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Diffrenciation

UNIT 6 DIFFERENTIATION

Structure

6.1 Introduction Objectives

6.2 Definition of Derivative, its Meaning and Geometrical Interpretation 6.3 Derivative at a Point

6.4 Derivative by First Principle 6.5 Chain Rule

6.6 Derivatives of Exponential, Logarithmic, Parametric and Implicit Functions

6.7 Derivatives of Higher Orders 6.8 Concept of Maxima and Minima 6.9 Summary

6.10 Solutions/Answers

6.1 INTRODUCTION

In the preceding unit, we have discussed concept of limit and continuity. In fact, the definition of derivative involves these concepts. So, learner must go through the previous unit before starting this unit. Derivatives have large number of applications in the fields of mathematics, statistics, economics, insurance, industrial, health sector, etc.

In this unit, we will present this concept from a very simple and elementary point of view, keeping in mind that learner knows nothing about derivatives.

In this unit, we have discussed some examples basically based on the formulae for derivatives of a constant, polynomial, exponential, logarithmic, parametric and implicit functions. Product rule, quotient rule, chain rule have also been discussed. Finally, we close this unit by giving a touch to higher order derivatives and maxima and minima of functions.

Objectives

After completing this unit, you should be able to:

 find derivative of a function at a particular point and at a general point;

 find derivative by first principle;

 find derivative of some commonly used functions;

 apply the chain rule;

 find derivative of exponential, logarithmic, parametric and implicit functions;

 find higher order derivatives; and

 find maxima and minima of a function.

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Fundamentals of

Mathematics-II

6.2 DEFINITION OF DERIVATIVE, ITS

MEANING AND GEOMETRICAL INTERPRETATION

Definition

Let f:DR be a function, where DR, i.e. f is a real valued function defined on D.

Let aDthen derivative of f at x = a is denoted by f'(a)and is defined as h ,

) a ( f ) h a ( limf ) a ( ' f

0 h

 

provided limit exists … (1)

From definition (1), we see that f'(a)measures the rate at which the function f(x) changes at x = a. This is clear from the figure 6.1 given below.

Geometrical Interpretation

Fig. 6.1

Let PT be the tangent at point P of the curve of the function y = f(x).

Draw PLOX,QMOX,PR QM Let OL = a, OM = a + h

h a h a OL OM LM

PR      

and RQMQMRMQLPf(ah)f(a)

(1)

PR lim RQ h

) a ( f ) h a ( limf ) a ( ' f

0 h 0

h   



   

PR

RQ Base

lar perpendicu tan

, PQR in tan

lim

0

h

Now as h0,chord PQ tends to coincide with the tangent at point P, i.e. as h    0

f'(a) lim tan tan

0 h

i.e. f'(a)tan

i.e. (derivative at point x = a) = (tangent of the angle which the tangent line at x = a makes with +ve direction of x-axis)

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Diffrenciation

In fact, if a line makes an angle  with position direction of x-axis, then value oftan is known as slope of the line.

Thus in mathematical language we can say

. int po that at gent tan the of slope the is int) po a at Derivative

( xa … (2)

i.e. we can say that derivative measures the rate at which the tangent to the curve at point x = a is changing

Meaning

Rewriting (1) h

) a ( f ) h a ( lim f ) a ( ' f

0 h

 

… (3) From the knowledge of previous unit, we know that limit in R.H.S. of (1) or (3) exists if

h ) a ( f ) h a ( lim f h and

) a ( f ) h a ( lim f

0 h 0

h

both exist and are equal.

In view of (2), we have, limit in (1) exists if





 





 ofthe pointx a

right the to gent tan the of Slope a

x int po the of

left the to gent tan the of Slope

i.e. limit in (1) exists if x = a is not a corner point.

i.e. f'(a)does not exists at corner points. … (4) For example, consider the function

x ) x (

f 

See the graph of this function in Fig. 6.2 .We observe that x = 0 is a corner point in its graph.

Fig . 6.2

So f(x) xis not differentiable at x = 0. but there is no other corner point in its graph, so it is differentiable at all points of the domain except x = 0.

Remark1:

(i) The last paragraph is very useful to understand the concept of derivative for those learners, who do not have mathematical background.

(ii) However, the units have been written keeping in mind that the learner has no mathematical background after 10th standard.

