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Exercise 2.5 (Solutions)
MathCity.org
Calculus and Analytic Geometry, MATHEMATICS 12
Merging man and maths
Available online @ http://www.mathcity.org, Version: 2.1.5
Some Important Derivative Formulas
d
•
c=0
where c is constant
dx
d
 d
•
tan x = sec2 x
 • dx sin x = cos x
dx

d
 • d cos x = − sin x
•
cot x = − csc2 x
 dx
dx
1
 d
−1
d
1
•
Tan −1x =
• dx Sin x =
2
1− x

dx
1+ x2

d
−1
• d Cos −1 x = −1
•
Cot −1 x =
2
 dx
dx
1 + x2
1− x
•
d n
x = nx n −1
dx
d
•
csc x = − csc x cot x
dx
d
•
sec x = sec x tan x
dx
d
1
Sec −1 x =
dx
x x2 −1
d
−1
•
Csc −1 x =
dx
x x2 −1
•
Question # 1(i)
Suppose y = sin 2 x
⇒ y + δ y = sin 2( x + δ x)
⇒ δ y = sin 2( x + δ x) − y
= sin 2( x + δ x) − sin 2x
Dividing both sides by δ x
δ y sin(2 x + 2δ x) − sin 2 x
=
δx
δx
 2 x + 2δ x + 2 x 
 2 x + 2δ x − 2 x 
2cos 
sin 


2
2




=
δx
2cos ( 2 x + δ x ) sin (δ x )
=
δx
Taking limit as δ x → 0
2cos ( 2 x + δ x ) sin (δ x )
δy
lim
=
lim
δ x →0 δ x
δ x →0
δx
sin (δ x )
dy
= 2 δlim
cos
2
x
+
δ
x
⋅
(
)
x →0
dx
δx
sin (δ x )
= 2 δlim
cos
2
x
+
δ
x
⋅
lim
(
)
x →0
δ x →0
δx
sin θ
= 2 cos ( 2 x + 0 ) ⋅ (1)
∵ lim
=1
θ →0
⇒
dy
= 2cos 2 x
dx
θ
FSc-II / Ex- 2.5 - 2
Question # 1(ii)
y = tan3x
Let
⇒ y + δ y = tan 3 ( x + δ x )
⇒ δ y = tan ( 3x + 3δ x ) − tan 3x
=
sin ( 3x + 3δ x ) sin 3x
−
cos ( 3 x + 3δ x ) cos3 x
=
sin ( 3 x + 3δ x ) cos3x − cos ( 3 x + 3δ x ) sin 3 x
cos ( 3x + 3δ x ) cos3 x
sin ( 3x + 3δ x − 3x )
sin ( 3δ x )
=
cos ( 3x + 3δ x ) cos3x
cos ( 3x + 3δ x ) cos3x
Dividing by δ x
sin ( 3δ x )
δy
1
=
⋅
δx
δ x cos ( 3x + 3δ x ) cos3x
Taking limit as δ x → 0
sin ( 3δ x )
δy
lim
= lim
δ x →0 δ x
δ x →0 δ x cos ( 3 x + 3δ x ) cos3 x
=
sin ( 3δ x )
dy
1
3
= lim
⋅
⋅
δ x →0
dx
cos ( 3x + 3δ x ) cos3x 3
δx
× ing and ÷ ing 3 on R.H.S
sin ( 3δ x )
1
⋅ lim
δ x →0
δ x→0 cos ( 3 x + 3δ x ) cos3 x
3δ x
1
= 3(1) ⋅
cos ( 3 x + 3(0) ) cos3 x
3
3
=
=
cos 2 3x
cos3 x cos3 x
= 3 lim
⇒
dy
= 3sec2 3x
dx
Question # 1(iii)
y = sin 2 x + cos 2 x
Let
⇒ y + δ y = sin 2 ( x + δ x ) + cos 2 ( x + δ x )
⇒ δ y = sin 2 ( x + δ x ) + cos 2 ( x + δ x ) − y
= sin 2 ( x + δ x ) + cos 2 ( x + δ x ) − sin 2 x − cos 2 x
= sin ( 2 x + 2δ x ) − sin 2 x  + cos ( 2 x + 2δ x ) − cos 2 x 

