UNIT- III
DIFFERENTIATION METHODS
3.1 Differentiation of function of functions and Implicit functions.
Simple Problems.
3.2 Differentiation of inverse trigonometric functions and parametric
functions Simple problems.
3.3 Successive differentiation up to second order (parametric form
not included) Definition of differential equation, formation of
differential equation. Simple Problems.
DIFFERENTIATION METHODS
3.1 DIFFERENTIATION OF FUNCTION OF FUNCTIONS
Function of Functions Rule:
If ‘y’ is a function of ‘u’ and ‘u’ is a function ‘x’ then
dy dy du
=
.
It is called Function of function Rule. This rule
dx du dx
can be extended.
Chain Rule:
If ‘y’ is a function of ‘u’ and ‘u’ is a function of ‘v’ and ‘v’ is a
function of ‘x’ then
dy dy du dv
=
. .
dx du dv dx
3.1 WORKED EXAMPLES
PART - A
Find
dy
if
dx
1) y = Sin(3 x + 4)
3) y = tan(4x + 3 )
(
2
5) y = tan x + 7x + 3
2) y = Cos (2x + 3 )
4) y = log(sec x + tan x )
)
11
206
Solution:
1.) y = Sin(3x + 4)
Put u= 3x+4
∴ y = sin u u = 3x + 4
dy
du
= Cos u
=3
du
dx
∴
dy dy du
=
.
= Cosu(3) = 3Cos(3 x + 4)
dx du dx
2.) y = Cos (2x + 3)
Put u = 2x + 3
∴ y = Cos u u = 2x + 3
dy
du
= −Sin u
=2
du
dx
dy dy du
=
.
= (− sin u)2
dx du dx
= −2Sin(2x + 3)
3.) y = tan(4x + 3 )
Put u = 4x + 3
∴ y = tan u u = 4x + 3
dy
du
= Sec 2u
=4
du
dx
∴
dy
= 4 sec 2 u = 4 sec 2 (4x + 3)
dx
4.) y = log(sec x + tan x )
Put u = sec x + tan x
∴ y = log u
u = sec x + tan x
du
= sec x tan x + sec 2 x
dy 1
=
dx
du u
= sec x (tan x + sec x )
dy 1
sec x (sec x + tan x )
= [sec x (sec x + tan x )] =
= sec x
dx u
sec x + tan x
207
(
5.) y = x 2 + 7 x + 3
)
11
2
Put u = x + 7 x + 3
∴ y = u11
u = x 2 + 7x + 3
dy
du
= 11 u10
= 2x + 7
du
dx
10
dy
= 11 u10 (2x + 7 ) = 11 x 2 + 7 x + 3 (2x + 7 )
dx
(
)
PART - B
Differentiate the following w.r.t.to ‘x’
(
)
1.) Sin x 2 + 1
2.) eSinx
3.) Sin 3 x
4.) logsinx
5.) e
Sin 2 x
6.) log(sin5x)
tan −1 2 x
7.) e
9.) log sec2 x
Solution:
(
8.) sin4 3x
10.) cos e5 x
)
(
( )
)
1.) Let y = sin x 2 + 1
Put u = x + 1
2
∴ y = sin u u = x 2 + 1
dy
du
= cos u
= 2x
du
dx
dy dy du
=
= cos u(2x )
.
dx du dx
= 2xCos( x 2 + 1)
2.) y = eSinx
Put u = sin x
∴ y = eu
u = sin x
dy
du
= eu
= cos x
du
dx
dy dy du
=
= eu . cos x = esin x . cos x
dx du dx
208
3.) y = sin3 x
Put u = Sinx
∴ y = u3
u = sin x
dy
du
2
= 3u
= cos x
du
dx
dy
= 3u2 .Cosx = 3 Sin 2 x.Cosx
dx
4.) y = log(sin x )
Put u = Sinx
∴ y = log u u = sin x
dy 1
du
=
= cos x
du u
dx
dy 1
Cosx
= Cosx =
= Cotx
dx u
sin x
5.) y = e sin
2
x
2
Put u = Sin x
u
y =e
dy
du
d
= eu .
