# Differentiation

Document Sample

```					LOCUS                                                                 1

Differentiation

NOTES
CONCEPT NOTES

01.    Introduction to differentiation

02.    Differentiation of Standard Functions

03.    Rules for differentiation

04.    Differentiation of Parametric / Implicit Functions

05.    L' Hospital' Rule

Maths / Differentiation
LOCUS                                                                                                                        2

Differentiation

Now that we have seen the concepts of continuity and differentiability, we will start developing applications of
these concepts. These applications include differentiation (developing techniques and formulae to differentiate a
given function), maxima and minima, drawing tangents and normals (to some given curve at some given point),
studying monotonicity of functions, and so on. In this chapter, we will study differentiation which is just an extension
of the differentiability concept (in fact, we already have studied almost everything pertaining to differentiation). Our
purpose here is to find out the derivatives of some standard functions by first principles, and then use these results
whenever we want to differentiate any arbitrary given function.

Section - 1                               INTRODUCTION TO DIFFERENTIATION

In the previous chapter, we saw in great detail the meaning of evaluating the derivative (differentiating) of a given
function at any given point.
We summarize that discussion briefly here:
If f(x) is a differentiable function for a given x, this means that we can draw a unique tangent to f(x) for that given
x. The slope of this unique tangent is called the derivative of f(x) for that given x. The process of finding the
derivative is known as differentiation.

For example, for f ( x ) = x 2 , the derivative at any given x has the value 2x (we evaluated this by first principles in

the previous chapter.) This means that the slope of the tangent drawn to f ( x ) at any given x has the numerical
value 2x.

Equivalently stated, we can differentiate f ( x ) = x 2 to get f ' ( x ) = 2 x .

{ f ' ( x ) represents the derivative of   f ( x )}.

y

The tangent drawn to
f(x) at any x has the
slope f'(x) = 2x,
i.e tan θ = 2x.

θ
x
x

Fig - 1

Maths / Differentiation
LOCUS                                                                                                                                    3

Recall that we can differentiate a function at a given point only if the LHD and RHD at that point have equal values.
If they do, then
d ( f ( x ))   f ( x + h) − f ( x)        f ( x − h) − f ( x)
f '( x) =                  = lim               = lim
dx        h →0          h            h →0          −h
d f ( x )  dy 
The notation that we use to signify the derivative of y = f ( x ) is either f ' ( x )( or y ') or            or 
dx  dx 
dy
We need to understand here the significance of the notation       (the derivative of y with respect to the variable x)
dx
Recall that to evaluate the derivative (slope of the tangent) of y = f(x) at a given point, we first drew a secant
passing through that point and then let that secant tend to a tangent as follows:
Secant
y
y = f(x)          Secant

Secant

Tangent

x

Fig - 2

The slope of any secant can be written easily:
y

y2 = f(x2 )                            C
y2 − y 1
tan θ = BC   =
AB       x 2 − x1
∆y
∆y
y1 = f(x1 )
θ               B
= ∆x
∆x
A

x
x1              x2

Fig - 3
The slope is obvious from the figure:
BC y2 − y1 ∆y
tan θ =     =       =
AB x2 − x1 ∆x
Now, to evaluate the derivative at x1, we need to make the secant AC tend to a tangent at A by letting x2 approach

Maths / Differentiation
LOCUS                                                                                                                         4

x1 ( x2 → x1 ) or equivalently, by letting ∆x → 0.
As ∆x becomes an infinitesmally small quantity (approaches 0), the corresponding ∆y will also become infinitesmally
∆y
small (will approach 0), but the ratio    will become an increasingly accurate representation of the slope of the
∆x
tangent at A.

An infinitesmally small change in the x value is represented by dx instead of ∆x.

Similarly, an infinitesmally small change in the y value would be represented by dy instead of ∆y.
Therefore,
∆y          ∆y dy
lim        = lim        =    = y'
x2 → x1 ∆x   ∆x → 0 ∆ x   dx

dy               dy d f ( x )
You should now be clear about the notation                . We will use    ,         , f ' ( x ) or y ' interchangeably to
dx               dx   dx
represent the derivative of y = f ( x ) at any given x.

Note:        From now on you should always keep in mind that d(variable) represents an infinitesmally small
change in the variable value while ∆ (variable) represents a finite change in the variable value.

Section - 2                         DIFFERENTIATION OF STANDARD FUNCTIONS

By now, the meaning and geometrical significance of differentiation should be pretty clear to you. We will use this
knowledge to evaluate the derivatives of some standard functions in this section.
You will notice that while differentiating these functions, we will only use the expression for the RHD; we could
equivalently use the LHD also since all the functions we will be concerned with in this section are differentiable
(except at discontinuous points); that the LHD and RHD are equal for each of these functions at a given x can be
easily verified.

1.            f (x) = k :
f ( x + h) − f ( x)        k −k
f ' ( x ) = lim                     = lim       =0
h →0          h            h →0   h
This is intuitively true also since the graph for a constant function is a horizontal line.

