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Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

13
...
99,1
...
9999, …
...
Similarly, if x take values 2
...
001,2
...
Even then the numerical difference
between x and 2 gets closure to 0
...

In general, x  a means that the variable x and x takes values either less than or greater than that
of a and the numerical difference between x and a can be made as small as we please
...
It can be symbolically written as: lim f ( x)  k
...

x a

xa

(ii) lim  Area of polygon of n sides   Area of circle
nx 

STANDARD RESULTS






Limit of a constant function is a constant
...
e
...
f ( x)  k
...
g ( x) = lim f ( x)  lim g ( x)

xa



xa

xa

xa

lim f ( x)
f ( x ) x a
=
, g x   0
lim g ( x)
xa g ( x)

lim

x a



lim n f ( x) = n lim f ( x)

xa

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lim 

E
...
:

 x 2  2x  5  2 2  2  2  5 4  4  5 3



3 2  5
65
11
x2  3x  5 
 x 2  3x  2   22  3   2  2 4  6  2 0

iii
...
lim 









iv
...

g
x a ( x ) 0
xa
xa
lim

The form

0
is called indeterminate form
...



We cannot find the limits such functions directly
...
Factorization Method:
a) Factorize the numerator and denominator and cancel the common factors from the numerator and
the denominator
...

c) Apply quotient rule of limit
...
g
...
Substitution Method:
f ( x)
f ( a  h)
= lim

...
As x  a , h  0
...

sin x sin  0


x   x    0

E
...
: Evaluate lim

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Rationalization Method:
a) Rationalize the expression, which involve square roots
...

1  x 1
1  0 1 1 1 0



x
0
0
0

E
...
i
...

x
x0 




x
 lim 
  lim 
x0  x 1  x  1 
x0 







1

  1  0  1



1
1
1


1  1 2
2



 1 x  1 x 

x0 

sin 1 x

ii) Evaluate lim 





1  x  1 
1





 1  x  1  x  1 1 0


0
0
x0 

sin 1 x

The expression  lim 

put x  sin 
...






 1  sin   1  sin  
 lim 


 0 




  1  sin  
 1  sin  




 1  sin   1  sin 
 lim 


 0 

1  sin  

1  sin  

 1  sin   1  sin   


2 sin 
 lim 
  lim 

 0  1  sin   1  sin    0  1  sin   1  sin  
 sin 

2
2
 lim 

  1
 0  
1  sin   1  sin  
1 0  1 0









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Please refer the class room note)
(1  x) n  1
x
x0

E
...
: i) Evaluate lim
The expn
...
1n1  n
(
1

x
)

1
1 x 1 

lim

 1n  1

9
 9
 x 2 

 9  2 91 9  2 8
x 9  512
x 9  29
 lim
 lim  x  2  

 9  2 85  9  2 3  72
ii) lim
x2 x 4  16
x2 x 4  2 4 x2  x 4  2 4  4  2 41 4  2 3
 x2 



If x is measured in radians, then

1
...


lim sin x  0

x0

lim cos x  1

x0

3
...


x
1
x0 sin x

5
...
lim

x

x 0 tan x

7
...
lim

sin ax a

sinbx b

9
...
in

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4

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

OTHER IMPORTANT THEOREMS
1
...


e x 1
 log e  1
x0 x

3
...
g
...


Meaning of x  a 
Similarly, If the variable x takes values, which are close to a constant a and always remains on
the right of a then we say that x approaches to a from right and we write as x  a 
...

Working rule:
1
...
in

h0

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5

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

2
...
If lim f ( x)  lim f ( x) , then lim f ( x) does not exist
...
If lim f ( x)  lim f ( x) , then lim f ( x) exists and is equal to lim f ( x) is equal to f (a)
...
g
...
 lim f ( x) does not exist
...

Suppose f ( x) 

1

...