= eu .
sin2 x
dx
dx
dx
(
)
= e sin x (2 sin x Cosx )
6.) y = log(sin5x)
2
1
d
(sin 5x )
sin 5 x dx
1
(5 cos 5x ) = 5 cot 5x
=
sin 5 x
−1
y = e tan (2 x )
dy
dx
7.)
=
(
)
−1
dy
d
tan −1 2x
= e tan 2 x
dx
dx
−1
1
(2)
= e tan 2 x .
2
1 + (2x )
−1
=
2e tan 2 x
1 + 4x 2
209
8.) y = Sin4 3 x = (sin 3x )4
dy
3 d
(sin 3x )
= 4(sin 3x )
dx
dx
= 4 sin3 3 x (3 Cos3x )
= 12 sin3 3 x Cos3x
(
9.) Y= log sec2 x
)
dy
1
d
=
(sec 2 x )
2
dx sec x dx
1
=
(2 sec x )(sec x tan x ) = 2 tan x
sec 2 x
( )
d
dy
= − sin(e ) (e )
dx
dx
= − sin(e ).5e
= −5e sin(e )
10.) y = Cos e5 x
5x
5x
5x
3.1.2.
5x
5x
5x
Differentiation of implicit functions
If the variables x and y are connected by the relation of the form
f(x,y)= 0 and it is not possible to express y as a function of x in the
dy
form y=f(x) then y is said to be an implicit function of x. To find
in
dx
this case, we differentiate both sides of the given relation with respect
to x. keeping in mind that derivative of φ(y )
w.r.t ‘x’ as
dφ dy
d
(φ(y )) = dφ . dy
.
i.e
dx
dy dx
dy dx
For Example
( )
d
(sin y ) = Cosy dy and d y 2 = 2y dy
dx
dx
dx
dx
210
WORKED EXAMPLES
PART – A
Find
dy
for the following functions
dx
2
1) xy = c
2) y = cos (x + y)
2
2
2
2
3) y = 4ax
4) x + y = a
2
5) xy = k
Solution:
1)
2)
2
xy = c
Differentiate both sides w.r.t ‘x’
dy
+y=0
x
dx
dy
= −y
x
dx
dy
y
=−
dx
x
y = cos (x + y)
Differentiate both sides w.r.t ‘x’
dy
§ dy ·
= − sin(x + y )¨1 +
¸
dx
© dx ¹
= − sin(x + y ) − sin(x + y )
3)
dy
dy
+ sin(x + y )
= − sin(x + y )
dx
dx
[1 + sin(x + y )] dy = − sin(x + y )
dx
dy
sin(x + y )
=−
dx
1 + sin(x + y )
2
y = 4ax
Differentiate both sides w.r.t ‘x’
dy
2y
= 4a
dx
dy 4a 2a
=
=
dx 2y
y
211
dy
dx
4)
5)
2
2
2
x +y =a
dy
=0
2x + 2y
dx
dy
= −2 x
2y
dx
dy
x
=−
dx
y
2
xy = k
dy
x 2y
+ y2 = 0
dx
dy
2xy
= −y 2
dx
− y2
2xy
−y
=
2x
dy
dx
=
PART - B
Find
dy
of the following
dx
1)
x2
a2
+
y2
b2
=1
3) x 3 + y 3 = 3axy
5) y = x
Solution:
1)
x2
4) ax 2 + 2hxy + by 2 = 0
y2
=1
a
b2
Differentiate both sides w.r.t ’x’
2
+
sin(a + y )
2) x 2 + y 2 + 2gx + 2fy + c = 0
2x
2y dy
+
=0
a 2 b2 dx
2y dy
2x
=− 2
b2 dx
a
212
dy
dx
=−
=−
2)
2x b2
.