2.            f (x) = x :
f ( x + h) − f ( x)        ( x + h) − x = 1
f ' ( x ) = lim                        = lim
h →0            h            h →0       h
This corresponds to the fact that the line f ( x ) = x is inclined at 45º to the x–axis (and tan 45º is 1).
Maths / Differentiation
LOCUS                                                                                                                     5

3.            f ( x ) = mx + c :

f ( x + h) − f ( x)        m ( x + h ) + c − ( mx + c )
f ' ( x ) = lim                        = lim                               =m
h →0            h            h →0              h
This is again a straight forward result: ‘m’ is the slope of f ( x ) = mx + c so it must equal f ' ( x ) .

4.            f ( x ) = x2 :
f ( x + h) − f ( x)       ( x + h ) − x 2 = lim h 2 + 2 xh = 2 x
2

f ' ( x ) = lim                     = lim
h→0          h            h→0        h          h→0      h
We have already obtained this result earlier.

5.            f ( x ) = xn :

f ( x + h) − f ( x)        ( x + h ) − xn
n

f ' ( x ) = lim                      = lim
h →0          h            h →0        h
 n  h n      
 x 1 +  − x 
n

               
= lim  
x
h→0

       h       

               

  h n 
 1 +  − 1
           
= x lim  
n           x
h→0

     h     

           


 nh
1+   +n
( n − 1) h 2 + .... − 1
                           
   x       2! x 2                          [By the binomial expansion]
= x lim
n
h →0              h

n
= xn ⋅     = nx n −1
x

d ( x2 )                     d ( x3 )
So, for example,                 = 2.x 2−1 = 2 x and          = 3.x 3−1 = 3x 2 and so on.
dx                           dx

Maths / Differentiation
LOCUS                                                                               6

6.            f ( x ) = sin x :

f ( x + h) − f ( x)        sin ( x + h ) − sin x
f ' ( x ) = lim                       = lim
h →0            h            h →0           h

    h h
2 cos  x +  sin  
= lim           2 2
h→0            h
                  h 
            sin   
         h        2 
= lim cos  x +  ⋅   
h→0
        2 h 
  

               2 
= cos x.

7.            f ( x ) = cos x :

f ( x + h) − f ( x)        cos ( x + h ) − cos x
f ' ( x ) = lim                        = lim
h →0            h            h →0           h
            h      h
 −2sin  x + 2  sin 2 
                       
= lim               
h→0

           h           

                       


                   h 
              sin   
         h        2 
= lim − sin  x +  ⋅   
h →0
         2 h 
  

                2 
= – sin x.

8.            f ( x ) = tan x :

f ( x + h) − f ( x)        tan ( x + h ) − tan x
f ' ( x ) = lim                        = lim
h →0            h            h →0           h

 tan h                              
= lim        ⋅ {1 + tan x ⋅ tan ( x + h )}
h→0
 h                                  
= 1 + tan2 x
= sec2 x.
Maths / Differentiation
LOCUS                                                                                           7

9.            f ( x ) = sec x :

f ( x + h) − f ( x)        sec ( x + h ) − sec x
f ' ( x ) = lim                        = lim
h →0            h            h →0           h

cos x − cos ( x + h )
= lim
h → 0 h cos x cos ( x + h )

           h      h
 2sin  x + 2  sin 2 
                       
= lim               

 h cos x cos ( x + h ) 
h →0


                       


            h       h 
 sin  x + 2   sin  

= lim                       2 
h → 0 cos x cos ( x + h )
 h 
                        

                     2 
= sec x tan x.

10.           f ( x ) = cosec x :

f ( x + h) − f ( x)        cosec ( x + h ) − cosec x
f ' ( x ) = lim                        = lim
h →0            h            h →0             h

sin x − sin ( x + h )
= lim
h → 0 h sin x sin ( x + h )

             h      h
 −2 cos  x + 2  sin 2 
                        
= lim                 

 h sin x sin ( x + h ) 
h →0


                        


             h    h  
 − cos  x + 2   sin  2   

= lim                     
h → 0 sin x sin ( x + h )

                     h 
                     2 
                            
= – cosec x cot x

Maths / Differentiation
LOCUS                                                                                                                         8

11.           f ( x ) = cot x :

f ( x + h) − f ( x)        cot ( x + h ) − cot x
f ' ( x ) = lim                        = lim
h →0            h            h →0           h
sin x cos ( x + h ) − cos x sin ( x + h )
= lim
h →0          h sin x sin ( x + h )

− sin h                     Notice how the numerator 
= lim                                                               
h → 0 h sin x sin ( x + h )
 was simplified           
= – cosec2 x
12.           f ( x ) = ex :
f ( x + h) − f ( x)        e x+h − e x        e x ⋅ eh − e x
f ' ( x ) = lim                      = lim              = lim
h →0          h            h →0      h        h →0        h
x         eh − 1
= e lim
h→0       h
= ex

13.           f (x) = ax :

f ( x + h) − f ( x)        a x+ h − a x
f ' ( x ) = lim                      = lim
h →0          h            h →0       h
ah −1
= a x ⋅ lim
h →0   h
= a x ln a

14.           f ( x ) = ln x :

f ( x + h) − f ( x)        ln ( x + h ) − ln x
f ' ( x ) = lim                        = lim
h →0            h            h →0          h
 h
ln  1 + 
= lim 
x
h →0       h
 h
ln  1 + 
= lim 
x
h →0        h
x⋅
x
1     
                 ln (1 + θ )    
=       Because lim                   = 1
x     

θ →0
     θ          

Maths / Differentiation
LOCUS                                                                                                                    9

15.           f ( x ) = log a x :

ln x d ( log a x )          1 d ln ( x )             1
Since log a x can be written as        ,              will be               {because        is a constant
ln a       dx               ln a dx                 ln a
so it can be taken outside the differentiation operator; we will prove the validity of this step later}.