(iii) In the definition of derivative of a function at a point given by (1), we see that in order to find the derivative at the said point, we have to evaluate the limit in R.H.S. But sometimes functions may have different values for h0 andh0. In such cases like modulus function or where there is break for function in order to evaluate the limit we have to

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Fundamentals of Mathematics-II

find out the L.H.L. and R.H.L. as we have done in the previous unit. But in the definition of derivative these L.H.L. and R.H.L. are known as left hand derivative (L.H.D.) and right hand derivative (R.H.D.) respectively.

A function is said to have derivative at a point if L.H.D. and R.H.D. both exist and are equal at that point, i.e. L.H.D. = R.H.D.

We denote L.H.D. of the function f(x)atx abyL(f'(a)) and R.H.D. of the function f (x) at xa by R(f '(a)). See Example 2 of this unit for more clarity.

6.3 DERIVATIVE AT A POINT

Here, we give some examples which will illustrate the idea as to how we calculate derivative of a function at a point.

Example 1: Find the derivative of the following functions at the indicated points:

(i) f(x)a, at x5, where a is a real constant (ii) f (x) axb, at x2 , a0

(iii) f(x)ax2 bxc, at x3, a 0

(iv) , atx 1

x ) 1 x (

f  

Solution:

(i) f(x)a, where a is real constant By definition

lim0 0

h lim 0 h

a lima h

) 5 ( f ) h 5 ( limf ) 5 ( ' f

0 h 0 h 0

h 0

h

 

 

 

(ii) f(x)axb, a 0 By definition

h ) 2 ( f ) h 2 ( limf ) 2 ( ' f

0 h

 

lima a

h limah h

) b a 2 ( b ) h 2 ( lim a

0 h 0 h 0

h

 

 

(iii) f(x)ax2 bxc, a0 By definition

h ) 3 ( f ) h 3 ( limf ) 3 ( ' f

0 h

 

h

) c b 3 a 9 ( c ) h 3 ( b ) h 3 ( lim a

2 0

h

 

(iv)

By definition

h 1 1 h 1

1 h lim

) 1 ( f ) h 1 ( limf ) 1 ( ' f

0 h 0

h

 

 

 

h(1 h)

lim h ) h 1 ( h

) h 1 ( lim1

0 h 0

h

 

 

1

0 1

1 h

1 lim 1

0

h 

 

 

b a 6 ) b a 6 ah ( h lim

bh ah 6 limah

0 h 2

0 h

 

 

x ) 1 x (

f 

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Diffrenciation

Here are some exercises for you.

E 1) Find the derivative of the following functions at the indicated points (i) f(x)x3x1, atx 1

(ii) f (x) 2 3x ,2 at x1/ 2

E 2) Find the value of a, if f'(2)3,where f(x)2x23ax5

Example 2: Find the derivative (if exists) of the following functions at the indicated points.

(i) f(x) x at x0 (ii)



 

1 x , x 2 9

1 x , x 2 ) 5

x (

f at x = 1

Solution:

(i) By definition

h 0 h lim 0 h

) 0 ( f ) h 0 ( lim f ) 0 ( ' f

0 h 0

h

 

 

h

lim h h

0 lim h

0 h 0

h  

We note that to deal with h we must know the sign of h in advance. So We must have to calculate L.H.D. and R.H.D. separately.

h lim h . D . H . L

0 h

Putting h = 0 – k as h0 k0 L.H.D.

k lim k k

k lim 1

k lim k k 0

k lim 0

0 k 0

k 0

k 0

k  

 

 

 

lim( 1) 1

0 k

… (1) R.H.D. = … (2) From (1) and (2), we have

L.H.D.R.H.D.

f'(0)does not exists.

(ii) By definition

h 7 ) h 1 ( limf h

) 1 2 5 ( ) h 1 ( limf h

) 1 ( f ) h 1 ( limf ) 1 ( ' f

0 h 0

h 0

h

 

 

 

We note that function have different values for x<1 and x >1, so we must have to calculate L.H.D. and R.H.D. separately.

h 7 ) h 1 ( 2 lim 9 h

) 1 ( f ) h 1 ( lim f . D . H . L )) 1 ( ' f ( L

0 h 0

h

 

 

lim( 2) 2 h

h lim 2

0 h 0

h

 

… (1)

h 7 ) h 1 ( 2 lim 5 h

) 1 ( f ) h 1 ( lim f . D . H . R )) 1 ( ' f ( R

0 h 0

h

 

 

lim (2) 2 h

h lim 2

0 h 0

h

… (2) From (1) and (2)

L(f'(1))R(f'(1)) f'(1)does not exists.