 2 x + 2δ x + 2 x   2 x + 2δ x − 2 x  
=  2cos 
 sin 

2
2

 



 2 x + 2δ x − 2 x   2 x + 2δ x − 2 x  
+  −2sin 
 sin 

2
2

 


FSc-II / Ex- 2.5 - 3
= 2cos ( 2 x + δ x ) sin (δ x ) − 2sin ( 2 x + δ x ) sin (δ x )
Dividing by δ x
δy
1
=
 2cos ( 2 x + δ x ) sin (δ x ) − 2sin ( 2 x + δ x ) sin (δ x ) 
δx
δx
Taking limit as δ x → 0
δy
1
lim
= lim
 2cos ( 2 x + δ x ) sin (δ x ) − 2sin ( 2 x + δ x ) sin (δ x ) 
δ x →0 δ x
δ x →0 δ x 
sin (δ x )
sin (δ x )
dy
= 2 lim cos ( 2 x + δ x ) lim
− 2 lim sin ( 2 x + δ x ) lim
δ x →0
δ x →0
δ x →0
δ x →0
dx
δx
δx
sin θ
= 2cos ( 2 x + 0 ) ⋅ (1) − 2sin ( 2 x + 0 ) ⋅ (1)
Since lim
=1
θ →0 θ
dy
⇒
= 2cos 2 x − 2sin 2 x
dx
Question # 1(iv)
Let
y = cos x 2
⇒ y + δ y = cos( x + δ x) 2
⇒ δ y = cos( x + δ y ) 2 − cos x 2
 ( x + δ x)2 + x 2 
 ( x + δ x)2 − x2 
= − 2sin 
 sin 

2
2




 x 2 + 2 xδ x + δ x 2 + x 2 
 x 2 + 2 xδ x + δ x 2 − x 2 
= − 2sin 
sin



2
2




 2 x 2 + 2 xδ x + δ x 2 
 2 xδ x + δ x 2 
= − 2sin 
⋅
sin



2
2





δ x2 
δx

= − 2sin  x 2 + xδ x +
 ⋅ sin  x +
δ x
2
2




Dividing by δ x

δy
1
δ x2 
δx

= −
⋅ 2sin  x 2 + xδ x +
 ⋅ sin  x +
δ x
δx
δx
2 
2 


δx

× ing and ÷ ing  x +
 on R.H.S
2 

δx

x+
2




 2
δy
2
δx 
δx

2 

⇒
= −  sin  x + xδ x +
 ⋅ sin  x +
δ x  ⋅
2
2
δx
δ
x





  x + δ x 
2 

FSc-II / Ex- 2.5 - 4

δx 

sin  x +
2
δ x

 2
δx 
δx
2   

= −  2sin  x + xδ x +
 ⋅ x +
⋅

δx
2  
2 

 

x
+
x
δ




2 

Taking limit as δ x → 0

δx 

sin  x +
2
δ x

 2
δy
δx 
δx
2   

lim
= − lim  2sin  x + xδ x +
⋅
 ⋅ x +


δ x →0 δ x
δ x →0
δx
2  
2 

 

x
x
+
δ




2 

δx

sin  x +
2
δ x
 2
dy
δx 
δx
2 


⇒
= − 2 lim sin  x + xδ x +
⋅ lim
⋅ lim  x +


δ x→0
δ x→0 
δx
dx
2  δ x →0 
2 

δ
x
x
+


2 

= − 2sin x 2 + (0) + (0) ⋅ (1) ⋅ ( x + (0) )
(
⇒
)
dy
= − 2 x sin x 2
dx
Question # 1(v)
Let y = tan 2 x
⇒ y + δ y = tan 2 ( x + δ x )
⇒ δ y = tan 2 ( x + δ x ) − tan 2 x
= ( tan ( x + δ x ) + tan x ) ⋅ ( tan ( x + δ x ) − tan x )
=
 sin x + δ x