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i
...
A function f (x) is said to be a derivable function on the closed interval a, b ,
a
...
it is derivable at x  a from right
c
...
Then corresponding to each point x  a, b , we get
a unique real number equal to the derivative of f (x) and are denoted by f (x) or
i
...
,

dy
or Dy y1 or y  , etc
...
The process of obtaining the derivative of
dx h0
h
dx h0
h

a function is called differentiation
...
Let f (x) is
differentiable at x  c
...
Then slope of the chord PQ 

f x   f c 

...
e
...
As Q  P , the chord PQ becomes tangent at P
...
r
...
t =  y  
dt
dt
d
dx
derivative of x w
...
t
...

dt
dt

Note: derivative of y w
...
t
...
Let  x be a small increment in x and  y be the
corresponding increment in y respectively
...
in

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7

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

Then y  y  f x  x
y  f x  x   f x 
f x  x   f x 
y
=
x
x

taking limits we have,
y
f x  x   f x 
= lim
x

x
x0
x0
lim

dy
 f  x 
dx

i
...
,

d
 f x   f x 
...
r
...
r
...
This
dx

method is called first principles or delta (  or  ) method or differentiation by definition or ab initio
...

dx

Note: If y  f (x) is a real function defined at a real constant ‘h’, then

f   x   lim

f  x  h  f h

h 0

h

Find the derivative of the following functions using the first principle:
1
...
e
...



  x  h 2  x 2
  lim 
  x  h  x   x  h   x



Let f  x   e x
f  x  h   e xh

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lim  e

e x eh  1

h 0

h

h

x

h 0



1
h



 x

 lim  e  1   log e  1
 x0 x 






dy
 e x 1  e x
dx

 

d x
e  ex
dx

3
...
lim  a

a x ah  1

h 0

h

dy
 a x 1  a x
dx

x

h 0

h



1

h



 ax 1 
 log a 
 lim 


x 0
 x 



 

d x
a  ax
dx


 x

 lim  a  1   log a 
 x0 x 






dy
 a x  log a  a x
dx

 

d x
a  a x
...


Let f  x   x
f  x  h  x  h
f  x  h  f  x  x  h  x 

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
 
x  h  x h

h





1



xh  x

dy
dx



1

 x



h 0

d
dx

[DIFFERENTIAL CALCULUS]



x0  x



1
xh  x






 

1
x x





1
2 x

1
2 x

Let f  x  

1
 x 2
2
x

f  x  h 

1

  x  h

2

 x  h
2
f  x  h   f  x    x  h   x 2
f  x  h  f h
f   x   lim
h 0
2

h

 lim

 x  h 2  x2

h 0

6
...


h

=

 x  h  n  x  n
n
 n 1
  n  x  n 1   n  x    n 1
 x  h  x  x  h   x
x
lim

Let f x   sin x
f x  h  sinx  h

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e
...



 h 
  2 x  h  sin  2  
 
 lim cos 
 h 
h 0 

 2 

2 


h


 as h  0,  0 
2



 cos x 1  cos x

sin x


 1
 lim
 x0 x


d
 sin x   cos x
dx

Let f  x   cos x

f  x  h   cos  x  h 
 xh x  xhx
 2x  h   h 
f  x  h   f  x   cos  x  h   cos x  2 sin 
 sin 
  2 sin 
 sin  
2
2

 

 2  2
f x 

 f x  h   f x  
 lim 

h
h0



 h 
 2x  h   h 
sin   
2 sin 
sin  


 2 
 2   2   lim  2 sin  2 x  h  
 lim

 h
h

0
h 0
h

 2 
2 


2



 h 
sin   

2x  h 

 2 
 lim   sin 


h 0 
h 
 2 

2 


 2x  0  
  sin
 1
 2  


i
...
,

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Let f  x   tan x
f  x  h   tan  x  h 
 f  x  h  f  x 
f   x   lim 

h 0
h



 sin  x  h  sin x 
 lim tan  x  h   tan x   lim 


h 0
h 0 cos  x  h 
cos x 

sin  x  h  cos x  cos  x  h  sin x

cos  x  h  cos x
 sin  x  h  x  

1
sinh 
 lim 

  lim 

h 0 cos  x  h  cos x  h
h 

 h0  cos  x  h  cos x

sin x


 1
 lim
x
 x0




1
 
1
 cos x  0 cos x

dy
dx



1
1
 sec 2 x

2
cos x cos x
cos x

d
tan x   sec 2 x
dx

10
...
in

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...


sin x


 1
 lim
 x0 x


12

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]



1
sinh  
1
 lim 

 1
  

h 0 sin  x  h  sin x
h   sin  x  0  sin x


dy
dx



1
1
  2  co sec2 x
sin x sin x
sin x

d
cot x    cos ec 2 x
dx

11
...

