Available online @ http://www.MathCity.org, Version: 1.1.0
Question # 1
Let y = x 4 + 2 x3 + x2
Differentiating w.r.t. x dy d 4
=
x + 2 x3 + x 2 dx dx d 4 d d
=
x + 2 x3 + x 2 dx dx dx 3
2
= 4x + 6x + 2 x
(
)
(
)
= 4 x 4−1 + 2 3 x3−1 + 2 x 2 −1
Question # 2
−
−3
3
2
Let y = x + 2 x + 3
Diff. w.r.t x
−3
dy d −3
= x + 2x 2 + 3
dx dx
d −3 d −3 d
=
x + 2 x 2 + ( 3) dx dx dx 5
− dy ⇒
= −3 x − 4 − 3 x 2 or dx 3 − 3 −1
= −3 x −3−1 + 2 − x 2 + 0
2
1 dy 1
= −3 4 + 5 2 dx x
x
Question # 3 a+x Let y = a−x Now
dy d a + x
=
dx dx a − x
=
d dx d dx (a − x ) ( a + x) − (a + x) (a − x )
(a − x)
( a − x )( 0 + 1) − ( a + x )( 0 − 1) = ( a − x )(1) − ( a + x )( − 1)
=
2
2
(a − x)
( a − x)
=
a−x+a+x
(a − x)
2
2
=
2a
(a − x)
2
Answer
Question # 4
2x − 3
Let
y=
2x + 1 dy d 2 x − 3
Now
=
dx dx 2 x + 1 d d
( 2 x + 1) ( 2 x − 3) − ( 2 x − 3) ( 2 x + 1) dx dx
=
2
( 2 x + 1)
( 2 x + 1)( 2 − 0 ) − ( 2 x − 3)( 2 + 0 )
2
( 2 x + 1)
2 ( 2 x + 1 − 2 x + 3)
2 ( 4)
=
=
2
2
( 2 x + 1)
( 2 x + 1)
=
=
=
( 2 x + 1)( 2 ) − ( 2 x − 3)( 2 )
2
( 2 x + 1)
8
( 2 x + 1)
2
Answer
FSc-II / Ex- 2.3 - 2
Question # 5
Let y = ( x − 5)( 3 − x )
= 3 x − x 2 − 15 + 5 x
= − x 2 + 8 x − 15
Now
(
)
dy dy
=
− x 2 + 8 x − 15 dx dx dy d d =
− x 2 + 8 ( x) − (15 ) = −2 x 2−1 + 8 (1) − 0 = −2 x + 8 Answer dx dx dx (
)
Question # 6
1
Let y = x −
x
( )
2
2
( )
1
1
1
−1
= x +
−2 x
= x+ x −2 = x+ x −2
x
x
Now diff. w.r.t x dy d d d d = x + x −1 − 2 = ( x ) + ( x −1 ) − (2) dx dx dx dx dx = 1 + (−1 ⋅ x −1−1 ) − 0 = 1 − x −2
1
x2 − 1
= 1− 2 =
Answer
x x2 2
(
Question # 7
)
(1 + x ) ( x − x )
3/ 2
Consider
y =
=
=
x
(1 + x )
( ) x 2
x 1− x 2
3
1+ 1
= x 1 + x 1− x
Since x 2 = x 2 x x
2
1 − x
1
1
3
= x (1 − x ) = x 2 (1 − x ) = x 2 − x 2 x (
1
)(
)
( )
Now
1
3
dy d 2
=
x − x2 dx dx
1
3
1
1
1 2 −1 3 2 −1
1 −2 3 2
= x − x
= x − x
2
2
2
2
=
1 1
−3 x
2 x
Question # 8
Let
y =
(x
2
)
+1
2
x2 − 1
Differentiating w.r.t. x
2
2
dy d x +1
=
dx dx x 2 − 1
2
2 d d 2 x2 − 1 x + 1 − x2 + 1 x2 − 1 dx dx