a 2 2y
b2 x
a2y
x 2 + y 2 + 2gx + 2fy + c = 0
Differentiate both sides w.r.t ‘x’
dy
dy
+ 2g + 2f
=0
dx
dx
(2y + 2f ) dy = −2x − 2g
dx
dy
2(x + g)
=−
dx
2(y + f )
2x + 2y
§ x + g·
¸¸
= −¨¨
© y+f ¹
3)
x 3 + y 3 = 3axy
Differentiate both sides w.r.t ’x’
3x 2 + 3y 2
§ dy
·
= 3a¨ x
+ y¸
© dx
¹
dy
= 3ax
+ 3ay
dx
dy
dx
dy
dy
− 3ax
= 3ay − 3x 2
dx
dx
dy
2
3y − 3ax
= 3ay − 3 x 2
dx
3y 2
(
dy
dx
)
=
=
(
3(y
)
− ax )
3 ay − x 2
2
ay − x 2
y 2 − ax
213
4)
ax 2 + 2hxy + by 2 = 0
Differentiate both sides w.r.t ‘x’
dy
§ dy
·
2ax + 2h¨ x
+ y ¸ + 2by
=0
dx
© dx
¹
dy
dy
+ 2hy + 2by
=0
dx
dx
(2hx + 2by ) dy = −2ax − 2hy
dx
dy − 2ax − 2hy
=
dx
2hx + 2by
(
ax + hy )
=−
hx + by
y = x sin(a + y )
5)
Differentiate both sides w.r.t ‘x’
dy
dy
= x cos(a + y )
+ sin(a + y )
dx
dx
dy
dy
− x cos(a + y )
= sin(a + y )
dx
dx
[1 − x cos(a + y )] dy = sin(a + y )
dx
dy
sin(a + y )
=
dx 1 − x cos(a + y )
3.2.1. Differentiation of Inverse Trigonometric Functions
2ax + 2hx
If x = siny then
-1
y = sin x
-1
-1
-1
-1
-1
-1
sin x, cos x, tan x, cosec x, sec x and cot x are inverse
trigonometric functions.
Example: (1)
If y = sin-1x, find
dy
dx
y = sin−1 x
∴ sin y = x
214
Differentiate both sides w.r.t ‘x’
cosy
dy
=1
dx
dy
1
=
=
dx cos y
1
cos 2 y
=
1
1 − sin2 y
=
1
1− x2
Example: (2)
-1
Find the differentiation of cos x
-1
Let y = cos x
∴cosy = x
Differentiate both sides w.r.t ‘x’
dy
- siny
=1
dx
dy
1
1
1
1
=−
=−
=−
=−
2
2
dx
sin y
sin y
1 − cos y
1− x2
Example: (3)
-1
Find the differentiation of tan x.
-1
Let y = tan x
∴tany = x
Differentiate both sides w.r.t ‘x’
sec 2 y
∴
dy
=1
dx
dy
1
1
1
=
=
=
2
2
dx sec y 1 + tan y 1 + x 2
Example: (4)
-1
Find the differentiation of cot x.