1 ) 1 ( h lim lim h h lim h

0 h 0 h 0 h

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Fundamentals of Mathematics-II

Remark 2: In part (i) x =0 is a corner point (see Fig. 6.2) that is why its

derivative did not exist at this point, which was indicated in equation (4) in Sec.

6.2. Same is the case in part (ii).

6.4 DERIVATIVE BY FIRST PRINCIPLE

In section 6.3 of this unit, we have discussed as to how we calculate the derivative of a function at a given point x = a (say). Suppose we want to calculate the derivative at 10 points, then using the definition 10 times is a very time consuming and lengthy procedure. To get rid of this difficulty, we will introduce a procedure in this section which will provide us the derivative of the function at a general point. After calculating the derivative at the general point we can replace this point by any number of points very quickly (provided derivative at the required point exists). Let us first describe the procedure as to how we calculate the derivative at a general point. After this we shall give some results to get a good understanding of the procedure. This process of finding derivative is known as derivative by first principle or by definition or by delta method or ab-intio.

Let us explain the procedure of first principle for the function y = f(x) … (1)

in the following steps.

Step I Let xbe the small increment (+ve or –ve) in the value of x and y be the corresponding increment in the value of y.

(1) becomes

yyf(xx) … (2) Step II (2) – (1) gives

yf(xx)f(x) … (3)

Step III First we simplify the expression in the R.H.S. of (3). After simplifying the expression, we divide both sides byxand get

x ) x ( f ) x x ( f x y

 

Step IV Proceeding limit as x on both sides

x ) x ( f ) x x ( lim f x lim y

0 x 0

x

 

… (4) Step V The term in L.H.S. of (4) is denoted by

dx

dyand limit in R.H.S. of (4) is evaluated using suitable formula discussed in the previous unit i.e.

x ) x ( f ) x x ( lim f dx dy

0

x

 

… (5)

The expression obtained after simplification of the R.H.S. of (5), is derivative of y w.r.t. x at a general point x.

If we want the derivative of the function y = f(x) at a particular point x = a (say), then replace x by a in the result.

Some Results

Result 1: Find the derivative of the constant function given by f (x) = k, where k is a real constant

by using first principle.

Solution: Let y = f(x) = k … (1) 0

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35

Diffrenciation

Step I Let xbe the small increment in the value of x and y be the corresponding increment in the value of y.

(1) becomes

y + y = k … (2) Step II (2) – (1) gives

yyykk

Or y0 … (3) Step III Dividing on both sides of (3) by x

0

x 0 x

y 

 

 0

x y 

 

Step IV Proceeding limit as x0,we get

lim 0

x lim y

0 x 0

x

Step V 0 dx dy 

i.e. (k) 0 dx

d  .

Result 2: Find the derivatives of the functions given by (i) f(x)x2 (ii) f(x)x3

by using first principle.

Solution: (i) Let yf(x) x2 … (1) Step I Let xbe the small increment in the value of x and y be the

corresponding increment in the value of y.

(1) becomes

yy(xx)2 = x2

 

x 2 2xx … (2) Step II (2) – (1) gives

yx2

 

x 2 2xxx2 =

 

x 22xx Step III Dividing on both sides by x, we get

 

x x x 2 x x

y 2

 

 x2x

Step IV Proceeding limit as x0,we get

0 x 0

x lim

x lim y

 

x2x

Step V

0 x

dx lim dy

 x+

0 x

lim

2x = 0 + 2x = 2x i.e. (x2) 2x 2x21

dx

d

Second Method

Let yf(x)x2 … (1)

Step I Let xbe the small increment in the value of x and y be the corresponding increment in the value of y.

So, derivative of a constant function is zero.

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Fundamentals of Mathematics-II

36

(1) becomes

yy(xx)2 … (2) Step II (2) – (1) gives

y(xx)2x2

Step III Dividing on both sides by x, we get

x

x ) x x ( x

y 2 2

 

x ) x x (

x ) x x

( 2 2

 

Step IV Proceeding limit as x0,we get x

) x x (

x ) x x lim ( x lim y

2 2 0

x 0

x  

 

Step V as x 0 x x x

x ) x x (

x ) x x lim ( dx

dy 2 2

x x x

 

 

i.e.