=
2
2
x −1
(
(
)
) (
) (
(
)
)
(
)
Answer
FSc-II / Ex- 2.3 - 3
⇒
dy
=
dx
=
(
) (
)
x2 − 1 2 x2 + 1
(x
2
) (
2 −1
(
)
(
) (
(
)
) ( 2x )
− 1 2 x2 + 1 ( 2x ) − x2 + 1
(x
(
)
−1
2
)
2 d 2 x + 1 − x 2 + 1 ( 2x ) dx 2 x2 − 1
2
2
) ( ) ( )
( )
2 x ( x + 1)( x − 3)
Answer
( x − 1)
(
)
(
2
2
2
2 x x2 + 1 2 x2 − 1 − x 2 + 1
= 2 x x + 1 2 x − 2 − x − 1
=
2
2
2
2
x −1 x −1
2
=
)
2
2
2
Question # 9 x2 + 1
Let y = 2 x −3
Differentiating w.r.t. x dy d x +1
=
= dx dx x 2 − 3
2
=
=
(x
2
−3
(x
) (2x) − ( x
( x − 3)
(x
2
−3
)
=
2
2
−3
d
) dx ( x
)
+ 1 ( 2x )
2
2
2 x ( − 4)
2
(x
− 8x
2
−3
)
2
2
=
) (
d
) dx ( x
+ 1 − x2 + 1
(x
2
−3
(
)
2
2
)
2 x x2 − 3 − x2 − 1
(x
2
−3
)
2
Answer
Question # 10
12
Let y =
1+ x
1+ x
=
1− x
1− x
12
dy d 1+ x
Now
=
dx dx 1 − x
1 −1
1 1 + x 2 d 1+ x
=
2 1 − x dx 1 − x d d
−1
1 − x ) (1 + x ) − ( 1+ x ) ( 1− x )
1 1+ x 2 ( dx dx
=
2
2 1− x
(1 − x )
1
1 1 − x 2 (1 − x )(1) − ( 1+ x )( − 1)
=
2
2 1+ x
(1 − x )
1
1
1 (1 − x ) 2 1 − x + 1+ x
(1 − x ) 2 2
=
=
1
2
2 (1 + x ) 1 (1 − x )2
2
2 (1 + x ) 2 (1 − x )
1
1
=
=
Answer
1
1
3
2−
2 1− x
2
2
1 + x (1 − x )
(1 + x ) ( )
−3
)
FSc-II / Ex- 2.3 - 4
Question # 11
2x −1
Let y = x2 + 1
Differentiating w.r.t. x
Question # 12
Do yourself as Question # 10
Question # 13
1
x2 + 1 2
= 2
x2 − 1
x −1
Now do yourself as Question # 11
x2 + 1
Let y =
Question # 14
1+ x −
1+ x +
1+ x −
1+ x +
Assume y =
=
=
=
(
(
1− x
1− x
1 − x 1+ x − 1− x
⋅
1 − x 1+ x − 1− x
1 + x − 1− x
1+ x
) −(
2
)
2
1− x
)
2
=
(
1+ x
1 + x + 1 − x − 2 (1 + x )( 1− x )
2 1 − 1 −
=
2x
Now differentiation w.r.t x
1
2 2 dy d 1 − 1− x
=
dx dx x
(
(
)
2x
2
= 1− 1− x x 1 x2 2
)
(
Rationalizing
) (
2
+
1− x
)
2
)
(
1 + x − 1+ x
2 − 2 1 − x2
=
2x
1
2
−2
1+ x
)(
1− x
)
FSc-II / Ex- 2.3 - 5
x
=
=
1 x2 =
1 x2 d
2
1 − 1 − x dx
1 d
− 1− 1− x 2 2 x
dx
2
x
1 −1 d
1
⋅ x 0 − 1 − x2 2
1 − x 2 − 1− 1− x 2 dx 2
1
1
1
2 −2
2 2
⋅ x − 1− x
( −2 x ) − 1 + 1 − x
2
(
)
(
1
2
(
)
(
(
(
(
(
(
)
)
)
) 2 (1)
1
)
1
2
x − 1 − x2 2 + 1 − x 2
1