-1
Let y = cot x
∴coty = x
Differentiate both sides w.r.t ‘x’
∴ -cosec 2 y
dy
=1
dx
dy
1
1
1
=−
=−
=−
2
2
dx
cos e y
1 + cot y
1+ x2
215
Example: (5)
-1
(Differentiation of sec x)
-1
Let y = sec x
∴secy = x
Differentiate both sides w.r.t ‘x’
secy tany
∴
dy
=1
dx
dy
1
1
1
=
=
=
dx sec y tan y sec y sec 2 y − 1 x x 2 − 1
Example: (6)
-1
Differentiation of cosec x
-1
Let y = cosec x
∴cosecy = x
Differentiate both sides w.r.t ‘x’
- cosec y cot y
dy
=1
dx
dy
1
1
1
=−
=−
=−
2
dx
cos ecy cot y
cos ecy cos ec y − 1
x x2 − 1
FORMULA
(
)
(
)
(
)
(
(
)
)
d
1
sin−1 x =
dx
1− x2
d
1
cos −1 x = −
2)
dx
1− x2
1)
d
1
tan −1 x =
dx
1+ x2
d
1
cot −1 x = −
4)
dx
1+ x2
d
1
sec −1 x =
5)
dx
x x2 − 1
d
1
cos ec −1x = −
6)
dx
x x2 − 1
3)
(
)
216
WORKED EXAMPLES
PART - A
Differentiate the following w.r.t ‘x’
( )
(4) (sin x )
( )
(1) sin−1 x
(2) cos −1 x
2
(5) x 2 sin−1 x
−1
(6) 1 − x 2 sin−1 x
Solution:
1)
( )
=
2)
1−
d
dx
( x)
2
1
1
=
1− x 2 x
( )
1
1−
( x)
2
1
2 x 1− x
d
dx
( x )= −
=
1
1
2 x (1 − x )
1
1− x 2 x
§ 1·
y = cot −1¨ ¸
©x¹
d § 1·
1
¨ ¸=−
1
dx
x
© ¹
§ 1·
1+ 2
1+ ¨ ¸
x
©x¹
2 ·
§
¨ − x ¸§¨ − 1 ·¸ = 1
¨ x 2 + 1 ¸© x 2 ¹ x 2 + 1
©
¹
4)
(8) tan −1 x
( x)
y = cos −1 x
dy
=−
dx
3)
1
( )
(7) e x tan−1 x
y = sin−1 x
dy
=
dx
§ 1·
(3) cot −1¨ ¸
©x¹
dy
=−
dx
1
(
)
y = sin−1 x
2
2
(
)
dy
d
= 2 sin−1 x
sin−1 x
dx
dx
§
1 ·¸ 2 sin−1 x
= 2 sin−1 x ¨
=
¨
2 ¸
1− x2
© 1− x ¹
217
§ 1·
¨− 2 ¸
© x ¹
=−
1
2 x (1 − x )
5)
y = x 2 sin−1 x
§
dy
1
= x 2¨
¨
2
dx
© 1− x
=
6)
x2
1− x
·
¸ + 2x sin−1 x
¸
¹
+ 2x sin−1 x
2
y = 1 − x 2 sin−1 x
§
dy
1
= 1− x2 ¨
¨
2
dx
© 1− x
7)
·
1
x sin−1 x
¸ + sin−1 x
(
− 2x ) = 1 −
¸
2 1− x2
1− x2
¹
y = e x tan−1 x
( )
dy
§ 1 ·
= ex ¨
¸ + tan −1 x e x
2
dx
© 1+ x ¹
º
ª 1
= ex «
+ tan −1 x »
2
¬1 + x
¼
8)
( )
y = tan −1 x
dy
1
=
dx 1 + x
( )
2
d
dx
( x ) = 1 +1 x 2 1x = 2
1
x (1 + x )
PART - B
Differentiate the following w.r.t ‘x’
§ 2x ·
1) sin−1¨
¸
2
© 1+ x ¹
§ 2x ·
2) tan −1¨
¸
2
© 1− x ¹
§ 1− x 2 ·
¸
3) cos −1¨
¨ 1+ x2 ¸
©
¹
Solution:
1.)
§ 2x ·
Let y = sin−1¨
¸
2
© 1+ x ¹
Put x = tan θ
∴ θ = tan-1x
§ 2 tan θ ·
∴ y = sin-1¨
¸
2
© 1 + tan θ ¹
2 sin A ·
§
¨∴ sin 2A =
¸
1
sin2 A ¹
+
©
218
= sin-1(sin 2θ )
= 2θ
y = 2 tan −1 x
dy
2
=
dx 1 + x 2
2.)
§ 2x ·
Let y = tan −1¨
¸
2
© 1− x ¹
Put x = tan θ
∴ θ = tan-1x
§ 2 tan θ ·
∴ y = tan-1¨
¸
2
© 1 − tan θ ¹
= tan-1(tan 2θ)
= 2θ
ª
2tan A º
»
« tan2A =
1 − tan2 A »¼
«¬
y = 2 tan −1 x
dy
2
=
dx 1 + x 2
3.)
