Exercise 2.7 (Solutions)Page 94
MathCity.org
Calculus and Analytic Geometry, MATHEMATICS 12
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Question # 1
Find y2 if
5
4
3
(i) y = 2 x − 3 x + 4 x + x − 2
Solution
(i)
y =
Diff. w.r.t x
dy
=
dx
⇒ y1 =
(ii) y = ( 2 x + 5 )
2 x5 − 3x 4 + 4 x3 + x − 2
d
2 x5 − 3x 4 + 4 x 3 + x − 2
dx
2 5 x 4 − 3 4 x3 + 4 3 x 2 + 1 − 0
(
)
( ) ( ) ( )
= 10 x 4 − 12 x3 + 12 x 2 + 1
Again diff. w.r.t x
dy1
d
=
10 x 4 − 12 x3 + 12 x 2 + 1
dx
dx
⇒ y2 = 10 4 x 3 − 12 3 x 2 + 12 ( 2 x ) + 0
(
( )
)
( )
= 40 x3 − 36 x 2 + 24 x
Ans.
3
2
(ii)
y = ( 2 x + 5)
Diff. w.r.t x
3
dy
d
=
( 2 x + 5) 2
dx
dx
3
3
−1 d
⇒ y1 = ( 2 x + 5 ) 2
( 2 x + 5)
2
dx
1
1
3
= ( 2 x + 5) 2 ( 2 ) = 3( 2 x + 5) 2
2
Again diff. w.r.t x
1
dy1
d
= 3 ( 2 x + 5) 2
dx
dx
1
1
−
⇒ y2 = 3 ⋅ ( 2 x + 5 ) 2 (2) ⇒ y2 =
2
(iii)
y =
x+
1
1
x
⇒ y = ( x)2 + ( x)
Diff. w.r.t x
−1
2
3
2x + 5
3
2
(iii) y =
x+
1
x
FSc-II / Ex- 2.7 - 2
dy
d  12
1 −1 1 −3
−1
x ) + ( x ) 2  ⇒ y1 = ( x ) 2 − ( x ) 2
=
(

2
2
dx
dx 
Again diff. w.r.t x
dy1
1 d  − 12
−3
=
x) − ( x) 2 
(

dx
2 dx 
1  1 −3 3
−5 
⇒ y2 =  − ( x ) 2 + ( x ) 2 
2 2
2

1 1
3 
1  −x + 3
3− x
= − 3 + 5  = 
or
y2 =
5
5 
4  x 2 x 2 
4  x 2 
4x 2
Question # 2
Find y2 if
 2x + 3 
(i) y = x 2e − x
(ii) y = ln 

 3x + 2 
Solution
(i)
y = x 2e − x
Diff. w.r.t x
dy
d 2 −x
=
xe
dx
dx
d
d
⇒ y1 = x 2 e− x + e − x x 2
dx
dx
2 −x
−x
= x e (−1) + e ( 2 x )
(
= e− x − x2 + 2 x
)
Again diff. w.r.t x
dy1
d −x
=
e − x2 + 2 x
dx
dx
d
d −x
y2 = e − x
− x2 + 2 x + − x2 + 2x
e
dx
dx
= e − x ( −2 x + 2 ) + − x 2 + 2 x e− x (−1)
(
)
(
) (
)
(
(
(x
= e − x −2 x + 2 + x 2 − 2 x
= e− x
2
− 4x + 2
)
)
)
 2x + 3 
y = ln 

 3x + 2 
⇒ y = ln ( 2 x + 3) − ln ( 3 x + 2 )
Diff. w.r.t x
dy
d
d
⇒
=
ln ( 2 x + 3) − ln ( 3 x + 2 )
dx
dx
dx
1
1
⇒ y1 =
( 2) −
( 3)
2x + 3
3x + 2
(ii)
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FSc-II / Ex- 2.7 - 3
−1
−1
= 2 ( 2 x + 3) − 3 ( 3 x + 2 )
Again diff. w.r.t x
dy1
d
d
−1
−1
= 2 ( 2 x + 3) − 3 ( 3 x + 2 )
dx
dx
dx
−2
−2
⇒ y2 = 2  − ( 2 x + 3) (2)  − 3  − ( 3 x + 2 ) (3) 