= 2⋅
1
x
2 2
1− x
(
1
2
)
(
)
)
1 x2
2
= 2⋅
1 −1 + 1− x
x
2 2
1− x
1
2 2
1 1 − 1− x
= 1−
= 2 ⋅
1 x x2 1 − x2 2
(
)
(
1− x2
1− x
)
)
Answer
2
Question # 15 x a+x
a+ x
Let y =
= x
a−x
a−x
Diff. w.r.t. x dy d a+x
=
x
dx dx a − x d = x dx d
Now
dx
1
1
2
2
2
a+x
a−x
a+x
a−x
1
1
2
a+x
+
a−x
1a+ x
=
2a − x
=
=
=
=
1 −1
2
1
2
d x ……….. (i) dx a+ x
a−x
d d
−1
(a + x) − (a + x ) (a − x)
2 (a − x)
1 a + x dx dx
2
2 a − x
(a − x)
1
1 a − x 2 ( a − x ) (1) − ( a + x )( − 1)
2
2 a + x a − x)
(
1
2a
1
1
1 (a − x)2 a − x + a + x
⋅
=
1
2 ( a + x ) 1 ( a − x )2
2 ( a + x ) 1 ( a − x ) − 2 ( a − x )2
2
2
a a
=
1
1
3
1
2−
(a + x )2 ( a − x) 2
(a + x) 2 (a − x) 2 d dx
Using in eq. (i)
1
dy a a+x 2
= x⋅
+
(1)
3
1 dx 2 (a − x) 2
a−x
(a + x)
1
=
=
ax
3
2
+
(a + x) (a − x) ax + ( a + x )( a − x )
3
1
(a + x) 2 (a − x) 2
1
2
(a + x) 2
1
(a − x) 2
=
ax + a 2 − x 2 a + x (a − x)
3
2
Answer
FSc-II / Ex- 2.3 - 6
Question # 16
Since
y =
1 x x−
1
2
−
1
2
= x −x
Diff. w.r.t. x
−1
dy d 1
=
x2 − x 2
dx dx
1 −1 1 −3
= x 2+ x 2
2
2
Multiplying by 2x
1
−1 dy 2x
= x2 +x 2 dx Adding y on both sides
1
−1 dy 2x + y = x 2 + x 2 + y dx 1
1
1
−1
−1
−1
dy
⇒ 2x + y = x 2 + x 2 + x 2 − x 2
Q y=x2 −x 2 dx 1 dy dy
⇒ 2x + y = 2x 2
⇒ 2x + y = 2 x
Proved
dx dx Question # 17
Since y = x 4 + 2 x2 + 2 dy d 4
=
x + 2x2 + 2
Now
dx dx dy
⇒
= 4 x 4−1 + 2 2 x 2−1 + 0 dx = 4 x3 + 4 x dy ⇒
= 4 x x 2 + 1 ……… (i) dx Now y = x 4 + 2 x2 + 2
(
)
(
(
)
)
⇒ y − 1 = x4 + 2 x 2 + 2 −1
( x + 1)
( x + 1) i.e. ( x + 1)
= x 4 + 2 x2 + 1 =
⇒
y −1 =
2
2
2
2
Using it in eq. (i), we have dy ⇒
= 4x y − 1 dx =
y −1
as required.
Error Analyst
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Calculus and Analytic Geometry Mathematic 12
Punjab T extbook Board, Lahore.
Made by: Atiq ur Rehman (Atiq@MathCity.org)
Available online at http://www.MathCity.org in PDF Format (Picture format to view online).
Page Setup used Legal ( 8′′ 1 2 × 14′′ )
First Printed: August 24, 2005, Updated: February 16, 2009.