sin x
( tan ( x + δ x ) + tan x ) ⋅  cos( x + δ x) − cos

x
(
)


 sin ( x + δ x ) cos x − sin x cos ( x + δ x ) 
= ( tan ( x + δ x ) + tan x ) ⋅ 

cos ( x + δ x ) cos x


 sin ( x + δ x − x ) 
= ( tan ( x + δ x ) + tan x ) ⋅ 

 cos ( x + δ x ) cos x 


sin δ x
= ( tan ( x + δ x ) + tan x ) ⋅ 

cos
x
+
x
cos
x
δ
(
)


Dividing by δ x


δy
1
sin δ x
=
tan ( x + δ x ) + tan x ) ⋅ 
(

δx
δx
 cos ( x + δ x ) cos x 
Taking limit when δ x → 0


δy
1
sin δ x
lim
= lim
tan ( x + δ x ) + tan x ) ⋅ 
(

δ x →0 δ x
δ x →0 δ x
 cos ( x + δ x ) cos x 
FSc-II / Ex- 2.5 - 5
⇒
⇒
 tan ( x + δ x ) + tan x 
dy
 sin δ x 
= lim 
⋅ lim 


δ x→0
δ x →0  δ x 
dx
 cos ( x + δ x ) cos x 
 tan ( x + 0 ) + tan x 
2 tan x
tan x + tan x
=
= 
 ⋅ (1) =
cos 2 x
cos x ⋅ cos x
 cos ( x + 0 ) cos x 
dy
= 2 tan x sec2 x
dx
Question # 1 (vi)
Let
y = tan x
⇒ y +δ y =
⇒ δy =
=
=
=
tan ( x + δ x )
tan ( x + δ x ) − tan x
(
 tan ( x + δ x ) + tan x 

tan ( x + δ x ) − tan x ⋅ 
 tan ( x + δ x ) + tan x 


tan ( x + δ x ) − tan x
)
tan ( x + δ x ) + tan x
 sin ( x + δ x ) sin x 
⋅
−

tan ( x + δ x ) + tan x  cos ( x + δ x ) cos x 
Now do yourself as above.
1
Question # 1 (vii)
Let
y = cos x
⇒ y + δ y = cos x + δ x
⇒ δ y = cos x + δ x − cos x
 x +δ x + x 
 x +δ x − x 
= − 2sin 
 sin 

2
2




Dividing by δ x
 x +δx + x 
 x +δ x − x 
2sin 
sin



2
2
δy




= −
δx
δx
Taking limit as δ x → 0
 x +δ x + x 
 x +δ x − x 
sin 
 sin 

2
2
δy




lim
= − 2 lim
δ x →0 δ x
δ x →0
δx
As δ x = ( x + δ x + x )( x + δ x − x ) , putting in above
FSc-II / Ex- 2.5 - 6
 x +δ x + x 
 x +δx − x 
sin 
 sin 

2
2
dy




⇒
= − 2 lim
δ x →0
dx
( x + δ x + x )( x + δ x − x )
 x +δx + x 
 x +δx − x 
sin 
sin 