d                  1
Therefore,        (log a x ) =
dx              x ln a

16.            f ( x ) = sin −1 x :

f ( x + h) − f ( x)        sin −1 ( x + h ) − sin −1 x
f ' ( x ) = lim                        = lim
h →0            h            h →0              h

                                                                             
The numerator can be simplified as follows:                                  
Let us consider a general expression sin –1 x − sin −1 y.                    
                                                                             
Let sin –1 x = p and sin −1 y = q, so that x = sin p and y = sin q           
                                                                             
 Now, sin ( p − q ) = sin p cos q − cos p sin q                              
                                                                             
                       = x 1− y − y 1− x                                     
2           2

Therefore,                                                                   
                                                                             
                                    (
 p − q = sin −1 x − sin −1 y = sin −1 x 1 − y 2 − y 1 − x 2
                                                                   )         


We use this relation now to simplify the numerator:

f ' ( x ) = lim
{
sin −1 ( x + h ) 1 − x 2 − x 1 − ( x + h )
2
}
h →0                               h

sin −1 y             we replaced the large 
= lim                                      –1     
h→0    h                argument of sin by y 
sin −1 y y
= lim           ⋅
h →0    y      h
sin −1 y
Notice that as h → 0, y → 0 so                         →1
y
( x + h) 1− x2 − x 1− ( x + h)
2
y
Also,                      lim = lim
h →0 h h →0                h

= lim
(x + h)
2
(1 − x ) − x
2        2
(1 − ( x + h ) )
2

h →0
{
h ( x + h) 1 − x2 + x 1 − ( x + h)
2
}
Maths / Differentiation
LOCUS                                                                                                                                   10

h 2 + 2 xh
= lim
h →0
{
h ( x + h) 1 − x2 + x 1 − ( x + h)
2
}
2x
=
2 x 1 − x2
1
=
1 − x2
Therefore,
1
f '( x) =
1 − x2

17.           f ( x ) = cos −1 x :

π
Notice that sin −1 x + cos −1 x =
2

Thus,
d                    d π
(sin x + cos x ) = dx2 = 0
−1      −1
( )
dx
d (cos −1 x )          −d (sin −1 x )        −1
⇒                      =                    =
dx              dx           1 − x2
Note that for the last step, we have used the fact that differentiation operation is distributive over
addition, i.e. (f + g)' = f ' + g'. We will justify this later.

18.           f ( x ) = tan −1 x :

f (x + h) − f (x)       tan −1 ( x + h ) − tan −1 x
f ' ( x ) = lim                   = lim
h→0         h           h→0              h

tan −1
h           we used tan −1 A − tan −1 B 
1 + x (x + h )                                     
= lim                                = tan −1  A − B  ; verify this 
h→0               h                                             
           1 + AB               

 −1         h        
 tan 1 + x ( x + h ) 
                            1
= lim                      ⋅
h             1+ x ( x + h)
h→0

 1+ x ( x + h) 
                     
1
=
1 + x2

Maths / Differentiation
LOCUS                                                                                                                         11

19.           f ( x ) = cosec − 1 x :

f ( x + h) − f ( x)       cosec−1 ( x + h ) − cosec −1 x
f ' ( x ) = lim                       = lim
h→0            h            h→0              h

 1        −1  1 
sin −1      − sin                Notice that          
= lim          x+h          x                                
h →0             h                        cosec −1θ = sin −1 1 
                      
                    θ

 1                       1  
2              2
             1 1              
sin −1         1−   −   1−      
 x + h 
              x  x     x+h 
= lim
h →0                         h

                          −1 
( x + h)
2
 x2 −1
−1                          
sin             −                
 x ( x + h)

x x+h 

= lim
h→0                 h

sin −1 y          where y is the argument of sin −1 ; 
= lim                                                        
h →0    h              Note that y → 0 as h → 0            

sin −1 y y
= lim            ⋅
h →0    y      h
Now, it can easily be verified by rationalization that
y   −1
lim     =
h→0   h x x2 − 1

Therefore,
−1
f '( x) =
x   x2 − 1

20.           f ( x ) = sec −1 x :

−1        −1                 π
Notice that sec x + cosec x =
2

d (sec −1 x )       − d (cosec −1 x )            1
Therefore,                     =                       =
dx                    dx               x   x2 −1

Maths / Differentiation
LOCUS                                                                                                               12

21.           f ( x ) = cot −1 x :

π
Notice again that tan −1 x + cot −1 x =
2
Therefore, as in the earlier cases,

d ( cot −1 x )       − d ( tan −1 x )         −1
=                      =
dx                      dx             1 + x2
It would be of help to you to get used to these differentiation on formulae as soon as possible, since they will be
widely used subsequently.

