Class Note-M101 Module-II PDF
Class Note-M101 Module-II PDF
Class Note-M101 Module-II PDF
The process of differentiating a given function again and again is called as Successive
Differentiation and the results of such differentiation are called successive derivatives.
The higher order differential coefficients will occur more frequently in spreading a function all
fields of scientific and engineering applications.
Solved Examples :
Solution:
Differentiating y = sin(sin x) .. (1) with respect to x, we get y1 = cos (sin x) cos x .(2)
Again differentiating (2) w.r.to x, we find
2
2. , show that 2y1y3 = 3 y2
Solution:
y1 = =
Now, 2y1y3 = 2. = 12
1
=3[- ]2 = 3 y2 2 .
Solution:
y1 = =- +
2
y2 = (-1)
y3 = (-1) 3 + (-1) 2 +
n
yn = (-1) + (-1) (n-1) +
n
= (-1) [ - + ]
(ii) D n[ ] =
Proceeding likewise
2
yn = (m log a )n y = mn ( log a )n amx
(v) D n ( e mx) = mn e mx
(vi) D n sin(ax + b) = an sin
Proof: y1 = a cos (ax+b) = a sin etc..
y1 = ea x {a sin (bx+c) + b cos (bx+c). Let a = r cos and b = r sin , thus y1 = e ax r sin (+bx+c)
y n = r n e ax sin ( n + bx + c), but r = (a2 + b2 ) 1/2 and hence the proof follows.
Leibnitzs Theorem
Solved Examples:
3
1. Find the nth derivative of the function x2 log 3x.
Solution. We take u = log 3x and v = x2 , then v1 = 2x, v2 = 2 and v3 , v4 etc.. are all zero.
By Leibnitzs theorem,
n n
( x2 log 3x.) n = ( log 3x) n x2 + C1 ( log 3x) n-1 2x + C2 ( log 3x) n-2 2
= (-1) n-1 +
= [ (n - 1)(n - 2) x2 - n(n - 3) x2 ]
2. If y =
x = 0 . Also find yn (0).
At x=0
More clearly,
4
y2k(0) = m2 (22 + m2) (42 + m2) . . . . . . . . . .{(2k -2)2 + m2}
Solution. i.e, y1 =
Solution. Let us take x = tan (log y) i.e, tan-1 x = log y or, y = . Now differentiating
with respect to x,
5. + = 2x or, y = [ x + ]m or, [ x - ]m
Prove that ( x2 1) yn+2 + (2n + 1) x yn+1 + ( n2 m2 ) yn = 0.
5
2
Solution. + = x so, 2x + 1 = 0. Solving this
quadratic equation we find
In each of the following problems, apply Leibnitzs theorem to get the results.
(ii) If y = a cos (log x) + b sin (log x), prove that x2 yn+2 + (2n + 1) x yn+1 + (n2 + 1) yn = 0
Mean-Value Theorems
In this topic we shall discuss a few important theorems of differential calculus- Rolles Theorem,
Lagranges & Cauchys Mean-Value Theorem. Expansions of functions by Taylors and
Maclaurins theorem.
6
Rolles theorem and its application
Statement: Let a function f be defined on a closed interval [a,b]. Suppose, further that
(i) f is continuous on [a,b]
(ii) f is differentiable in the open interval (a,b) and
(iii) f(a) = f(b)
then, there exists at least one point x=c lying within a < c < b, such that f(c) = 0.
Proof. Since f is continuous on [a,b], it must attains its lub M and glb m there, i.e, there exist
two points of [a,b] such that f(
I. If M=m, then f(x) must be constant for all x and hence f(x) = 0 at all points
of [a,b].
II. Next suppose M
Then since f(a) = f(b) and m either M or m, if not both must be different from f(a) or f(b).
Then f( f(a) and f( f(b) . Thus is neither a nor b.
By hypothesis f(x) is differentiable in the open interval (a,b), so f( ) exists. We shall prove that
f( ) = o.
Since f( ) exists,
R f( ) = L f( ) = a finite number or,
Since f( ) = M ( the greatest value of f(x) in [a,b],
f( - f( )
If h > 0, then
R f( ) and L f( ) .
Corollary. If a < b are two roots of the equation f(x)=0, then the equation f(x) = 0 will have at
least one root between a and b, provided
7
(i) f(x) is continuous on [a,b] and
(ii) f is differentiable in (a,b).
If f(x) be a polynomial, the conditions (i) and (ii) are satisfied. Hence
Between any two roots of a polynomial f(x) lies at least one zero of the polynomial
f(x)
f(x) =
f(x) is differentiable on [0,4]. So by Rolles theorem f(x) should have at least one zero within
(0,4), i.e, = 0 for some x (0,4). Equating = 0, we find x = -2 . Here x = -
2 lies within (0,4).