2
2




= − lim
⋅ lim
δ x →0
( x + δ x + x ) δ x →0  x + δ x − x 




2
 x+0 + x 
sin 

sin x
2
dy


= −
⋅ (1)
⇒
= −
dx
( x +0 + x)
2 x
( )
Question # 2(i)
Assume y = x 2 sec 4 x
Differentiating w.r.t x
dy
d 2
=
x sec 4 x
dx
dx
d
d
= x 2 sec 4 x + sec 4 x x 2
dx
dx
d
= x 2 sec 4 x tan 4 x (4 x) + sec 4 x (2 x)
dx
2
= x sec 4 x tan 4 x(4) + 2 x sec 4 x
= 2 x sec 4 x ( 2 x tan 4 x + 1)
Question # 2(ii)
Let y = tan 3 θ sec 2 θ
Diff. w.r.t θ
dy
d
=
tan 3 θ sec 2 θ
dθ
dθ
d
d
= tan 3 θ
sec2 θ + sec2 θ
tan 3 θ
dθ
dθ
d
d




= tan 3 θ  2secθ
secθ  + sec 2 θ  3tan 2 θ
tan θ 
dθ
dθ




= tan 3 θ ( 2secθ ⋅ secθ tan θ ) + sec 2 θ 3tan 2 θ ⋅ sec 2 θ
(
(
= sec 2 θ tan 2 θ 2 tan 2 θ + 3sec 2 θ
Question # 2(iii)
2
Let y = ( sin 2θ − cos3θ )
Diff. w.r.t θ
)
)
FSc-II / Ex- 2.5 - 7
dy
d
2
=
( sin 2θ − cos3θ )
dθ
dθ
d
= 2 ( sin 2θ − cos3θ ) ( sin 2θ − cos3θ )
dθ
d
d


= 2 ( sin 2θ − cos3θ )  cos 2θ ⋅
(2θ ) + sin 3θ ⋅
(3θ ) 
dθ
dθ


= 2 ( sin 2θ − cos3θ )( cos 2θ ⋅ (2) + sin 3θ ⋅ (3) )
= 2 ( sin 2θ − cos3θ )( 2cos 2θ + 3sin 3θ )
Question # 2(iv)
y = cos x + sin x
Let
1
1
= cos( x ) 2 + ( sin x ) 2
Diff. w.r.t x
1
1
dy
d 

=
 cos( x) 2 + ( sin x ) 2 
dx
dx 

1 d
1
1
−1 d
= − sin( x) 2
x 2 + ( sin x ) 2 ( sin x )
dx
2
dx
1 1 −1
−1

 1
= − sin( x ) 2  x 2  + ( sin x ) 2 ( cos x )
2
 2
1  cos x sin x 
= 
−

2  sin x
x 
Question # 3(i)
Since y = x cos y
dy
d
=
x cos y
dx
dx
d
dx
= x cos y + cos y
dx
dx
dy
= x (− sin y ) + cos y (1)
dx
dy
dy
⇒
+ x sin y
= cos y ⇒
dx
dx
dy
cos y
⇒
=
dx
1 + x sin y
Question # 3(ii)
Since x = y sin y
dx
d
=
( y sin y )
dx
dx
( 1 + x sin y )
dy
= cos y
dx
FSc-II / Ex- 2.5 - 8
d
dy
sin y + sin y
dx
dx
dy
dy
dy
= y cos y + sin y
= ( y cos y + sin y )
dx
dx
dx
dy
1
⇒
=
dx
y cos y + sin y
⇒ 1 = y
Question # 4(i)
Since y = cos
1+ x
1 + 2x
Diff. w.r.t x
dy
d
1+ x
=
cos
dx
dx
1 + 2x
1
1+ x d  1+ x 
1 + x d  1 + x 2
= − sin

 = − sin


1 + 2 x dx  1 + 2 x 
1 + 2 x dx  1 + 2 x 
1+ x 1  1+ x 
= − sin
⋅ 

1 + 2x 2  1 + 2x 
−1
d  1+ x 


dx  1 + 2 x 
1  1 + 2x d 1 + x − 1 + x d 1 + 2x 
) ( ) ( ) (
)
1 + x 1  1 + 2x  2  (
dx
dx
= − sin
⋅ 

 
2
1 + 2x 2  1 + x  
(1 + 2 x )