Let
§ 1− x2 ·
¸
y = Cos −1¨
¨ 1+ x2 ¸
©
¹
Put x = tan θ
∴ θ = tan-1x
§ 1 − tan2 θ ·
¸
∴ y = Cos -1¨
¨ 1 + tan2 θ ¸
©
¹
= Cos -1(Cos2θ)
= 2θ
ª
1 - tan2 A º
=
Cos2A
»
«
1 + tan2 A ¼»
¬«
∴ y = 2 tan −1 x
dy
2
=
dx 1 + x 2
219
3.2.2. Differentiation of parametric function:
Sometimes ‘x’ and ‘y’ are functions of a third variable called a
dy
as
parameter without eliminating the parameter, we can find
dx
follows.
Let x = f(θ) and y = F(θ) so that ‘θ ’ is a parameter.
dy
dy dy dθ dθ
=
.
=
dx dθ dx dx
dθ
WORKED EXAMPLES
PART-A
dy
1.) If x = a Cos θ and y = b Sin θ, find
dx
Solution:
dx
= − a sin θ
dθ
dy
= b cos θ
dθ
dy dy dθ
b cos θ
b
=
=
= − cot θ
dx dx dθ
− a sin θ
a
2.)
Find
dy
if x = at2 and y = 2at
dx
Solution:
dx
= 2at
dt
dy
= 2a
dt
dy dy dt 2a 1
=
=
=
dx dx dt 2at t
220
3.)
Find
dy
if x = a sec θ and y = b tan θ
dx
Solution:
dx
= a sec θ tan θ
dθ
dy
= b sec 2 θ
dθ
dy dy dθ
b sec 2 θ
b sec θ
=
=
=
dx dx dθ a sec θ tan θ a tan θ
§ 1 ·
b¨
¸
b
b
Cosθ ¹
=
= Co sec θ
= ©
§ sin θ · a sin θ a
a¨
¸
© Cosθ ¹
4.)
Find
dy
c
if x= ct and y =
dx
t
Solution:
dx
=c
dt
dy
c
=− 2
dt
t
dy
− c t2
1
=−
=− 2
dx
c
t
PART - B
1.)
Find
dy
if x= cost + tsint and y=sint- tcost
dx
Solution:
dx
= − sin t + t cos t + sin t = t cos t
dt
dy
= cos t + t sin t − cos t = t sin t
dt
dy t sin t
=
= tan t
dx t cos t
221
2.)
dy
if x = sin 2t and y = cos 2t
dx
1
cos 2t
cos 2t(2) =
=
sin 2t
2 sin 2t
1
(− sin 2t)2 = − sin 2t
=
2 cos 2t
cos 2t
Find
dx
dt
dy
dt
§ − sin 2t ·
¨
¸
dy ¨© cos 2t ¸¹
sin 2t sin 2t
=
=−
dx § cos 2t ·
cos 2t cos 2t
¨¨
¸¸
sin
2
t
©
¹
3
(
sin 2t ) 2
=−
(cos 2t)3 2
3.)
Find
= −(tan 2t )
3
2
dy
if x = a(θ + sin θ)andy = a(1 − cos θ)
dx
Solution:
dx
= a(1 + cos θ )
dθ
dy
= a(0 + sin θ ) = a(sin θ )
dθ
dy
a sin θ
sin θ
=
=
dx a(1 + cos θ ) 1 + cos θ
=
4.)