Section - 3                                            RULES OF DIFFERENTIATION

In this section, we will see certain general rules pertaining to differentiation that will help us in calculating the
derivative of an arbitrary function without using first principles.

In the discussion that follows, we assume that f ( x ) and g ( x ) are two differentiable functions

Rule 1:       d ( kf ( x ))       kd ( f ( x ))
=
dx                  dx
This rule says that a constant can be taken out from the argument of the differentiation operator. The
proof is very straight forward:

d ( kf ( x ))             kf ( x + h ) − kf ( x )
= lim
dx               h →0           h

f ( x + h) − f ( x)
= k lim
h→0            h

kd ( f ( x ))
=
dx

d ( f ( x ) ± g ( x ))       d ( f ( x ) ) d ( g ( x ))
Rule 2:                                =                ±
dx                      dx            dx
This rule says that the differentiation operator is distributive over addition and subtraction. The proof
for this is again quite straightforward:

Maths / Differentiation
LOCUS                                                                                                                                    13

d ( f ( x ) ± g ( x ))              { f ( x + h ) ± g ( x + h )} − { f ( x ) ± g ( x )}
= lim
dx                  h →0                                    h

= lim
{ f ( x + h ) − f ( x )} ± {g ( x + h ) − g ( x )}
h →0                                    h

f ( x + h) − f ( x)        g ( x + h) − g ( x)
= lim                           ± lim
h →0             h            h →0          h

d ( f ( x ))          d ( g ( x ))
=                     ±
dx                 dx

d { f ( x ) g ( x )}              d ( g ( x ))                   d ( f ( x ))
Rule 3:                              = f ( x)                  + g (x)
dx                             dx                             dx
This rule, called the Product rule, is of great help in evaluating the derivative of the product of two (or
more) functions.
The proof is as follows:

d { f ( x ) g ( x )}              f ( x + h) g ( x + h) − f ( x) g ( x)
= lim
dx                 h →0                     h
Introduction of an
extra term

f (x + h ) g (x + h ) − f (x + h ) g (x ) + f (x + h ) g (x ) −f (x ) g (x )
= lim
h→0
h
f ( x + h ){g ( x + h ) − g ( x )} + g ( x ){ f ( x + h ) − f ( x )}
= lim
h →0                                              h
g ( x + h) − g ( x)                f ( x + h) − f ( x)
= lim f ( x + h ) lim                                 + g ( x ) lim
h →0                   h →0            h                    h →0          h
d ( g ( x ))                   d ( f ( x ))
= f ( x)                     + g ( x)
dx                             dx

d ( f ( x ))                  d ( g ( x ))
g (x)                   − f ( x)
d  f (x)
         
Rule 4:                   =
dx                             dx        ; wherever g ( x ) ≠ 0
dx  g ( x ) 
                                  ( g ( x ))
2

This rule, called the Quotient rule, helps us evaluate the derivative of the ratio of two functions
f ( x ) / g ( x ) , wherever g ( x ) ≠ 0 .

Maths / Differentiation
LOCUS                                                                                                                                     14

f ( x + h) f ( x)
−
d  f (x)
                g ( x + h) g ( x)
          = lim
dx  g ( x )  h →0
                         h

g (x) f (x + h) − f (x) g (x + h)
= lim
h→0          h ⋅ g (x)⋅ g (x + h)

Introduction of an
extra term

g ( x ) f ( x + h ) −g ( x ) f ( x ) + g ( x ) f ( x ) − f ( x ) g ( x + h )
= lim
h→0
h ⋅ g ( x) ⋅ g( x + h)

1                g ( x ){ f ( x + h ) − f ( x )}         {g ( x ) − g ( x + h )}
= lim                           lim                                  + f (x)                        
h→0   g ( x ) ⋅ g ( x + h ) h→0 
                h                                   h           


1                   d ( f ( x ))         d ( g ( x )) 
=                     ⋅ g (x)              − f (x)              
( g ( x ))
2

          dx                   dx      

g ( x) f '( x) − f ( x) g '( x)
=
( g ( x ))
2

d f ( g ( x ))       d f ( g ( x )) d ( g ( x ))
Rule 5:                        =                 ⋅
dx               d ( g ( x ))      dx

This can be stated more conveniently as

( f ( g ( x ))) = f ' ( g ( x ) ) g ' ( x )
'

This rule, called the Chain rule, is extremely useful to differentiate composite functions, and will be
used extensively. It says that to differentiate f ( g ( x )) , we first differentiate f with respect to g ( x ) and
not x(i.e, we treat g(x) as a variable y and differentiate f with respect to y) and then we multiply this by
the derivative of g ( x )(or y ) with respect to x.