Mean-Value Theorem:
Lagranges Form.
Statement. If a function f is
a) f is continuous on [a,b]
b) f is differentiable in (a,b)
then there exists at least one value of x, say c, such that = f(c , for a < c
Geometrically, this is equivalent to stating that the tangent line to the graph of f at c parallel to the
chord joining the points (a, f(a)) and (b, f(b) ).
Example 7. Verify Lagranges Mean-Value Theorem for the function f(x) = 2x2 7x 10 over
( 2 , 5 ) and find c of the Lagranges Mean-Value Theorem.
8
Again f(x = 4x 7, so f(c = 4c 7 (ii)
Combining (i) and (ii) we get 4c 7 = 7 so, c = and lies within (2,5).
Example 9 Estimate
Solution. We take f(x) = , x [27,28]. So that f(x) = . Then using Lagranges
Mean-Value Theorem
= f , x0 (27,28)
= +
= 3+ , 27 < x0 <28
9
Cauchys Mean-Value Theorem.
Statement. If f(x) and g(x) are continuous in [a,b] and differentiable in (a,b) and g(x) for
any x in (a,b), then there exists at least one point x= c in (a,b) such that
Here g(b) g(a) 0 since g(c) , because otherwise g(b) = g(a) and by Rolles theorem g(c)
= 0. Now F(x) is continuous in [a,b] since f(x) and g(x) are continuous in [a,b]. Again since f(x)
and g(x) exist in (a,b), F(x) exists in (a,b) and that
Clearly, F(a) = F(b). Thus F(x) satisfies all the conditions of Rolles theorem in the interval [a,b].
Therefore there should have at least one point c of x between a and b, such that
F(c) = 0
x -x
Example 10. If in the Cauchys Mean-Value Theorem we take f(x) = e and g(x) = e ,
then prove that c is the arithmetic mean between a and b.
Solution. f(x) and g(x) satisfies the conditions of Cauchys Mean-Value Theorem, so there exists
at least one point x=c in (a,b) such that
Example11. Using Cauchy Mean Value Theorem, show that 1 - < cos x for x
Solution. Applying Cauchys Mean-Value Theorem f(x) = 1 cos x and g(x) = on the
2. Use the mean value theorem (MVT) to establish the following inequalities.
(i) ex > 1 + x for x R (ii) < log x < x 1 for x > 1.
3.Does there exist a differentiable function f:[0,2] R satisfying f(0) = -1, f(2) = - 4
and f(x) < 2 for x [0,2] ?
Statement. Let f(x) be a function defined in the closed interval [a,a+h] such that (i) (n-1)th
derivative f(n-1) is continuous on [a,a+h] and (ii) nth derivative f(n) exists in (a,a+h). Then there
exists at least one number , where 0 < < 1 such that
where Rn = f (n)
(a + is called the Lagranges Form of
Remainder after n terms.
Statement. Statement. Let f(x) be a function defined in the closed interval [a,a+h] such that (i) (n-
1)th derivative f(n-1) is continuous on [a,a+h] and (ii) nth derivative f(n) exists in (a,a+h). Then there
exists at least one number , where 0 < < 1 such that
where Rn = f (n)
(a + is called the Cauchys Form of
Remainder after n terms.
11
III. Maclaurins Theorem
Statement. Let f(x) be a function defined in the closed interval [0,x] such that
then there exists at least one number , where 0 < < 1 such that
(d)
Example12. Find the Maclaurins theorem with Lagranges form of remainder for f(x) = sin x.
Solution. f(n)(x) = = sin and f(2n)(x) = (-1)n sin x, f(2n+1)(x) = (-1)n sin
f(n)(0) = sin . Therefore f(0) = 0, f(0) = 1, f(0) = 0 , . . . . . , f(2n)(0) = 0 and f(2n+1)(0) = (-1)n.
Substituting these values in the Maclaurins theorem with Lagranges form of remainder, i.e,
f(x) = f(0) + x f (0) + f (0) + . . . . . .+ (-1)n-1 f(2n-1) (0) + f (2n)( ), 0 < < 1.
We find
12
sin x = 0 + x + 0 - - . . . . . . . + (-1)n-1 + (-1)n sin 0< <1
Simplifying, =- . =-
Here, lies within (0,1), hence Maclaurins theorem with Lagranges form of
remainder is verified.
Example14. Express log (1+x) with Maclaurins theorem with Lagranges form of remainder.
we find ,
(a+x)n = an + x. n an-1 + .n(n-1) an-2 + . . . . . . . . + . n(n-1)(n-2)..{n-(n-2)}a + . n!