Find
2 sin θ 2 cos θ 2
2 cos θ 2
2
= tan
θ
2
dy
if y = log(sec θ + tan θ ) and x = sec θ
dx
Solution:
dy
1
sec θ tan θ + sec 2 θ
=
dθ sec θ + tan θ
sec θ(tan θ + sec θ)
=
= sec θ
sec θ + tan θ
(
222
)
dx
= sec θ tan θ
dθ
dy
sec θ
1
=
=
= cot θ.
dx sec θ tan θ tan θ
3.3.1 Successive differentiation:
dy
, the derivative of y w.r. to x is a function of x
If y = f( x ), then
dx
dy
and can be differentiated once again. To fix the idea, we shall call
dx
as the first order derivative of y with respect to x and the derivative of
dy
w.r. to ‘x’ as a second order derivative of y w.r. to x and will be
dx
denoted by
d2 y
dx 2
. Similarly the derivative of
third order derivative and will be denoted by
Note: Alternative notations for
dy
= y1 = y ' = f '(x ) = D( y )
dx
d2 y
dx 2
dn y
dx n
= y 2 = y '' = f ''(x ) = D 2 (y )
= y n = y (n ) = f (n ) ( x ) = D n ( y )
223
d2 y
dx 2
d3 y
dx 3
w.r.t ’x’ will be called
and so on.
WORKED EXAMPLES
PART - A
1.)
If y = tan x, find
d2 y
dx 2
y = tan x,
dy
= sec 2 x
dx
d2 y
dx 2
= 2 sec x(sec x tan x )
= 2 sec 2 x tan x
2.)
If y = log(sin x ) , find y2
y1 =
1
cos x = cot x
sin x
y 2 = − cos ec 2 x.
3.)
If y = a cos x + b sin x, find y 2
y 1 = −a sin x + b cos x
y 2 = −a cos x − b sin x.
4.)
If y = Ae3 x + Be −5 x , find D 2 ( y )
D( y ) = 3 A e3 x − 5 B e−5 x
D 2 ( y ) = 9 A e3 x + 25 B e− 5 x
5.)
1
, find y 2
x
1
y1 = − 2
x
2
y2 = 3
x
If y =
224
PART - B
1.)
(
)
2
2
If y = x sin x, prove that x y 2 − 4xy1 + x + 6 y = 0
2
Solution:
y = x 2 sin x
y 1 = x 2 (cos x ) + 2x sin x = x 2 cos x + 2x sin x
y 2 = x 2 (− sin x ) + 2x cos x + 2x cos x + 2 sin x
2
= -x sinx + 4xcosx + 2sinx
[
∴ x 2 y 2 − 4xy1 + ( x 2 + 6)y = x 2 − x 2 sin x + 4x cos x + 2 sin x
]
− 4x(x cos x + 2x sin x ) + ( x + 6)x 2 sin x
2
2
= − x 4 sin x + 4x 3 cos x + 2x 2 sin x − 4x 3 cos x
− 8x 2 sin x + x 4 sin x + 6x 2 sin x
=0
2
2.) If y = a cos log x + b sin log x , show that x y 2 + xy1 + y = 0,
Solution:
y = a cos log x + b sin log x
§ 1·
§ 1·
y1 = −a sin log x ¨ ¸ + b cos log x ¨ ¸
©x¹
©x¹
Multiply by x on both sides
∴ xy1 = −a sin log x + b cos log x
Differentiate both sides w.r.to x
§ 1·
§ 1·
xy 2 + y1 = −a cos log x ¨ ¸ − b sin log x ¨ ¸
x
© ¹
©x¹
Again multiply by x on both sides
∴ x 2 y 2 + xy1 = −a cos log x − b sin log x
= −(a cos log x + b sin log x )
= −y
∴ x 2 y 2 + xy1 + y = 0
225
3.)
If y = esin
y = e sin
−1
y1 = e sin
−1
x
(
)
, prove that 1 − x 2 y 2 − xy1 − y = 0
x
−1
x
.
1
1− x 2
1 − x 2 y1 = esin
−1
x
Differentiate both sides w.r.t x
−1
1
1
(
1 − x 2 y 2 + y1
− 2x ) = esin x
2
2 1− x
1− x2
1− x 2 on both sides
Multiply by
(1 − x )y − xy = y
If y = (tan x ) , prove that (1 + x ) y
2
2
4.)