 sec x
...
cos x cos x cos x
dx
d
sec x   sec x tan x
dx

12
...
in

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13

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

 f  x  h  f  x 
f   x   lim 

h 0
h




 sin x  sin  x  h  
1
1 
 lim cos ec  x  h   cos ecx   lim 

  lim 

h 0
h 0 sin  x  h 
sin x  h0  sin  x  h  sin x 


 x  x  h   x   x  h  

 2 x  h   h  
 2 cos 
2 cos 
sin

 sin 

2
2

 
2   2  





 lim 
  lim
h 0
sin  x  h  sin x  h

 h0  sin  x  h  sin x x 











 2x  h 
 h 
 2x  h 
 h 
sin   
 2 cos  2   sin  2  
 cos  2 

 
    lim  

 
 2 
 lim 
h 0  sin  x  h  sin x
h
h

0
h
 sin  x  h  sin x

2 




2
2






 2x  h 
 h  
 2x  0 
sin     cos 

 cos  2 

2
2

 
    

  1
 lim  
h 0  sin  x  h  sin x
h   sin  x  0  sin x 


2  



sin x


 1
 lim
 x0 x


cos x
1 cos x
dy



...
cot x
sin x
...
cot x

STANDARD RESULTS
f (x)

f x 

sin x

cos x

cos x

 sin x

tan x

sec 2 x

cos ecx

 cos ecx cot x

sec x

sec x tan x

cot x

 cos ec 2 x

xn

nx n1

ex

ex

Hsslive
...
com
...
log a
1

x

2 x

log x

1
x

x

1

x2

2x

1



x

n

1
x

1
x



1



x2

n 1

1
x2
2
x3

y

dy
dx

y2

2y

dy
dx

Note: Derivative of any trigonometric function starting with ‘co’ is negative
...
Differential coefficient of a constant is zero
...
e
...
g
...

dx

d
5  0 , d  10  0 , etc
...
If u and v are functions of x , then

d
d
d
(u  v ) 
( u)  ( v )
dx
dx
dx

d
5 sin x  log x   d 5 sin x   d log x   5 d sin x  d log x   5 cos x  1
dx
dx
dx
dx
dx
x





 

 

d
d
d
d x
d
2e x  tan x 
2e x  tan x   2
e  tan x   2e x  sec 2 x
dx
dx
dx
dx
dx

Hsslive
...
com
...
Product rule: If u and v are functions of x , then derivative of the product of two functions is equal to
first function x derivative of the second function + (plus) second function x derivative of the first
function
...
e
...
(v )  v
...
g
...


 

d
d 3x
= e3x sin 4 x   sin 4 x
...
cos 4 x
...
e3x
...


dy
dx

 x2

 

d
tan x   tan x d x 2
dx
dx

 x 2 sec 2 x  tan x
...
( w )  vw
...
(v )
dx
dx
dx
dx

y  x 2 e x tan x

E
...
:

 

 

dy
d
tan x   e x tan x d x 2  x 2 tan x d e x
 x 2e x
dx
dx
dx
dx

 x 2 e x sec 2 x  e x tan x
...
e x









 xe x x sec 2 x  2 tan x
...
 xe x x sec 2 x  2  x  tan x
...
QUOTIENT FORMULA: If u and v are any two functions of x , then quotient of two functions is
equal to (2nd function x derivative of the 1st function minus 1st function x derivative of the 2 nd
function) divided by square of the 2nd function
...
e
...
g
...
in

v
...

(v )
dx
dx
v2

Sin x  Cos x

...
d sin x  cos x   sin x  cos x  d sin x  cos x 
=

dx

sin x  cos x 2

dx

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...


16

Remesh’s Maths Centre

[DIFFERENTIAL CALCULUS]

=

sin x  cos x cos x  sin x   sin x  cos x cos x  sin x 
sin x  cos x 2

=

sin x  cos x 
...
cos x  cos 2 x  (sin 2 x  2 sin x
...
cos x  cos 2 x  sin2 x  2 sin x
...
cos x  2 sin x
...
2 sin x
...


Note:





d
f   x    f    x     x 
dx 

E
...
: f  x  =
f  x 

ii
...
y  sin x

dy
d
 2 sin x   sin x   2 sin x  cos x  sin 2 x
dx
dx
n
iv
...
R
...
in

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...


17



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