13
Solution. We use Taylors theorem with Lagranges form of remainder taking a=1 and h = x-1
and obtain
Taylors Series
Statement.Let f(x), f(x), f(x), . . . . . . ., f(n)(x) exist finitely however large n may be in any
interval (x- enclosing the point x and let Rn 0 as n . Then Taylors series of
finite form can be extended to an infinite series of the form
Maclaurins Series
Statement. Let f(x), f(x), f(x), . . . . . . ., f(n)(x) exist finitely however large n may be in any
interval (- and Rn 0 as n . Then Maclaurins series of finite form can be
extended to an infinite series of the form
f(x) = f(0) + x f(0) + f(0) + . . . . . . . + f(n) (0) + . . . . . , . [Putting a=0 in (4)] (5)
The infinite series (5) is the expansion of f(x) in the neighbourhood of the point x=0.
Example16. Expand the function f(x) = ex in the form of Maclaurins series in the neighbour-
hood of the point x=0.
14
f(x) = f(0) + x f(0) + f(0) + . . . . . . . + f(n) (0) + . . . . .
x
i.e, e = 1 + x + +.......+ + .....
sin x = x - - . . . . . . . + (-1)n + .. ..
(ii)f(x) = cos x
cos x = 1 - - . . . . . . . + (-1)n + .. ..
Example19.(a) Find the Maclaurins series for the function f(x) = (1+x)m where m is not
necessarily an integer and hence show that the formula for the binomial series works for non-
integral exponents as well. (b) Use your answer to find the expansion of up to the term in
6
x.
Solution. f(x) = (1+x)m , so f(0) = 1, f(x) = m(1+x)m-1 f(0) = m. Likewise f(0) = m(m-1), . .,
15
(1+x)m = 1 + mx + x2 + x3 + . . . . . .. + xk + . . . . .
.
This is an infinite series. If m is a positive integer the series will stop when K = m and will agree
with the standard binomial expansion.
2 3
= 1+ (- + + +.....
= 1 + x2 + x4 + 6
+......
In =
=-
=-
= - - (n-1)
= - In-2 - (n - 1) In
In = In2 (6)
I.
Jn =
= + , using (6)
=0+ Jn-2
16
= Jn-2
II.
In =
= In-2 ( n 1 ) In
In = In2 (7)
II.
Jn =
= +
= Jn-2
J7 =
= J5 = =
= [ sinx =
III.
Im,n =
17
=
= dx
= +
= +
= +
= + Im,n-2 - Im,n
= + Im,n-2 (8)
Alternately,
III.
Jm,n =
= Jm,n-2 (10a)
= [ - + Jm-2,n
18
= Jm-2,n (10b)
IV.
Im,n =
= --
= -
= -
= Im,n + Im-1,n-1
IV.
Jm,n =
= + Jm-1,n-1 (13)
19
Solution. Using (13)
Jm,n =
= + Jm-1,n-1
Jm,m = + Jm-1,m-1
= + [ + Jm-2,m-2 ]
= Jm-3.m-3 ]
= J1,1 ]
= J1,1
= + ...... +
[using J1,1 = = ]
= [2+ + +........ + ]
Hence proved.
V.
In =
= + 2n
20
= + 2n
= +2n In 2n a2 In+1
Solution. In = = =
= = -
= - In-2 +
= - In-2 + =2 - In-2
21
Module-III
(b) z g(x,y) = is defined over the entire x-y plane excluding the only point (0,0).
(c) z is defined over the region x+y , which include the points
of the line x+y = 1.
(d) z is defined over the region x+y > 1, which does not include points
of the line x+y = 1.
(h) f(x,y,z) = x2y + y2z + z2x is a function of three variables x, y, z. It is defined for all points
(x, y, z) R3 .
Limit
22
Let z = f(x,y) be given in a domain D, and let (x1,y1) be a point of D or a boundary point of D .
Then the equation
(14)
means the following:
Given any > 0 , a D and within the
neighbourhood of (x1,y1) of radius , except possibly for (x1,y1) itself, one has
. (15)
Examples on Limit
Solution. Let be given. We are to find a such that in some neighbourhood of (0,0)
for all points (x,y)
< or,
Now, clearly
2
so that < = < holds if, 0 < i.e when
Therefore whenever
23
Example24. To prove does not exist.
Solution. The domain of f is the whole xy-plane punctured at the origin (0,0). For existence of
limit we are to examine the values of f near (0,0). If we allow the limit through the straight line
y= mx, we observe that
f(x,y) = f(x,mx) = =
so ,
Continuity
Then f(x,y) is said to be continuous at (x1,y1). If this holds for every point (x1,y1) of D, then
f(x,y) is said to be continuous in D.
A function f(x,y) is said to be bounded when (x,y) is restricted to a set E, if there is a number M
such that when (x,y) is in E. For example, z = x2 + y2 is bounded with M = 2, if
24