 

x2 2x2 1

dx

d

 = 2x 

 

1 n n n a

x na

a x

a limx

(ii) Let y f(x) x3 … (1) Step I Let xbe the small increment in the value of x and y be the

corresponding increment in the value of y.

(1) becomes

yy(xx)3 = x33x2x3x

 

x 2

 

x 3 … (2) Step II (2) – (1) gives

y = x33x2x3x

 

x 2

 

x3x3 = 3x2x3x

 

x 2

 

x 3

Step III Dividing on both sides by x, we get

   

x

x x

x 3 x x 3 x

y 2 2 3

 

 = 3x2 3xx

 

x 2 Step IV Proceeding limit as x0,we get

0 x 0

x lim

x lim y

 

3x23xx

 

x 2

Step V

0 x

dx lim dy

 3x2+

0

lim

x

 

2

0

x x

lim x x

3   

= 3x200 = 3x2 i.e. (x3) 3x2 3x31

dx

d

Second Method

Let yf(x)x3 … (1)

Step I Let xbe the small increment in the value of x and y be the corresponding increment in the value of y.

(1) becomes )3

x x ( y

y   … (2)

Step II (2) – (1) gives

3

3 x

) x x (

y  

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37

Diffrenciation

Step III Dividing on both sides by x, we get

x

x ) x x ( x

y 3 3

 

x ) x x (

x ) x x

( 3 3

 

Step IV Proceeding limit as x0,we get

(x x) x

x ) x x lim ( x lim y

3 3 0

x 0

x  

 

Step V as x 0 x x x

x ) x x (

x ) x x lim ( dx

dy 3 3

x x

x     

 

i.e. (x3) 3x31 dx

d

 

 

1 n n n a

x na

a x

a limx

Result 3: Find the derivative of the function given by

ax b

2

) x (

f   , where a, b are real constants and a0 by using first principle.

Solution: Let us use second method here.

Let yf(x)(axb)2 … (1)

Step I Let xbe the small increment in the value of x and y be the corresponding increment in the value of y.

(1) becomes

yy[a(xx)b]2 (axaxb)2 … (2) Step II (2) – (1) gives

y(axaxb)2(axb)2 Step III Dividing on both sides by x, we get

x

) b ax ( ) b x a ax ( x

y 2 2

 

 

 

x a

) b ax ( ) b x a ax a (

2 2

 

) b ax ( ) b x a ax (

) b ax ( ) b x a ax a (

2 2

Step IV Proceeding limit as x0,we get

 

(ax a x b) (ax b) ) b ax ( ) b x a ax lim ( x a

lim y

2 2

0 x 0

x

as x 0 a x 0

) b ax ( ) b x a ax (

) b ax ( ) b x a ax lim ( a

2 2

0 x

a      

 

Step V

 

(ax a x b) (ax b) ) b ax ( ) b x a ax lim (

dx a

dy 2 2

b ax b x a ax

a 2(axb)21

 

1 n n n a

x na

a x

a limx

Similarly, (xn) nxn 1

dx

d

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Fundamentals of Mathematics-II

38

i.e.

ax b

2 2a(ax b)2 1 dx

d

Similarly, we can easily obtain

n n 1

d (ax b) na(ax b) dx

  

Result 4: Find the derivative of the exponential function f(x)eax by using first principle.

Solution: Let yf(x)eax … (1)

Step I Let x be the small increment in the value of x and y be the corresponding increment in the value of y.

(1) becomes

yyea(xx)eaxax … (2) Step II (2) – (1) gives

ax x a

ax e

e

y 

eaxeaxeax [amn aman] = eax

eax1

Step III Dividing on both sides by x, we get x

) 1 e e ( x

y ax a x

 





 

x a

1 ae e

x a ax

Step IV Proceeding limit as x0,we get





 

a x

1 ae e

x lim lim y

x a ax 0 x 0

x

Step V

x a

1 lim e

dx ae

dy ax

0 x a ax

 

as x0ax0 aeax (1) 

  

1

x 1 lime

x 0 x

 i.e. (eax) aeax

dx

d 

Result 5: Find the derivative of the logarithm function f(x)logax by using first principle.