1
2
−1
Solution:
(
y = tan−1 x
2 2
)
2
§ 1 ·
y1 = 2 tan −1 x ¨
¸
2
© 1+ x ¹
(
)
Multiply by 1+ x 2 on both sides
(1 + x )y
2
1
= 2 tan−1 x
Differentiate both sides w.r.t.x
(1 + x )y
2
2
+ y1(2x ) = 2
(
)
1
1+ x2
Multiply by 1+ x 2 on both sides
(1 + x ) y
2 2
2
(
)
+ 2x 1 + x 2 y 1 = 2
226
2
(
)
+ 2x 1 + x 2 y 1 = 2
3.3.2 Formation of differential equation:
Definition: An equation involving differential coefficients is called
differential equation.
Examples:
dy
= 2xy
dx
dy
+ 2xy = x 2
2.)
dx
1.)
d2 y
3.)
dx
2
−5
dy
+ 6y = x 2
dx
2
3
§ d3 y ·
dy ·
§
4.) ¨¨ 3 ¸¸ + ¨1 + dx ¸ = 0
©
¹
© dx ¹
Order of the differential equation:
The order of the differential equation is the order of the highest
derivative appearing in the equation, after removing the radical sign
and fraction involved in the equation.
d2 y
e.g: 1.)
2.)
dx 2
+3
dy
+ 2y = e x (order is 2)
dx
2
d3 y
§ dy ·
+ 6¨ ¸ − 4y = 0 (order is 3)
3
dx
© dx ¹
Degree of the differential equation:
The degree of the differential equation is the power (or) degree of
the highest derivative occuring in the differential equation.
Example:
1.) 5
d2 y
dx 2
+7
dy
+ 5y = e x
dx
(degree 1)
2
5
§ d2 y ·
§ dy ·
¨
¸
7
5
+ 7 y = sin x
+
¨
¸
2.) ¨ 2 ¸
© dx ¹
© dx ¹
227
(degree 2)
1.)
WORKED EXAMPLES
PART - A
2
If xy = c , form the differential equation by eliminating the
Constant ‘c’
xy = c2
x
1.)
dy
+y=0
dx
PART - B
If y= A Cos 2x+ Bsin2x, form the differential equation by
eliminating the constants ‘A’ and ‘B’.
y=A Cos 2x + B sin 2x
dy
= −2 A sin 2x + 2 BCos2x
dx
d2 y
dx 2
d2 y
dx 2
2)
= −4 A Cos2x − 4 B sin 2x
= −4[ ACos 2x + B sin 2x ]
= −4y
+ 4y = 0 is the required differential equation
Form the differential equation by eliminating the constant ‘a’ and
‘b’ from the equation.
xy = ae x + be − x
Solution:
Given xy = ae x + be − x
Differentiate both sides w.r.to x
x y1 + y = ae x − be − x
Differentiate again w.r. to x,
x y 2 + y1 + y1 = ae x + be − x
x y 2 + 2y1 = xy
x y 2 + 2y1 − xy = 0
is the required differential equation.
228
I.
EXERCISE
PART - A
Differentiate the following w.r. to ‘x’
1.) (3x + 5)
(2.) sin2 (5x + 3 )
10
(
4.) tan−1 3 x
II.
III.
Find
)
2
(5.) e sin
−1
(3.) e cos
3
(6.) log(tan2 x )
2x
dy
if
dx
1.) x 2 + y 2 = 0
(2.) x sin y = 0
3.) xy = k
(4.) xy2 = x + y
1.) sin−1 5 x
(2.) sin−1(cos x )
3.) sin−1 3 x
(4.) tan−1 4x
(
)
3
PART - B
I)
Differentiate the following w.r.t ‘x’
3.) e sin
−1
x
)
Find
2x
tan 5 x
6.) xCos ( x )
−1
7.) 1 + x tan x
II)
−1
−1
5.) x tan x
(
)
4.) e tan
cos 6 x
2
2
(
2.) x 2 + 1 Cos5x
5x
1.) e cos 2x sin 3 x
8.)