Maths / Differentiation
LOCUS                                                                                                                                           15

d  f ( g ( x ) )
                            f ( g ( x + h )) − f ( g ( x ))
= lim
dx                h→0                        h

f ( g ( x + h ) ) − f ( g ( x )) g ( x + h ) − g ( x )
= lim                                      ⋅
h→0        g ( x + h) − g ( x)                  h
Notice carefully that the first ratio in the limit above is actually the derivative of f, but with respect to
{
g ( x ) as the variable as h → 0, ( g ( x + h ) − g ( x )) → 0 .                          }
You can view this graphically as follows:
f(g(x))

f(g(x+h))                                             B
tan θ = ∆f
∆g
∆f    As ∆g → 0, tan θ
will tend to the instantaneous
A        θ                     value of slope at A.
f(g(x))
∆g

g(x)
g(x)                  g(x+h)

Fig - 4
Therefore,

d  f ( g ( x ) )
                        d f ( g ( x )) d ( g ( x ))
=                 ⋅
dx                  d ( g ( x ))      dx

= f ' ( g ( x )) ⋅ g ' ( x )
For example,

(
d sin ( x 2 )      )=       cos ( x 2 ) ⋅
# %
"\$#
2x
!
dx                          ↑
Derivativeof
↑
Derivativeof
sin( x 2 ) w.r .t ( x 2 ) ( x 2 ) w.r .t x

d ( log (sin x ))                 1
=                         ⋅ cos x
!
dx                   sin x
!                      ↓
↓           Derivative of
Derivative of log sin x w.r .t x
(sin x ) w.r .t sin x

and so on. We will encounter lots examples of the application of this rule soon.
You must note here that for this rule to be applicable at any x = a the function f(x) must be differentiable
at x = g(a), since the variable that is the argument (input) to f is g(x) {and not x}.

Maths / Differentiation
LOCUS                                                                                                         16

Example – 1

Evaluate the derivatives of the following functions by first principles:
(a) f ( x ) = log (sin x )                 (b) f ( x ) = sin ( log x )

(c) f ( x ) = 3 sin x                      (d) f ( x ) = e x
2

Solution: These functions are all compositions of multiple functions and can easily be differentiated using the
Chain rule. However, our purpose here is to differentiate them using first principles. This exercise,
instead of being considered futile, should be seen as an application of your algebraic manipulation
skills.
log (sin ( x + h )) − log (sin x )
(a) f ' ( x ) = lim
h →0                   h
 sin ( x + h ) 
log                
= lim         sin x 
h →0          h
 sin x cos h + cos x sin h 
log                            
= lim                 sin x           
h →0                 h
log ( cos h + cot x sin h )
= lim
h →0               h
log {cos h (1 + cot x tan h )}
= lim
h→0                   h
log ( cos h )       log (1 + cot x tan h )
= lim               + lim
h→0       h         h→0           h

This limit is 1

log {1+ ( cos h − 1)}      log (1+ cot x tan h) cot x tan h
= lim                      + lim                     ⋅
h→0          h             h→0      cot x tan h          h

This limit is 1

log {1+ (cos h − 1)} cos h -1
= lim                                   + cot x
h→0          cos h − 1          h

cos h − 1
= lim           + cot x
h→0     h
= 0 + cot x
= cot x

Maths / Differentiation
LOCUS                                                                                                                       17

sin ( log ( x + h )) − sin ( log x )
(b) f ' ( x ) = lim
h →0                     h

 log ( x + h ) + log ( x )   log ( x + h ) − log ( x ) 
cos                             sin                        
= 2 lim                  2                           2              
h →0                                 h

 1           h  
 sin  2 log 1 + x   
                    
= 2 cos (log ( x )) lim                        
h→0             h
                        

                        


  1         h   1       h 
  sin  log 1 +     log 1 +  
            x   2
= 2 cos (log ( x )) lim                              x 
2
h →0 
⋅
 1        h             h      
  2 log 1 + x                   

                                 


        h 
 1 log 1 + x  

= 2 cos (log ( x )) lim              

h →0 2
      x⋅h
x 

                


cos ( log ( x ))
=
x

3   sin ( x + h ) − 3 sin x
(c) f ' ( x ) = lim
h →0                 h

sin ( x + h ) − sin ( x )                             1
= lim                                ⋅
sin ( x + h ) + (sin x ) + sin x ⋅ sin ( x + h )
h→0                                                  2/3       2/3                       1/ 3
h
                                                
{This step was accomplished by rationalization}
cos x
=
3 ⋅ (sin x )
2/3

Maths / Differentiation
LOCUS                                                                                                                 18

e( x + h ) − e x
2                2

(d) f ' ( x ) = lim
h→0           h
+ h 2 + 2 xh
− ex
2                            2
ex
= lim
h→0                            h
+ 2 xh
−1
2
eh
= e lim
x2
h →0                  h

 e h2 + 2 xh − 1 h 2 + 2 xh 
                            
= e lim  2
x2
⋅          