Solution: Let yf(x)logax … (1) Step I Let xbe the small increment in the values of x

and y be the corresponding increment in the value of y (1) becomes

yyloga(xx) … (2) Step II (2) – (1) gives

yloga(xx)loga x 



  



 

 

 n

logm n log m x log

x

loga x 

 

 

 x

1 x loga

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39

Diffrenciation

Step III Dividing on both sides by x, we get

x

x 1 x log x

y a



 

 

 





 

 

 x)

1 x xlog x x 1

a

x

x

a x

1 x x log

1



 

 

 [nlogmlogmn] Step IV Proceeding limit as x0,we get

x x 0 a

x 0

x x

1 x xlog lim 1 x

lim y

 

 

 

Step V 0

x 0 x x x as

1 x log x lim

1 dx

dy xx

a x 0

x  

 

 

 



 

 

x x

x 0 a x

x 1 x lim x log

1 



domain its

on function

continuous a

is arithm

log

log e x 1

a

 





  

1 x e

lim 1/x

0 x

i.e. log e

x ) 1 x dx(log

d

a

a

In particular, if base of the logarithmic is e in place of a, then

 

x e 1 xlog x 1 dx log

d

e

e   as logee1

Remark 3: Keep all these formulae put in the rectangular boxes always in mind, as we will use these formulae later on as standard results.

Some more Formulae of Finding Derivatives:

If u and v are functions of x, then (i)

 

dx cdu dx cu

d  , where c is a real constant

(ii)

 

dx dv dx v du dx u

d   

(iii)

 

dx vdu dx udv v . dx u

d   (Known as Product Rule)

(iv) 2

v dx udv dx vdu v u dx

d 



 

 (Known as Quotient Rule)

Remark 4: Aim of this unit from learners point of view is not to focus on the derivations of the formulae. But main aim of this unit is able to make the learners user friendly as to how these results can be used whenever we encounter a situation where derivative is involved. That is why we will not provide the derivations of more formulae.

If a function is continuous then it respects limit i.e. if a function f is continuous and a is point of its domain, then

x a x a

lim f (x) f (lim x) f (a)

i.e. limit can be taken inside the function.

i.e. Role of limit and function can be interchanged.

(12)

Fundamentals of Mathematics-II

40

Various formulae which are used in many practical situations are listed below:

S.

No

Function Derivative of the function 1 k (constant)

0 ) k dx(

d 

2 xn n n 1

nx ) x dx(

d

3 (axb)n (ax b)n na(ax b)n 1

dx

d

 4 Exponential function

(i) abx (ii) ebx

(i) (a ) ba loga dx

d bx bx

 (ii) (ebx) bebx

dx

d 

5 Logarithmic function (i) loga x

(ii) logex

(i) log e

x ) 1 x dx(log

d

a

a

(ii)

x ) 1 x dx(log

d

e

6 cu, where

c is constant and u is a function of x

 

u dx c d ) cu dx(

d 

7 (i) u v (ii) uv (iii)

v u

where, u, v are functions of x.

(i) dx

dv dx ) du v u dx(

d   

(ii) d dv du

(uv) u v (Pr oduct Rule)

dx  dx dx

(iii) (QuotientRule)

v dx udv dx vdu v u dx

d

2



 

 8

f(x)

n,n is +ve or –ve

real number

   

(f(x))

dx ) d x ( f n ) x ( dx f

d n n1

 9

) x ( f

1

 

 

dx(f(x))

d ) x ( f ) 1 x ( dx f

d ) x ( f

1 dx

d

2

1





10 y = f(u) u = g(w) w = h(x)

) Rule Chain dx (

dw dw du du dy dx dy 

11 Parametric functions x = f(t)

y = g(t) dt

dx dt dy dx dy 

12 Polynomial function

n 1 n

1 n 1 n 0

a x a

...

x a x a ) x ( f

n1

2 n 1 1

n

0x (n 1)a x ... a na

)) x ( f dx(

d

   

Now we take some examples. We will write “Diff. w.r.t x” in place of

“differentiating with respect to x”.