dy
of the following
dx
1.) 2x 2 + 6xy + y 2 = 1
2.) x 2 sin y = C
3.) y = a + xey
4.) x 2 + 3 xy + 2y 2 = 4
5.) x m y n = am+ n
229
x
x sin−1 x
III)
Differentiate the following w.r.t ‘x’
§ 3x − x 3 ·
¸
1.) y = tan −1¨
¨ 1 − 3x 2 ¸
©
¹
§
x
2.) y = sin−1¨
¨ 2
2
© x +a
3.) y = sin−1 3x − 4x 3
4.) y = cos −1 4x 3 − 3x
(
5.) y =
(
)
1− x 2 − 4
x
6.) y = sin−1§¨ 2x 1 − x 2 ·¸
©
¹
§ 1 ·
7.) y = sec −1¨ 2
¸
© 2x − 1 ¹
IV) Find
§ 1 ·
8.) y = sin−1¨
¸
2
© 1 − 2x ¹
dy
if
dx
1.) x = 3 sin t − sin3 t
2.) x = a(θ − sin θ)
3.) x = cos3 θ
V)
and y = cos t - cos3 t
and y = a(1 − cos θ )
and y = sin3θ
(1 − x )y − 4x y + (x + 6)y = 0
x ) , Show that (1 − x )y − x y − 2 = 0
, Pr ove that (1 − x )y + (2x − 1)y = 0
1.) If y = x 2 cos x, Pr ove that
(
3.) If y = e tan
−1
2
2
2
2
2.) If y = sin−1
2
x
2
5.) If y = ea cos
−1
x
9
3.) − 3ecos 3 x Cos 2 x sin x
2e sin
(1 − x )y
1
(1 + x )y
2
2
, Pr ove that
1.) 30(3 X + 5)
5.)
1
2
2
−1
1
2
4.) If y = logª x + x 2 + 1º , Show that
«¬
»¼
I
)
·
¸
¸
¹
2
+ xy1 = 2
− x y1 − a 2 y = 0
ANSWER
PART - A
2.) 10 sin(5x + 3 ) cos(5x + 3 )
4.)
6 tan −1 3x
1 + 9x 2
(2x )
1 − 4x 2
2
6.) Cosec x secx
230
II.
1.)
III.
1.)
4.)
−x
y
− tan y
x
2.)
−5
1 − 25 x
2.) -1
2
(
12 tan −1 4x
1 + 16 x 2
3.)
−y
x
3.)
)
4.)
1− y2
2xy − 1
1
x 9x 2 − 1
2
PART - B
I
1.) e5 x [3 cos 2x cos 3x ] − 2 sin 2x sin 3x + 5 cos 2x sin 3x
(
)
2.) 2x cos 5 x − 5 x 2 + 1 sin 5 x
ª
cos 6 x
3.) esin−1x «− 6 sin 6x +
«¬
1− x 2
4.) e tan
5.)
−1
2 tan 5x º
2
»
«5 sec 5x +
1 + 4x 2 ¼
¬
2x ª
x2
1+ x
º
»
»¼
2
+ 2x tan −1 x
6.)
7.) 1 + 2x tan−1 x
II.
III.
1.) −
2x + 3 y
3x + y
2.) −
2
tan y
x
4.) −
2x + 3 y
3x + 4y
5.) −
my
nx
1.)
5.)
IV.
8.)
3
1+ x
(
2.)
2
1
2 1+ x
2
3
1.) − tan t
)
6.)
a
a +x
2
2
2
1− x2
2.) cot
θ
2
3.)
7.)
−x
1− x
2
x
1− x2
+
sin−1 x
2 x
3.)
3
1− x
2
−2
1− x2
3.) − tan θ
231
+ cos −1 x
4.)
8.)
ey
1 − xey
−3
1− x2
2
1− x2