cos 2 x
=  5 . tan 5x
= cos 2 x – x.sin2x
STEP 2 :
STEP 2 :
d (1 + x) 3
y
= log (cos 5x)
x 2 + 3x – 1
dx
= 3 (1 + x) 2
d (1 + x)
d yd x
=
dx
( x 2 + 3x– 1 ) d l og( cos 5 x )  l og( cos 5 x ) d (x 2 + 3 x–1 )
= 3 (1 + x) 2
dx
dx
( x 2 + 3x– 1 ) 2
STEP 3 :
y
= ( x 2 + 3x– 1 ) ( 5 .tan 5 x )  l og (cos 5 x ) . ( 2 x+3 )
= x.cos 2 x
( x 2 + 3x– 1 ) 2
(1 + x) 3
dy =
( 1+ x ) 3 d x .co s 2 x  x .co s 2 x d (1+ x ) 3
dx
dx
dx
(1 + x)3
=
=  5 ( x 2 + 3x– 1 ) .ta n 5x  ( 2 x+3 ). l o g( co s5x )
( x 2 + 3x– 1 ) 2
2
( 1+ x ) 3 ( co s 2 x – x .si n2 x )  x.co s 2 x . 3 (1+ x ) 2
(1 + x)6
05.
y
=
x4 + 4x
8 + sinx
A R R A N G I N G T H E T E RM S
=
dy = (8 + sinx) d (x 4 + 4 x )  (x 4 + 4 x ) d (8+si n x)
( 1+ x ) 3 ( co s 2 x – x .si n2 x )  3( 1+ x ) 2 .x .co s 2 x
dx
(1 + x)6
=
( 1+ x ) 2
dx
(8 + sinx) 2
(1+ x ) (cos 2 x – x .sin 2 x )  3 x .cos 2 x
= (8 + sin x)(4x 3 + 4 x .log 4 ) + (x 4 + 4 x ).cosx
(1 + x)6
=
dx
(8 + sin x) 2
( 1+ x ) ( cos 2 x – x .si n2 x )  3x .cos 2 x
(1 + x)4
2
06.
y
= log (si ne x ) +
07.
5 + x 6 . secx
sec 3 x
=
e 4 x .(1+x) 5
STEP 1 :
STEP 1 :
d log (si ne x )
d sec 3 x
dx
=
y
= 3sec 2 x . d sec x
dx
dx
d sin e x
1
si n e x
= 3sec 2 x . sec x . tanx
dx
= 3sec 3 x.tan x
=
1
. cos
ex
si n e x
=
.
d
dx
STEP 2 :
d e 4 x .( 1+x ) 5
. cos e x . e x
1
dx
si n e x
=
= e x . cot e x
=
dx
( 1+ x ) 5 . d e 4 x
dx
e 4 x . 5( 1+x ) 4 d ( 1+ x ) + (1+x ) 5 . e 4 x d 4 x
dx
=
e 4 x . 5( 1+x ) 4 + ( 1+ x ) 5 . e 4 x . 4
=
5 .e 4 x . ( 1+x ) 4 + 4 .e 4 x . ( 1+x) 5
=
e 4 x . ( 1+ x ) 4
5 + 4 .( 1+ x )
=
e 4 x . ( 1+ x ) 4
(9 + 4 x )
d sec x + sec x d 5 + x 6
dx
=
+
dx
5 + x 6 . sec x
= 5 + x6 .
e 4 x . d (1+ x ) 5
dx
STEP 2 :
d
x
ex
dx
5 +x 6 . se cx .ta nx + se cx
1
2 5 +x 6
d (5+ x 6 )
dx
STEP 3 :
=
5 +x 6 . se cx .ta nx + se cx
1
2
6x5
dy =
5 +x 6
e 4 x .( 1+x ) 5 d sec 3 x  se c 3 x d e 4 x .(1+x ) 5
dx
=
5 +x 6 . se cx .ta nx + se cx
dx
dx
e 4 x .( 1+x ) 5
3x5
2
5+x 6
=
=
s ec x
5 +x 6 . ta nx +
e 4 x .( 1+x ) 5 3 s ec 3 x tan x  sec 3 x . e 4 x . ( 1+x ) 4 ( 9+ 4 x )
3x5
e 4 x .( 1+x ) 5
2
5+x 6
=
3 e 4 x .(1+ x ) 5 s ec 3 x tan x  e 4 x .( 1+x ) 4 ( 9+ 4 x ). s e c 3 x
STEP 3 :
y
e 4 x . 2 ( 1+ x ) 1 0
= log (si ne x ) +
5 + x 6 . secx
=
e 4 x .( 1+x ) 4 s e c 3 x
3 (1+ x )ta n x  ( 9+4 x )
e 4 x . 2 ( 1+ x ) 1 0
dy =
dx
e x .cote x
+ s ec x
5+ x 6 . ta nx +
3x 5
5+ x 6
=
s ec 3 x
3( 1+x )ta n x  ( 9+4x )
e 4 x . ( 1+ x ) 6
3
08.
y
=
si n 3 3x . e  x
+
l og
x +1
STEP 3 :
x 2 +1
dy =
STEP 1
dx
si n 3 3x . e  x
d
dx
=
si n 3 3x . d e  x
+ ex d
dx
=
si n 3 3x
dx
si n 3 3x . e  x d x + e  x 3 si n 2 3 x d si n 3 x
dx
=
dx
si n 3 3x . e  x 1
+ e  x 3 si n 2 3x .co s3 x d 3 x
2 x
=
si n 3 3x . e  x 1
dx
+ e  x 3 si n 2 3 x . cos 3 x 3
2 x
=
e  x .si n 3 3 x . + 9 e  x si n 2 3x . co s 3x
2 x
=
e  x .si n 2 3 x
si n 3x + 9 . cos 3 x
2 x
STEP 2 :
d
l og
x 2 +1
dx
=
d
x+1
l og ( x+ 1 )  l og
x2+1
dx
=
d
l og ( x+ 1 )  l og ( x 2 + 1 ) 1  2
dx
=
=
d
l og ( x+ 1 )  1
dx
2
1
d ( x+ 1 )

x +1 d x
=
1

x +1
=
1
x +1

1
2
1
1
2
x 2 +1
l og ( x 2 + 1 )
1
d
(x2+1)
x 2 +1 dx
2x
x
x 2 +1
4
e  x .si n 2 3 x si n3 x + 9 co s3 x +
1
2 x
x +1
 x
x 2 +1