(13)

41

Diffrenciation

Example 3: Find derivative of the following functions:

(i) 5 (ii)  (iii) 11

1 (iv) 17

 (v) x11

(vi) x5/2 (vii)

3 x

1 (viii) x7

1 (ix) x

1 (x) x

(xi) (2x5)3 (xii) (43x)8 (xiii)

9 / 4

2x

5 3 

 

  (xiv) 23x

(xv) x5x2 1 (xvi) x

1 x x32

(xvii)

2 2 2

x

x 1 

 

  (xviii)

100

x x 1

 

 

(xix) (x2 1)(x1) (xx) x3(1x2x5x8) (xxi) (x2 1)(x3x2 1) (xxii) (4x1)3(7x1)4 (xxiii) (x2)2(x3)4(x1)5

Solution:

(i) Let y = 5 Diff. w.r.t. x 0

dx

dy  5 is a constant and derivative of a constant function is zero.

 

 

 

(ii) Let y =  Diff. w.r.t. x 0

dx

dy  and both are constants is constant and derivative of a constant function is zero.

     

 

 

 

(iii) Let y = 11 1 Diff. w.r.t. x 0

dx

dy 

1 is a cons tan t and derivative of a constant function is zero.

 

  

 

 

 

(iv) Let y = 17

Diff. w.r.t. x 0

dx

dy  is a cons tan t and derivative 17

of a constant function is zero.

  

 

 

 

 

(v) Let y = x11 Diff. w.r.t. x 11x111 11x10

dx

dy   



 (xn)nxn1 dx

 d (vi) Let y = x5/2

Diff. w.r.t. x

2

1 3 2 5

2x x 5 2 5 dx

dy   



 (xn)nxn1 dx

 d

(14)

Fundamentals of Mathematics-II

42

(vii) Let 1/3

3 / 1

3 x

x 1 x

y 1  

Diff. w.r.t. x x 4/3

3 1 dx

dy

 



 (xn)nxn1 dx

 d

(viii) Let y = 7

7 x

x

1

 Diff. w.r.t. x 7x 8

dx

dy

 =

x8

 7 



 (xn)nxn1 dx

 d

(ix ) Let x 1 x y 1  Diff. w.r.t. x

 

2 2

x x 1

dx 1

dy    



 (xn)nxn1 dx

 d (x) Let y x x1/2

Diff. w.r.t.x

x 2 x 1

2 1 dx

dy 1/2





 (xn)nxn1 dx

 d (xi) Let y(2x5)3

Diff. w.r.t. x 3(2x 5) (2)

dx

dy 2

 = 6(2x5)2

 



 (axb) n(axb) a dx

d n n 1

 (xii) Let y(43x)8

Diff. w.r.t. x 8(4 3x) ( 3)

dx

dy 7

 24(43x)7

 



 (axb) n(axb) a dx

d n n1

(xiii) Let y =

9 / 4

2x

5 3 

 

  Diff. w.r.t. x

 





 

 

2 x 3

2 5 3 9 4 dx

dy 5/9

 



 (axb) n(axb) a dx

d n n 1

 =

9 / 5

2x 5 3 3

2



 

 

(xiv) Let y 23x (23x)1/2 Diff. w.r.t. x

(2 3x) (3) 2

1 dx

dy 1/2

 



 (axb) n(axb) a dx

d n n 1

x 3 2 2

3

(xv) Let y x5x21 Diff. w.r.t. x

(15)

43

Diffrenciation

5x 2x 0 dx

dy 4

 5x42x Using formula written at serial number12 of the table of formulae

 

 

 

(xvi) Let

x x 1 x x

1 x x x x x

1 x

y x 2

2 3 2

3

 

 

x x 1 x y 2   Diff. w.r.t. x

 

 

 x

1 dx ) d x dx( ) d x dx(

d dx

dy 2

x2

1 1 x 2  



 (xn)nxn1 dx

 d

(xvii) Let

2 2 2

x x 1

y 

 

 

 Diff. w.r.t. x

 

 



 

 

2

2 2

2

x x 1 dx

d x x 1 dx 2

dy

 



  (f(x))

dx )) d x ( f ( n )) x ( f dx (

d n n 1

 

 



 

 

2 3

2

x x 2 2 x x 1

2 

 

 



 

   

5

3 5

3

x x 1 4 x

1 x 1 x x 1 4

(xviii) Let

100

x x 1

y 

 

 

 Diff. w.r.t. x

 

 



 

 

 x

x 1 dx

d x x 1 dx 100

dy 99

[Same reason as given in (xvii)]

 

 



 

 

2

99

x 1 1 x x 1 100

(xix) Let y(x21)(x1) x3x2x1 Diff. w.r.t. x

3x 2x 1 0 dx

dy 2

 Using formula written at serial

number12 of the table of formulae.