2
1
1 + x (1 + 2 x ) 2  (1 + 2 x )(1) − (1 + x )( 2 ) 
= − sin
⋅


2

1 + 2 x 2 (1 + x ) 12 
(1 + 2 x )

1
1 + x (1 + 2 x ) 2  1 + 2 x − 2 − 2 x 
= − sin
⋅


1 + 2 x 2 (1 + x ) 12  (1 + 2 x )2 
1
1 + x (1 + 2 x ) 2  −1 
= − sin
⋅


1 + 2 x 2 (1 + x ) 12  (1 + 2 x )2 
1
1
1+ x
(1 + 2 x ) 2
= sin
⋅
2
1 + 2 x 2 (1 + x ) 12 (1 + 2 x ) 2− 12
⇒
dy
1
1+ x
=
3 sin
dx
1 + 2x
2 1 + x (1 + 2 x ) 2
Question # 4(ii)
Do yourself as above.
FSc-II / Ex- 2.5 - 9
Question # 5(i)
Let y = sin x and u = cot x
Diff. y w.r.t x
dy
d
=
sin x
dx
dx
= cos x
Now diff. u w.r.t x
du
d
=
cot x
dx
dx
= − csc2 x
dx
1
⇒
= −
du
csc 2 x
= − sin 2 x
Now by chain rule
dy
dy dx
=
⋅
du
dx du
= ( cos x ) − sin 2 x = − sin 2 x cos x
(
)
Question # 5(ii)
Let y = sin 2 x and u = cos 4 x
Diff. y w.r.t x
dy
d
=
sin 2 x
dx
dx
d
= 2sin x ( sin x ) = 2sin x cos x
dx
Now diff. u w.r.t x
du
d
=
cos 4 x
dx
dx
d
= 4cos3 x ( cos x ) = 4cos3 x ( − sin x )
dx
= − 4sin x cos3 x
dx
1
⇒
= −
du
4sin x cos3 x
Now by chain rule
dy
dy dx
=
⋅
du
dx du
1


= ( 2sin x cos x )  −
3 
 4sin x cos x 
FSc-II / Ex- 2.5 - 10
1
= − sec 2 x
2
Question # 6
Since tan y (1 + tan x ) = 1 − tan x
1 − tan x
⇒ tan y =
1 + tan x
tan π4 − tan x
1 − tan x
=
=
1 + 1 ⋅ tan x
1 + tan π4 ⋅ tan x
π