 h + 2 xh
h→0                        h
                            

&
The limiting value of this ratio will be 1

= ex ⋅ 2x
2

Example – 2

Evaluate the derivatives of the following functions using the techniques developed in section - 3:

 x +1 
(a) f ( x ) = ( x + 3 x ) sin x                                             (b) f ( x ) = tan 
2           −1

 x+2

(c) f ( x ) = cos    −1
( 1− x )              2
+
x2 + 2x + 3
x+2
(d) f ( x ) = 4 − tan −1 x 2

Solution: (a) This function can be differentiated by the Product Rule (Rule - 3)

d ( f ( x ))                                      d (sin −1 x )             d ( x2 + 3x )
= ( x + 3x )
2                                           −1
+ sin x
dx                                                   dx                    dx

=
(x        2
+ 3x )
+ (sin −1 x ) ⋅ ( 2 x + 3)
1− x             2

 x +1 
(b) This function will be differentiated by the Chain rule (Rule 5); the argument        can be
 x+2
differentiated by the Quotient rule (Rule 4):
d ( f ( x ))
dx
  x + 1   x + 1 
d  tan        d     
=     x + 2  ⋅  x + 2 
 x +1         dx
d        
 x+2

Maths / Differentiation
LOCUS                                                                                                                                19

 x + 1  ( x + 2 ) ⋅ (1) − ( x + 1) ⋅ (1)
= sec 2        ⋅
 x+2               ( x + 2)
2

1           x +1 
⋅ sec 2       
( x + 2)          x+2
2

(c) Here, we will have to use a combination of these techniques as follows:

                       
d ( f ( x ))                        −1       −x       ( x + 2 )( 2 x + 2 ) − ( x 2 + 2 x + 3) ⋅ (1)
=                      2
⋅      +
(                   )                   ( x + 2)
2
dx             1 −            1 − x2  1 − x
2

                       

1                x2 + 4 x + 1
=                     +
( x + 2)
2
x 1 − x2
You are urged to work out the solution on your own.

d ( f ( x ))                     1                      −1
(d)                  =                                  ⋅          ⋅ 2x
dx             2 4 − tan −1 x 2                   1 + x4

−x
=
(1 + x )  4
4 − tan −1 x 2

Example – 3

If f '(x) = g(x), find the derivative of f–1 (x)

Solution: To evaluate the required derivative, we can apply the chain rule on the relation:
f ( f −1 ( x )) = x

⇒
d
dx
{
f ( f −1 ( x )) = ( x ) = 1
d
dx
}
⇒ f ' ( f −1 ( x ) )               ( f −1 ( x ) ) = 1
d
dx

⇒
d
( f −1 ( x )) f ' f 11 ( x ) = g f −11 ( x )
dx                  ( − ) (                     )

Maths / Differentiation
LOCUS                                                                                                               20

d
For example, we know that         (sin x ) = cos x
dx

⇒
d
(sin −1 x ) = cos sin −1 ( x )
1
dx                  (             )
1
=
(
cos cos −1 1 − x 2   )
1
=
1 − x2
In this way, we can evaluate the derivative of any inverse function, given the derivative of the
original function.

Section - 4           DIFFERENTIATION OF PARAMETRIC/IMPLICIT FUNCTIONS

(A)          PARAMETRIC FUNCTIONS
Sometimes, when expressing y as a function of x, one might not use a direction relation between x and
y; instead, one might express both x and y as functions of a third variable, say t:
x = f (t )
y = g (t )
dy
In that case, how would      be evaluated?
dx
dy
One option is to eliminate the parameter t and obtain a relation involving only x and y, from which
dx
may be obtained; however, this could lead to cumbersome expressions.
dy
Another alternative can be taken as follows; we rearrange              to involve t also:
dx
dy
dy dt
=
dx dx
dt
This relation says that for evaluating the derivative of y w.r.t x, we evaluate the derivative of y and x
w.r.t the parameter t, and then take their ratio.

Maths / Differentiation
LOCUS                                                                                                            21

Let us try this on some examples:
(i) x = r cos θ             y = r sin θ     ;        r is a constant
This parametric relation represents a circle of radius r. We will follow both the approaches to
dy
determine      :
dx
⇒       ELIMINATION:
Square and add the two relations for x and y to obtain:
x2 + y 2 = r 2
⇒ y = ± r 2 − x2
dy       x             For each x, we obtain two y '       
⇒      =±                                                      
dx    r − x2
2
 values because the curveis a circle 

⇒       PARAMETRIC DIFFERENTIATION
dx                    dy
= − r sin θ           = r cos θ
dθ                    dθ
dy dy / dθ
⇒     =        = − cot θ
dx dx / dθ

(ii) x = a cos θ              y = b sin θ     ;       a, b are constants
This parametric relation represents an ellipse with major and minor axis 2a and 2b respectively.

⇒       ELIMINATION
θ can easily be eliminated to obtain:
x2 y 2
+   =1
a 2 b2
b 2
⇒y=±      a − x2
a
dy    b           x
⇒      =±
a − x2
dx    a       2

⇒       PARAMETRIC DIFFERENTIATION
dx                        dy
= − a sin θ               = b cos θ
dθ                        dθ
dy dy / dθ −b
⇒     =        =   cot θ
dx dx / dθ   a
In later examples, we will observe that in many cases, parametric differentiation turns out to be
much more convenient than differentiation after elimination.