 

 

 

3x22x1

Alternatively: Using Product Rule

(x 1)

dx ) d 1 x ( ) 1 x dx( ) d 1 x dx (

dy 2 2

(x21)(10)(x1)(2x0) x212x22x 3x22x1 (xx) Let yx3(1x2x5x8)

yx3x1x2 x5 Diff. w.r.t. x

3x 4 1x 2 2x 5x4 dx

dy    



 (xn)nxn1 dx

 d  3x4x22x 5x 4

(xxi) Let y(x21)(x3x2 1) Diff. w.r.t. x [Using Product Rule]

(16)

Fundamentals of Mathematics-II

44

(x 1)

dx ) d 1 x x ( ) 1 x x dx( ) d 1 x dx (

dy 2 3 2 3 2 2

(x21)(3x22x)(x3x21)(2x) 3x42x33x22x2x42x32x 5x44x33x2 4x

(xxii) Let y(4x1)3(7x1)4

Diff. w.r.t. x [Using Product rule]

3 4 4 (4x 1)3

dx ) d 1 x 7 ( ) 1 x 7 dx( ) d 1 x 4 dx (

dy      

(4x1)34(7x1)37(7x1)43(4x1)24 4(4x1)2)(7x1)3

7(4x1)3(7x1)

4(4x1)2(7x1)3(49x10) (xxiii) Let y(x2)2(x3)4(x1)5 Diff. w.r.t. x

4 5

2 5

4

2 (x 3)

dx ) d 1 x ( ) 2 x ( ) 1 x dx( ) d 3 x ( ) 2 x dx (

dy        

2 5

4 (x 2)

dx ) d 1 x ( ) 3 x

(   

if u, v, w, are functions of x, then

d d d d

(uvw) uv (w) uw (v) vw (u)

dx dx dx dx

 

 

    

 

(x2)2(x3)45(x1)4(x2)2(x1)54(x3)3 (x3)4(x1)52(x2)

(x2)(x3)3(x1)4

5(x2)(x3)4(x2)(x1)2(x3)(x1)

(x2)(x3)3(x1)4[5(x25x6)4(x23x2)2(x24x3)]

(x2)(x3)2(x1)4(11x245x44)

Example 4: Find the derivative of the following functions:

(i) x 1 1 x

 (ii) x 5 6

3 x 8

 (iii) 2 2

2

a x

a

(iv) x 1 1 x2

Solution:

(i) Let y = 1 x

1 x

 Diff. w.r.t. x

)2

1 x (

) 1 x dx( ) d 1 x ( ) 1 x dx( ) d 1 x ( dx dy

 [Using Quotient Rule]

)2

1 x (

1 ).

1 x ( 1 ).

1 x (

 

2

2 (x 1)

2 )

1 x (

1 x 1 x

 

 

(ii) Let y = x 5 6

3 x 8

 Diff. w.r.t. x

(17)

45

Diffrenciation

)2

x 5 6 (

) x 5 6 dx( ) d 3 x 8 ( ) 3 x 8 dx( ) d x 5 6 ( dx dy

 [Using Quotient Rule]

)2

x 5 6 (

) 5 )(

3 x 8 ( 8 ).

x 5 6 (

 

)2

x 5 6 (

63

  (iii) Let y =

2 2

2

a x

a

1 2 2 2(x a )

a 

 Diff. w.r.t. x

(x a )

dx ) d a x )(

1 ( dx a

dy 2 2 2 2 2 2



. numerator in

x of function no

is there

because here

rule quoient use

t ' Don

.2x

) a x (

a

2 2 2

2

2 2 2

2

) a x (

x a 2

 

(iv) Let y =

1 x

1 x2

 Diff. w.r.t. x





 x 1

1 x dx

d 1 x

1 2 x

1 dx

dy 2

2

 



  (f(x))

dx )) d x ( f ( n )) x ( f dx (

d n n 1

=





2 2 2

2 (x 1)

) 1 x dx( ) d 1 x ( ) 1 x dx( ) d 1 x ( 1 x

1 x 2

1 [Using Quotient Rule]

2 / 3 2

2

) 1 x ( 1 x 2

) 1 )(

1 x ( ) 1 x 2 )(

1 x (

 

2 / 3 2

2 2

) 1 x ( 1 x 2

1 x x 2 x 2

 

 

1 x )

1 x ( 2

1 x 2 x

2 2 / 3 2

1 x ) 1 x ( 2

) 1 x (

2 2 / 3

2

Here, are some exercises for you.

