Differentiation and Geometry | Questions and Answers

QUESTION ONE
(a) From the given graphical plot of the function f ( x )= x+1
x2−1 ,
The limit as x approaches 1 from the right is + ∞ i.e. lim
x→ 1+¿
[ x+1
x2−1 ]=+∞ ¿
¿
Whereas
The limit as x approaches 1 from the right is −∞ i.e. lim
x→ 1−¿
[ x +1
x2−1 ] =−∞ ¿
¿
Since
lim
x→ 1+¿
[ x+1
x2−1 ] ≠ lim
x→1− ¿
[ x+ 1
x2
−1 ] ¿
¿¿
¿
, the limit does not exist
(b)
lim
x→−1 [ x +1
x2 −1 ] = lim
x →−1 [ x +1
( x−1)( x+1) ]
¿ lim
x→−1 [ 1
x−1 ]
¿ 1
−1−1 =−1
2
∴ lim
x →−1 [ x+1
x2−1 ]=−1
2
(c) For f ( x) to be continuous at x=−1 :
i) f ( x) must be defined at x=−1 ; however, f (−1) is undefined at x=−1
ii) The limit must exist i.e.
lim
x→−1 [ x +1
x2 −1 ]=−1
2 as shown in part (b). Thus the limit exists
iii) lim
x→−1
f ( x )=f (−1 ) ; however, lim
x→−1 [ x +1
x2 −1 ]≠ f (−1 ) as the function is not defined at
x=−1
∴ Since not all the continuity conditionshave been met , the function is discontinous at

x=−1 . However , thediscontinuity is a removable one .
QUESTION TWO
From the definition of differentiation, d
dx f ( x ) =lim
h →0
f ( x+ h ) −f ( x)
h
d
dx [−2 x2 +3 x +58 ]=lim
h →0
[−2 ( x +h )2 +3( x +h)+58 ] − [−2 x2 +3 x+ 58 ]
h
Expanding the numerator,
d
dx f ( x ) =lim
h →0
[ −2(x2+2 hx +h2 )+ 3 x +3 h+58 ] − [ −2 x2 +3 x +58 ]
h
¿ lim
h→ 0
[−2 x2−4 hx−2 h2 ¿+ 3 x +3 h+58 ]− [−2 x2 +3 x +58 ]
h
¿ lim
h→ 0
[ −2 x2+2 x2+ 3 x −3 x−4 hx −2h2+3 h+58−58 ]
h
¿ lim
h→ 0
[−4 hx−2 h2 +3 h ]
h
¿ lim
h→ 0
h [−2 h−4 x+ 3 ]
h
¿ lim
h→ 0
[−2 h−4 x +3 ]
¿−2 ( 0 )−4 x+ 3=−4 x +3
∴ d
dx [−2 x2+ 3 x +58 ]=−4 x +3

QUESTION THREE
(a) Given p ( x ) = ( 2 x2 +3 x +1 ) ( x3 −x−1 ),
By Product rule i.e.
p' ( x )= [u ( x ) . v ( x ) ]'
=u' ( x ) . v ( x )+ u ( x ) . v' ( x )
For u ( x )= ( 2 x2+3 x +1 )∧v ( x ) = ( x3−x−1 )
p' ( x )= ( x3 −x−1 ) ∙ d
dx [ 2 x2 +3 x +1 ]+ ( 2 x2 +3 x+ 1 ) d
dx [ x3−x−1 ]
Applying power rule,
p' ( x )= ( x3 −x−1 ) ( 4 x +3 ) + ( 2 x2+ 3 x +1 ) (3 x2−1)
(b) Given q ( x )= 2 x2 +3 x +1
x3−x−1
Applying the quotient rule i.e.
q' ( x )= [ u( x )
v( x) ]'
=u' ( x ) . v ( x )−u ( x ) . v ' (x)
v (x)2
For u ( x )= ( 2 x2+3 x +1 )∧v ( x ) = ( x3−x−1 )
q' ( x ) =
( x3−x−1 ) . d
dx [ 2 x2+ 3 x +1 ] − ( 2 x2 +3 x+1 ) d
dx [ x3−x−1 ]
( x3−x−1 ) 2

¿ ( x3 −x−1 ) ( 4 x +3 ) − ( 2 x2 +3 x+ 1 ) (3 x2−1)
( x3−x−1 )
2
∴ q' ( x ) = ( x3−x −1 ) ( 4 x +3 )− ( 2 x2 +3 x +1 ) ( 3 x2−1)
( x3 −x−1 )2
(c) Given f ( x )=cos ( ax ) sin (ax ),where a is constant
By Product rule i.e.
f ' ( x )= [ u ( x ) . v ( x ) ]'
=u' ( x ) . v ( x ) +u ( x ) . v' ( x )
For u ( x )=cos (ax )∧v ( x )=sin(ax) ,
f ' ( x )=sin (ax)∙ d
dx [ cos ( ax ) ]+ cos ( ax ) ∙ d
dx [ sin(ax) ]
¿ sin ( ax ) ∙−a sin ( ax ) +cos ( ax ) ∙ a cos (ax)
¿ a cos2 (ax )−a sin2 ( ax)
∴ f ' ( x ) =a [ cos2 ( ax )−sin2 ( ax ) ]
(d) Given h ( x ) = tan ( √ x )
cos x
Applying the quotient rule,
h' ( x )=
[ u( x )
v ( x ) ]'
= u' ( x ) . v ( x )−u ( x ) . v ' ( x )
v (x)2
For u ( x ) =tan( √ x)∧v ( x ) =cos x