CHAPTER 7 SUCCESSIVE DIFFERENTIATION
[Pages:17]
CHAPTER 7 SUCCESSIVE DIFFERENTIATION
TOPICS: 1 . Successive differentiation-nth derivative of a function ? theorems. 2. Finding the nth derivative of the given function. 3. Leibnitz's theorem and its applications.
SUCCESSIVE DIFFERENTIATION
Let f be a differentiable function on an interval I. Then the derivative f is a function of x and
if f is differentiable at x, then the derivative of f at x is called second derivative of f at x. It is denoted by f(x) or f(2)(x).similarly, if f" is differentialble at x , then this derivative is called the 3rd derivative of f and it is denoted by f(3)(x). Proceeding in this way the nth derivative of f is the derivative of the function f(n-1)(x) and it the denoted by f(n)(x).
If y = f(x) then
f (n)(x)
is denoted by
dny dx n
or Dn y or y(n) or yn
and f (n) ( x) = lim f (n-1) ( x + h) - f (n-1) ( x)
h0
h
THEOREM
If f(x) = (ax + b)m, m R, ax + b > 0 and n N then f (n) (x) = m(m -1)(m - 2)...(m - n +1)(ax + b)m-n an
Note : If y = (ax + b)m then yn = m(m ? 1)(m ? 2) ...(m ? n + 1)(ax + b)m?n an.
COROLLARY If f(x) = (ax + b)m, m Z, m > 0, n N then (i) m < n f(n)(x) = 0, (ii) m = n f(n)(x) = n! an
(iii) m > n f(n)(x) = m! (ax + b)m-n an . (m - n)!
COROLLARY If f(x) is a polynomial function of degree less than n where n N then f(n)(x) = 0.
THEOREM
If f (x) = 1 ax + b
then f(n)(x) =
(-1)n (ax +
n !a n b)n+1
.(i.e.,
If y = 1 ax + b
yn =
(-1)n (ax +
n !a n b)n+1
)
THEOREM
If
f(x)
=
log
|ax
+
b|
and
n
N
then
f (n)(x)
=
(-1)n-1(n -1)!an (ax + b)n
.
i.e., y = log |ax+b|
yn
=
(-1)n-1(n -1)!an (ax + b)n
THEOREM
If
f(x)
=
sin(ax
+
b)
and
n
N
then
f (n)(x)
= an
sin
ax
+
b
+
n 2
.
THEOREM
If
f(x)
=
cos(ax
+
b)
and
n
N
then
f (n)(x)
=
an
cos
ax
+
b
+
n 2
.
THEOREM
If f(x) = eax+b and n N then f (n) (x) = aneax+b .
THEOREM If f(x) = cax+b, c > 0 and n N then f (n) (x) = ancax+b (log c)n .
THEOREM
If f(x) = eax sin(bx + c) and n N then f (n)(x) = rneax sin(bx + c + n) where a = r cos , b = r sin
and r =
a2
+
b2
,
=
Tan -1
b a
THEOREM
If f(x) = eax cos(bx + c) and n N then f (n) (x) = rneax cos(bx + c + n) where a = r cos , b = r sin
and r =
a2
+
b2
,
=
Tan -1
b a
.
Note: If f, g are two functions in x having their nth derivatives then
(f ? g)(n)(x) = f (n) (x) ? g(n) (x) .
Note: If f is a function in x having nth derivative and k R then (kf )(n) (x) = kf (n) (x) .
EXERCISE ? 7 (a)
1. Find the nth derivative of sin3x.
Sol: we know that sin 3x = 3sin x - 4 sin3 x sin3 x = 3sin x - sin 3x
4
Differentiate n times w.r.t x,
( ) dn
dx n
sin3 x
=
1 4
dn dx n
(3sin
x
- sin 3x )
=
1 4
-3n.sin
3x
+
n 2
+
3sin
x
+
n 2
n
z
2. Find the nth derivative of sin 5x. sin 3x.?
Sol: let y = sin 5x.sin 3x = 1 (2sin 5x.sin 3x)
2
y= 1 (cos 2x - cos8x)
2
y = 1 (cos 2x - cos8x)
2
Differentiate n times w.r.t x,
yn
=
1 2
dn dx n
(cos 2x
- cos8x)
yn
=
1 2
2n
cos
2x
+
n 2
-
8n
.
cos
8x
+
n 2
n
z
3. Find nth derivative of ex .cos x.cos 2x
Sol: cos x.cos 2x = 1 (2cos 2x.cos x) = 1 (cos3x + cos x)
2
2
Let y = ex (cos3x + cos x)
2
Differentiate n times w.r.t x,
( ) yn
=
1 2
dn dx n
ex cos 3x + ex cos x
( ) ( ) ( ) ( ) ( ) yn
=
ex 2
10
n
cos
3x + n tan-1 3
n
+
2
n
+ cos
x
+
n
tan-1 1
n
z
= ex 2
n 10
2
cos
3x + n tan-1 3
+
2n
/
2
cos
x
+
n 4
4.