E 3) Differentiate the following functions w.r.t. x (i) e (ii)

x7

1 (iii) x (iv) (43x)8 (v)

2 3 3

x

x 1 

 

  E 4) Find the derivative of the following functions:

(i) 

 

 



 

 

x x 1 x

x 1 (ii) 

 

 



 

 

x x 1 x x 1

3

3 (iii)

1 x

x

3 2

(iv) a x x2

6.5 CHAIN RULE

Sometimes variables y and x are connected by the relations of the form y = f(u) , u = g(w), w = h(x)

and we want to differentiate y w.r.t. x. then chain rule is used, which gives

(18)

Fundamentals of Mathematics-II

46

dx dw dw

du du dy dx

dy 

Following example will illustrate the rule more clearly.

Example 5: Find dx

dy in the following cases.

(i) y3u, uv , v2 4x25 (ii)

1 x v x , v 3 u , u

y 2

 

Solution:

(i) y3u, u v2, v4x25

Diff. w.r.t. u Diff. w.r.t. v Diff. w.r.t. x 3

du

dy  2v

dv

du  8x dx dv  by chain rule

3(2v)(8x) dx

dv dv du du dy dx

dy  

48xv48x(4x25) [Replacing the value of v in terms of x]

(ii)

1 x v x ,

v 3 u ,

u

y 2

 

Diff. w.r.t. u Diff. w.r.t. v Diff. w.r.t. x 2u

du

dy  3 dv

du 

 

)2

1 x (

) 1 .(

x 1 . 1 x dx dv

  =

)2

1 x (

1

 by chain rule

dx dv dv du du dy dx

dy 

 



2

) x 1 ( ) 1 3 ( u 2

2 2 (1 x)

) v 3 ( 6 ) x 1 (

u 6

 

 

3 2 (1 x)

x 18 )

x 1 (

v 18

 

 

6.6 DERIVATIVES OF EXPONENTIAL, LOGRITHMIC, PARAMETRIC AND IMPLICIT FUNCTIONS

Let us first take up some examples on derivatives of exponential and logarithmic functions as given in example 12.

Example 6: Find the derivative of the following functions:

(i) 2x (ii) 5ax (iii) e3x (iv) e9x2 (v) log(1x2) (vi) logxx2 (vii) log2x (viii) logx2 (ix)

) x 1 log(

1

Solution:

(i) Let y2x Diff. w.r.t. x 2 log2

dx

dy x

 



 (a )a loga dx

d x x

 (ii) Let y5ax

Diff. w.r.t. x

(19)

47

Diffrenciation

(ax)

dx 5 d log dx 5

dy ax

 



 (a )ba loga dx

d bx bx

 5ax alog5

(iii) Let ye3x Diff. w.r.t. x e (3)

dx dy 3x

 3e3x 



 (eax)aeax dx

 d

(iv) Let ye9x2 Diff. w.r.t. x

(9x ) e (18x) dx

e d dx

dy 9x2 2 9x2

 d f ( x ) f (x ) d

Using (e ) e (f (x))

dx dx

 

  

 

18xe9x2 (v) Let ylog(1x2) Diff. w.r.t. x

(1 x )

dx d x 1

1 dx

dy 2

2

2

x 1

x 2



  (f(x))

dx d ) x ( f )) 1 x ( f dx(log

 d

(vi) Let

x log

x x log log y

2 2

x

 [Using base change formula]

2

log log 2

x x

  log mn n log m Diff. w.r.t. x

0 dx dy 

(vii) Let y = log2x Diff. w.r.t. x

 



 

 log e

x x 1 dx log e d

xlog 1 dx dy

a a

2

(viii) Let ylogx2 x log

2

 log [Using base change formula]

y(log2)(logx)1 Diff. w.r.t. x

(logx)

dx ) d x )(log 2 dx (log

dy 2

 x

1 ) x (log

) 2 (log

2

2

) x (log x

2

 log

(ix) Let

) x 1 log(

y 1

  y

log(1x)

1

Diff. w.r.t. x

 

(log(1 x))

dx ) d x 1 log(

dx 1

dy 2

 

1 x

1 )

x 1 log(

1

2  

 

log(1 x)

2

) x 1 (

1

References

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