Maths / Differentiation
LOCUS                                                                                                                  22

(B)          IMPLICIT FUNCTIONS
Sometimes, the relation between the variables x and y is specified in the form f(x, y) = 0 that is, y is not
explicitly specified in terms of x, since this explicit expression is either not possible or not convenient.
In such a case, y is said to be an implicit function of x.
dy
How do we find        in such a case?
dx
dy
We simply differentiate the relation f(x, y) = 0 with respect to x, using       for the derivative of the
dx
dy
variable y. Then we solve for    .
dx
This will become clear from some examples:
⇒       x2 + y 2 = 1
Differentiating both sides w.r.t x:
dy
2x + 2 y       =0
dx
dy − x
⇒      =
dx   y

⇒ x3 + y 3 + 2 xy = 2
Differentiating both sides w.r.t x:
dy     dy
3x 2 + 3 y 2      + 2x + 2 y = 0
dx     dx

dy − (3x + 2 y )
2

⇒    =
dx   3y2 + 2x

⇒ y = cos ( x + y )
Differentiating both sides w.r.t x:
dy                    dy 
= − sin ( x + y )  1 + 
dx                    dx 
dy − sin ( x + y )
⇒     =
dx 1 + sin ( x + y )
dy
Observe that in case of differentiation of implicit functions, the expression for the derivative
dx
will generally not be independent of y.

Maths / Differentiation
LOCUS                                                                                                                23

Example – 4

1                      1                   dy −1
If x 2 + y 2 = t + and x 4 + y 4 = t 2 + 2 , then prove that   =
t                     t                    dx x3 y
Solution: We first try to use the two given relations to get rid of the parameter t, so that we obtain a (implicit)
relation between x and y.
1
x2 + y 2 = t +
t
Squaring, we get
1
x4 + y 4 + 2x2 y 2 = t 2 +       +2                      ... (i)
t2
Using the second relation in (i), we get
2 x2 y2 = 2
1
⇒ y2 =
x2
Differentiating both sides w.r.t x, we get
dy −2
2y =
dx x 3
dy −1
⇒   =
dx x 3 y

Example – 5

If y 2 = a 2 cos 2 x + b2 sin 2 x, then prove that
d2y     a 2b2
+y= 3
dx 2     y

Solution: The final relation that we need to obtain is independent of sin x and cos x; this gives us a hint that using
the given relation, we must first get rid of sin x and cos x:
y 2 = a 2 cos 2 x + b 2 sin 2 x

=     {                                 }
a ( 2 cos 2 x ) + b 2 ( 2 sin 2 x )
1 2
2

=     {a (1 + cos 2 x ) + b2 (1 − cos 2 x )}
1 2
2

=
1
2
{
(a 2 + b2 ) + (a 2 − b 2 ) cos 2 x    }
⇒ 2 y 2 − ( a 2 + b 2 ) = ( a 2 − b 2 ) cos 2 x           ... (i)

Maths / Differentiation
LOCUS                                                                                                                             24

Differentiating both sides of (i) w.r.t x, we get

= −2 ( a 2 − b 2 ) sin 2 x
dy
4y
dx
= ( a 2 − b 2 ) sin 2 x
dy
⇒ −2 y                                                   ... (ii)
dx
We see now that squaring (i) and (ii) and adding them will lead to an expression independent of the
trig. terms:
2

(2 y − (a                    ))          dy 
+ 4 y   = ( a2 − b2 )
2                          2
2          2
+b   2               2

 dx 
A slight rearrangement gives:
2
 dy                         a 2b 2
     + y 2 − (a 2 + b2 ) = − 2                                                  ... (iii)
 dx                          y
Differentiating both sides of (iii) w.r.t x:

 dy   d y 
2
dy 2a 2b 2 dy
2  2  + 2y     =
 dx   dx   dx   y 3 dx

d2y   a 2b2
⇒ 2 +y= 3
dx     y

Example – 6

g(x)
If the derivatives of f(x) and g(x) are known, find the derivative of y = { f ( x )}
Solution: We cannot directly differentiate the given relation since no rule tells us how to differentiate a term pq
where both p and q are variables.