If
y
=
(
x
2
- 1) (
x
-
2)
find
y n
Sol:
Given
y=
2
(x -1)(x - 2)
=
x
1 -
2
-
x
1 -1
(
partial
fractions)
Differentiate n times w.r.t x,
yn
=
2
(-1)n n! ( x - 2)n+1
-
(-1)n n! ( x -1)n+1
=
2
( -1)n
n!
(x
1
) - 2 n+1
-
(
x
1
- 1)n
+1
5.
If
y=
2x +1 ,
x2 - 4
find
y n
Sol:
Let
2x +1 x2 - 4
=
A x-2
+
B x+
2
2x +1 = A(x + 2) + B(x - 2) -----(1)
In (1) ,Put x = 2 5 = A(4) A = 5
4
In (1) , x = -2 -3 = B(-4) B = 3
4
Therefore,
y
=
2x +1 x2 - 4
=
5
4(x -
2)
+
3
4(x +
2)
Differentiate n times w.r.t. x,
yn
=
dn dx n
5
4(x - 2)
+
3
4(x + 2)
yn
=
5 4
(-1)n n! ( x - 2)n+1
+
3 4
(-1)n n! ( x + 2)n+1
=
( -1)n
4
n !
(x
5
) - 2 n+1
+
(x
3
) + 2 n+1
1.
Find the nth derivative of (i)
x
(x -1)2 (x +1)
(ii)
1
(x -1)(x + 2)2
(iii)
x3
(x -1)(x +1)
(iv) x
x2 + x +1
Sol: i)
(v) x +1
x2 - 4
(vi) Log (4x2 - 9)
Let
y=
x
(x -1)2 (x +1)
Resolving into partial fractions
(x
x
-1)2 (x
+ 1)
=
A x -1
+
(x
B
- 1)2
+
C x +1
x = A (x -1)(x +1) + B(x +1) + C(x -1)2 ----- (1
In (1 ), put x = 1 1 = B(1+1) = 2B B = 1
2
In (1 ), put x = -1 -1 = C(-1-1)2 = 4C C = - 1
4
Equating the co . efficient of x2 A + B = 0 A = - 1
2
Therefore,
y
=
-
1
2(x -1)
+
1
2(x -1)2
-
1
4(x +1)
Differentiate n times w.r.t. x,
yn
=
dn dx n
-
2
(
1 x-
1)
+
1
2(x -1)2
-
4(
1 x+
1)
yn
=
(-1)n n! 2( x -1)n+1
++1
2
(-2)(-3).....(-2 - n +1) ( x -1)n+2
-
1 4
(-1)n n! ( x +1)n+1
=
(-1)n n! 2.( x -1)n+1
+
(
-1)n 2(x
(n +1)! ) -1 n+2
-
1 4
(-1)n n! ( x +1)n+1
= (-1)n
n!
2(x
1
) -1 n+1
+
n +1
2( x -1)n+2
-
4(x
1
) +1 n+1
(ii)
y
=
(
x
-
1)
1
(x
+
2)2
Resolving
into partial fractions
1
(x -1)(x + 2)2
=
A x -1
+
B x+2
+
(x
C
+ 2)2
1=A (x + 2)2 + B(x -1)(x + 2) + C(x -1) ---(1)
In (1) put x = 1 1 = A (1+ 2)2 = 9A A = 1
9
In (1) put x = -2 1 = C(-2 -1) = -3C C = - 1
3
Equating the co ? efficient of x2 In (1)
A + B = 0 B = -A = - 1 9
y
=
9(
1
x -1)
-
9
(
1 x+
2)
-
3(x
1 +
2)2
Differentiate n times w.r.t. x ,
yn
=
dn dx n
9
(
1
x -1)
-
1
9(x + 2)
-
1
3(x + 2)2
yn
=
(-1)n n! 9( x -1)n+1
-
(-1)n n! 9 ( x + 2)n+1
-
1 3
(
-1)n (x +
(n +1) )2 n+2
= (-1)n
n!