= tan  − x 
4

⇒ y =
π
4
−x
Diff. w.r.t x
dy
d π

=
 − x
dx
dx  4

⇒
= 0 −1
dy
= −1
dx
Question # 7
Since y = tan x + tan x + tan x + ...∞
Taking square on both sides
y 2 = tan x + tan x + tan x + ...∞
= tan x + tan x + tan x + tan x + ...∞
⇒ y = tan x + y
Diff. w.r.t x
d 2
d
y =
( tan x + y )
dx
dx
dy
dy
dy dy
⇒ 2y
= sec 2 x +
⇒ 2y −
= sec 2 x
dx
dx
dx dx
dy
⇒ ( 2 y − 1)
= sec2 x
dx
2
Question # 8
x = a cos3 θ , y = b sin 3 θ
Diff. x w.r.t θ
dx
d
=
a cos3 θ
dθ
dθ
(
)
FSc-II / Ex- 2.5 - 11
d
( cosθ ) = 3a cos2 θ ( − sin θ )
dθ
dx
dθ
−1
⇒
= − 3a sin θ cos 2 θ ⇒
=
dθ
dx
3a sin θ cos 2 θ
Now diff. y w.r.t θ
dy
d
=
b sin 3 θ
dθ
dθ
d
= b ⋅ 3sin 2 θ
( sin θ ) = 3b sin 2 θ cosθ
dθ
Now by chain rule
dy
dy dθ
=
⋅
dx
dθ dx
1
= 3b sin 2 θ cosθ ⋅ −
3a sin θ cos 2 θ
b
= − tan θ
a
dy
dy
⇒ a
= − b tan θ
⇒ a
+ b tan θ = 0
dx
dx
= a ⋅ 3cos 2 θ
(
)
Question # 9
x = a ( cos t + sin t ) and y = a ( sin t − t cos t )
Do yourself
Derivative of inverse trigonometric formulas
(i)
1
d
Sin −1 x =
dx
1 − x2
See proof on book page 76
(ii)
d
−1
Cos −1 x =
dx
1 − x2
Proof
Let y = cos −1 x
where x ∈ [ 0, π ]
⇒ cos y = x
Diff. w.r.t x
d
dx
dy
cos y =
⇒ − sin y
= 1
dx
dx
dx
dy
1
=−
dx
sin y
−1
=
Since sin y is positive for x ∈ [ 0, π ]
1 − cos 2 y
−1
=
1 − x2
FSc-II / Ex- 2.5 - 12
(iii)
d
1
Tan−1x =
dx
1+ x2
See proof on book at page 77
(iv)
d
−1
Cot −1 x =
dx
1+ x2
Proof
Let y = cot −1 x
⇒ cot y = x
Diff. w.r.t x
d
d
dy
cot y =
x ⇒ − csc2 y
= 1
dx
dx
dx
dy
−1
⇒
=
dx
csc 2 y
−1
=
∵ 1 + cot 2 y = csc 2 y
2
1 + cot y
dy
1
⇒
= −
dx
1 + x2
d
1
Sec −1 x =
dx
x x2 − 1
Proof
⇒ sec y = x
Let y = sec −1 x
Diff. w.r.t x
d
d
dy
sec y =
x
⇒ sec y tan y
= 1
dx
dx
dx
dy
1
⇒
=
dx
sec y tan y
1
=
∵ 1 + tan 2 y = sec 2 y
2
sec y sec y − 1
d
1
∵ sec y = x
⇒
Sec −1 x =
dx
x x2 − 1
(v)
(vi)
d
1
Csc −1 x = −
dx
x x2 − 1
See on book at page 77
Question # 10(i)
x
Let y = Cos −1
a
Diff. w.r.t x
FSc-II / Ex- 2.5 - 13
dy
d
x
=
Cos −1
dx
dx
a
−1
1 d
−1
d  x
=
⋅
x
=


2
2 dx a
a
dx


x
 x
1− 2
1−  
a
a
 
−1
1
−a
1
−1
=
⋅ (1) =
⋅
=
a2 − x2 a
a2 − x2 a
a2 − x2
a2
Ans
Question # 10(ii)
x
Let y = cot −1
a
Diff w.r.t x
dy
d
x
=
cot −1
dx
dx
a
−1
d x
−1
1 d
=
⋅   = 2
⋅
( x)
2
2
a
+
x
dx
a
a
dx


x
 
1+  
a2
a
−a 2 1
−a
= 2
⋅ (1) = 2
.
2
a +x a
a + x2
Question # 10(iii)
1
a
Let y = Sin −1
a
x
Diff. w.r.t x
dy
1 d
a
=
Sin −1
dx
a dx
x
1
1
d −1
1
1
d a
=
⋅a
x
=


2
2
2 dx x
a
dx
a


x −a
a
1−  
x2
x
x
x
1
 1 
− 2 = −
Ans
=
− x −2 =

2
2
2
2
2
2
x


x −a
x −a
x x −a
( )
(
)
Question # 10(iv)
Let y = Sin −1 1 − x 2
Diff. w.r.t x
dy
d
=
Sin −1 1 − x 2
dx
dx
FSc-II / Ex- 2.5 - 14
1
=
(
1−
1
=
x2
⋅
1− x
2
)
1
1
2 1 − x2
(
⋅
2
)
1
2
1
d
1 − x2 =
dx
( −2 x )
= −
1 − 1 + x2
1
x
⋅
x 1 − x2
⋅
1
1 − x2
2
(
= −
)
−1
d
1 − x2
dx
(
2
)
1
1 − x2
Question # 10(v)
 x2 + 1 
Let y = Sec −1  2