What we can instead do is take the logarithm of both sides of the given relation:
g(x)
y = { f ( x )}
⇒ ln y = g ( x ) ln ( f ( x ))
Now we differentiate both sides w.r.t x:

⋅ f ' ( x ) + ln ( f ( x )) ⋅ g ' ( x )
1 dy            1
⇒         = g (x)⋅
y dx          f (x)

 f '( x) g (x)

+ g ' ( x ) ln ( f ( x ))
dy
⇒       = y
dx    
     f (x)

g(x)    f '( x) g ( x)
                                         

= ( f ( x ))                           + g ' ( x ) ln ( f ( x ))

     f (x)                               


Maths / Differentiation
LOCUS                                                                                                                25

As a simple example, suppose we have to differentiate y = xx :
ln y = x ln x

1 dy      1
⇒          = x ⋅ + ln x ⋅1
y dx      x
= 1 + ln x
dy
⇒        = y (1 + ln x )
dx

= x x (1 + ln x )

Section - 5                                               L'HOSPITAL' RULE

We had made a mention of the L'Hospital's Rule (abbreviated as the LH rule) in the unit on limits. We had deferred
the introduction of this rule to this chapter since it requires the use of differentiation. The LH rule can be used to
0    ∞
evaluate limits that are of the form     or
0    ∞

Consider two functions f ( x ) and g ( x ) which are differentiable in the neighbourhood of the point x = a (except

possibly at the point x = a itself). Let g ' ( x ) ≠ 0 in this neighbourhood.
The LH rule says that
if       lim f ( x ) = lim g ( x ) = 0
x→a             x→a

or       if       lim f ( x ) = lim g ( x ) = ∞
x→a             x→ a

f (x)        f '( x)
lim           = lim
then              x→a    g ( x ) x →a g ' ( x )

f (x)                                                                 0  ∞
provided that the limit lim          exists. Although the LH rule is applicable to limits of the form or , you
x→a   g (x)                                                                 0  ∞

should be able to understand that other indeterminate forms like 0.∞, ∞ − ∞, 1∞ , ∞ 0 or 00 can be reduced to
these two indeterminate forms using appropriate algebtain manipulations.
You are urged to think of some (non-rigorous) justifiction for this rule.
Lets apply this rule on some examples.

Maths / Differentiation
LOCUS                                                                                                              26

Example – 7

Evaluate:
ln x
(a)        lim x ln x         (b)           lim
x →0                             x →∞     x
Solution: We have encountered both these limits in the unit on Limits. Here, we’ll re-evaluate them using the LH
rule.

(a)                           L = lim x ln x
x →0
(0 × −∞ form )
ln x            −∞      
= lim
x →0     1                 form 
x            ∞       

1
= lim         x
x→0
− 1 2          (By applying the LH rule)
x
= lim ( − x )
x →0

=0
This is what we got earlier
ln x           ∞      
(b)                           L = lim                        form 
x →∞      x            ∞      
1
= lim x                    (By applying the LH rule)
x →∞ 1

=0
Example – 8

sin x
1
Evaluate lim+  
x→0  x 

0   ∞
Solution:   This limit is of the indeterminate form ∞ 0 . Lets first convert it into the form     or .
0   ∞
sin x
1
L = lim+  
x→0  x 

Doing this is justified since 
1                                               
= lim sin x ⋅ ln                    x > 0 ln  1  is defined    
x →0 +
 x                                            
           x                

Maths / Differentiation
LOCUS                                                                                                            27

= − lim sin x ⋅ ln x
+
x →0

= − lim
ln x                            ∞     
 form 
∞
+
x →0   cosec x                                  

1
= − lim         x                            (By applying the LH rule)
+
x →0   −cosec cot x

sin 2 x
= lim+
x →0   x cos x

 sin x 
= lim          ⋅ tan x
x→0 
+
x 
=0

Example – 9

 1       x 
Evaluate lim       −      

x →1 ln x   ln x 

1− x              0     
Solution:                           L = lim                      form 
x →1
 ln x             0     

= lim 
 −1 

x →1 1/ x


(By applying the )
LH rule

= lim ( − x )
x →1

= −1

Example – 10

 x 2 + sin x 
Evaluate lim               
x →∞
      x2     

∞
Solution: The limit is of the indeterminate form     , so we apply the L.H. rule:
∞

Maths / Differentiation
LOCUS                                                                                                                 28

 x 2 + sin x             ∞     
L = lim                            form 
x →∞
      x2                 ∞     

 2 x + cos x             ∞          
= lim                            form agin            ...(1)
x →∞
     2x                  ∞          

 2 − sin x 
= lim                                                   ...(2)
x →∞
     2     

1
= 1 − lim (sin x)
2 x →∞
Now, we know that lim(sin x ) does not exist since sin x is an oscillating function and does not
x →∞

converge to any particular value. What does this imply for our current limit? Does it not exist ? Think
 x 2 + sin x 
about the expression               carefully:
      x2     

 x 2 + sin x          sin x 
lim                = lim 1 + 2 
   x 
x →∞          2        x →∞
      x      

 sin x 
= 1 + lim  2 
x →∞
 x 

= 1+ 0            (∵ sin x is bounded )
=1
Thus, a limit does infact exists while the LH rule says that it does not exist. Why?
This is because the LH rule is not applicable here. Go back to the definition of the LH rule which says
f ( x)       f ′( x)
that lim            = lim         if the latter limit exists.
x→a    g ( x) x→a g ′( x)
In this example, you cannot apply the LH rule on the expression in (1) since the limit for the expression
obtained after differentiation (the one in (2)) does not exist.
Thus, the LH rule must be used with care.

Maths / Differentiation

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 6 posted: 8/2/2012 language: pages: 28