9 ( x
1
) -1 n+1
-
9(x
1
) + 2 n+1
-
n +1
3(x + 2)n+2
(iii)
y
=
(
x
x3
-1) (
x
+
1)
Ans:
( -1)n
2
n!
( x
1
) -1 n+1
+
(x
1
) +1 n+1
( iv)
x
x2 + x +1
Ans:
yn
=
( -1)n
r n +1
n!
cos
(n
+ 1)
-
1 3
sin
(
n
+
1)
(v)
y
=
x +1 x2 - 4
Ans:
( -1)n
4
n!
( x
3
) - 2 n+1
+
(x
1
) + 2 n+1
(vi) y = log (4x2 - 9)
Given y = log (4x2 - 9) = log[2x - 3][2x + 3]
= log (2x - 3) + log (2x + 3)
Differentiating n times,
yn
=
dn dxn
( log ( 2x
- 3)
+
log ( 2x
+ 3))
yn
=
(
) -1 n-1 2n (2x -
(n 3)n
-
1)!
+
( ) -1 n-1 2n (n -1)! (2x + 3)n
=
(
) -1 n-1
2n
(
n
-
1)!
(
2x
1 -
3)n
+
(
2x
1 +
3)n
2.
If y =
a + bx c + dx
then show that
2y1y3
=
3y
2 2
Sol: Given y = a + bx
c + dx
Differentiate w.r.t.x ,
dy dx
=
(c
+
dx) b - (a + (c + dx)2
bx ) .d
y1=
bc + bdx - ad - bdx = (c + dx)2
bc - ad
(c + dx)2
Again diff. w.t.t x,
y2
=
(bc - ad)(-2) d (c + dx)3
=
-2d (bc - ad) (c + dx)3
Diff.wrt.x, we get
y3
=
-2d (bc - ad)(-3).d = (c + dx)4
6d2 (bc - ad) (c + dx)4
L.H.S.= 2y1y3
=
2 ( bc (c +
- ad
dx )2
)
.
6d2 (bc - ad (c + dx)4
)
=
12d2 (bc - ad)2 (c + dx)6
=
3
-2d (bc - ad (c + dx )3
)
2
=
3y
2 2
=R.H.S.
3. If y = sin (sinx ), then show that y2 + (tan x) y1 + y cos2 x = 0
Sol: Given y = sin (sinx) Diff. wrt x,
y1 = cos(sin x) cos x
Diff. wrt x, y2 = cos x - sin (sin x ) cos x - cos (sin x )sin x
= - cos2 x.sin (sin x) - sin x.cos (sin x) LHS= y2 + (tan x) y1 + y cos2 x
= - cos2 x sin (sin x ) - sin x.cos (sin x ) + sin x cos x (sin x ) + sin x.cos(sin x) = 0 = RHS.
cos x
4. If y = axn+1 + bx-n , then show that x2y2 = n (n +1) y .
Sol: y = axn+1 + bx-n
Diff. wrt. X,
y1 = a (n +1) xn - bnx-(n+1) Diff. wrt x, y2 = a.n (n +1) xn-1 + bn.(n +1) x-(n+2) x2.y2 = n (n +1).x2 a.xn-1 + b.x-n-2
( ) = n (n +1) axn+1 + bx-n = n (n +1) y
5. If y = aenx + be-nx , then show that y2 = n2y Sol: y = aenx + be-nx
y1 = a.n.enx - bn.e-nx
( ) y2 = an2enx + bn2 .e-nx = n2 aenx + be-nx
y2 = n2y
6.
If
- kx
y = e 2 (a cos nx + bsin nx) then show that
y2
+
ky1
+
n2
+
k2 4
y
=
0
.
- kx
Sol. y = e 2 (a cos nx + bsin nx)
Differentiating w.r.to x.
y1
=
- kx
e2
[-an sin
nx
+
bn
cos
nx]
+
-
k 2
e
-lx 2
[a
cos
nx
+
b sin
nx]
y1
=
-kx
+e 2 n (-a sin nx + bcos nx) -
k 2
y
y1
+
k 2
y
=
- kx
+n.e 2
(-a sin nx + b cos nx) - (1)
Differentiating w.r.to x.
y2
+
k 2
y1
=
- kx
+n.e 2
-
k 2
(-a sin nx
+
b cos nx)
- kx
+n.e 2
[-an.conx
- bn sin nx]
=
-
k 2
y1
+
k 2
y
-
n2
y
=
-
k 2
y1
-
k2 4
y - n2y
................
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