 x −1
Diff. w.r.t x
 x2 + 1 
dy
d
=
Sec −1  2 
dx
dx
 x −1
1
=
2
 x2 + 1   x2 + 1 
 2
  2
 −1
 x −1  x −1
d  x2 + 1 
⋅  2

dx  x − 1 
=
 x +1
 2

 x −1
2
(
d 2
d 2
 2
2
 x − 1 dx x + 1 − x + 1 dx x − 1
⋅
2
x2 − 1


(
1
2
) (
)
x2 + 1 − x2 − 1
2
( x − 1)
2
2
) (
(
 x2 + 1 
 2
⋅
−
1
x


(
(
) (
)
)
(
(
)
(
2
)
−1
2
(
(
)
)
Question # 10(vi)
Do yourself as above.
Question # 10(vii)
Do yourself as above.
Question # 11
y
x
= Tan−1
Since
x
y
) 
(
)
⇒ y = x Tan −1
x
y
)
)
 2 x x2 − 1 − x2 − 1
=
⋅
2
2
4
2
4
2
2
x + 1 ⋅ x + 2x + 1 − x + 2x − 1 
x
−
1

− 4x
−2
1
=
=
⋅ ( 2 x ( −2 ) ) =
2
2
x + 1 ⋅ 2x
x +1
x2 + 1 ⋅ 4 x2
(x
(
) (
 x2 − 1 ( 2 x ) − x2 + 1 ( 2 x ) 

⋅
2
2
4
2
4
2


x −1
x + 2x + 1 − x + 2x + 1 

( x 2 − 1)
1
=
) (
)
(
)


Ans
) 



FSc-II / Ex- 2.5 - 15
Diff. w.r.t x
dy
d 
−1
=
 x Tan
dx
dx 
x

y




d 
d
1
d  x 
−1 x 
−1 x

−1 x
= x  Tan
⋅ ( x) = x
+
Tan
⋅ (1)
 + Tan


   2 dx  y  
y
dx 
y
y dx
x
1+  



y
  


dy  

 1  y (1) − x  
x 
dy  y
dx  + Tan −1 x =
=x  2
y−x +

2 
2
2
2 
y +x 
dx  x
y
y
 y +x 

 y2 



dy
xy
x2
dy y
⇒
= 2
− 2
⋅ +
2
2
dx
y +x
y + x dx x
dy
x2
dy
xy
y
⇒
+ 2
⋅
= 2
+
2
2
dx y + x dx
y +x
x
dy
y
Proved
⇒
=
dx
x
Question # 12
Since y = tan pTan −1 x
(
)
⇒ Tan −1 y = pTan −1 x
Differentiating w.r.t x
d
d
Tan −1 y = p Tan −1 x
dx
dx
1 dy
1
⇒
= p⋅
⇒
2
1 + y dx
1 + x2
⇒


x 2  dy
y  x2
⇒ 1 + 2
⋅
=
+
1



y + x 2  dx
x  y 2 + x2


dy
(1 + x ) dx
2
(1 + x ) y − p (1 + y ) = 0
2
2
1
(
= p 1 + y2
Since
)
dy
= y1
dx
Error Analyst
Muzammil Ahsan 2009-11
Govt. Post Graduate Collage Jauharabad Distt. Khushab
Nain Malik
2012-14
Muhammad Usman Saleem (2013-15)
DPS & IC Jauharabad Distt. Khushab
Book:
Exercise 2.5
Calculus and Analytic Geometry Mathematic 12
Punjab Textbook Board, Lahore.
Edition: May 2013.
Made by: Atiq ur Rehman ([email protected])
Available online at http://www.MathCity.org in PDF Format (Picture format to view online).
Page Setup used: A4.
Printed: September 25, 2014.