4
9
⇒ y2 = −
+
Ans.
2
2
2
x
+
3
3
x
+
2
(
) (
)
2
OR y2 =
=
=
(iii)
y =
− 4 ( 3x + 2 ) + 9 ( 3x + 2 )
2
( 2 x + 3) ( 3 x + 2 )
(
2
2
) (
− 4 9 x 2 + 12 x + 4 + 9 4 x 2 + 12 x + 9
2
( 2 x + 3) ( 3 x + 2 )
)
2
− 36 x 2 − 48 x − 16 + 36 x 2 + 108 x + 81
2
( 2 x + 3) ( 3 x + 2 )
2
2
=
60 x + 65
2
( 2 x + 3) ( 3 x + 2 )
1− x
1+ x
Diff. w.r.t x
1
dy
d  1− x 2
=


dx
dx  1 + x 
By solving, you will get (differentiate here)
−1
−1
−3
⇒ y1 =
= − (1 − x ) 2 (1 + x ) 2
3
1
(1 − x ) 2 (1 + x ) 2
Again diff. w.r.t x
dy1
d
−1
−3
= − (1 − x ) 2 (1 + x ) 2 

dx
dx 
−1 d
−3
−3 d
−1
⇒ y2 = − (1 − x ) 2 (1 + x ) 2 − (1 + x ) 2 (1 − x ) 2
dx
dx
−1  3
−5
−3  1
−3


= − (1 − x ) 2  − (1 + x ) 2 (1)  − (1 + x ) 2  − (1 − x ) 2 (−1) 
 2

 2

3
1
=
5 −
1
3
3
2 (1 − x ) 2 (1 + x ) 2 2 (1 + x ) 2 (1 − x ) 2
3(1 − x) − (1 + x)
3 − 3x − 1 − x
=
=
3
5
3
5
2 (1 − x ) 2 (1 + x ) 2
2 (1 − x ) 2 (1 + x ) 2
2 − 4x
1 − 2x
=
=
Ans.
3
5
3
5
2
2
2
2
2 (1 − x ) (1 + x )
(1 − x ) (1 + x )
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2
Ans.
FSc-II / Ex- 2.7 - 4
Question # 3
Find y2 if
(i) x 2 + y 2 = a 2
(ii) x 3 − y 3 = a 3
(iv) x = at 2 , y = bt 4
Solution
(i)
x2 + y2 = a2
Diff. w.r.t x
d 2
d 2
x + y2 =
a
dx
dx
(
)
⇒ 2 y y1 = − 2 x
(iii) x = a cosθ , y = a sin θ
(v) x 2 + y 2 + 2 gx + 2 fy + c = 0
dy
= 0
dx
x
y1 = −
y
⇒ 2x + 2 y
⇒
Again diff. w.r.t x
⇒
dy1
d x
= −  
dx
dx  y 
⇒
dy 
 dx
y
−
x


y2 = −  dx 2 dx 
y





 x
 y (1) − x  −  
 y
⇒ y2 = − 


y2





 y 2 + x2
x2 
y
+



y
y
 = −
= −
2
2
 y 
 y






 x2 + y 2 
= −
 Ans.
3
y


a2
y2 = − 3
y
OR
(ii)
x3 − y3 =
Diff. w.r.t x
d 3
x − y3 =
dx
dy
3x 2 − 3 y 2
=
dx
(
)
Again diff. w.r.t x
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dy
x
=−
dx
y






∵ x2 + y 2 = a2
a3
d 3
a
dx
0
⇒ − 3 y y1 = − 3 x
2
∵
2
x2
⇒ y1 = 2
y
FSc-II / Ex- 2.7 - 5
dy1
d  x2 
⇒
=
 
dx
dx  y 2 
d
d
y 2 ( x2 ) − x2 ( y 2 )
dx
dx
⇒ y2 =
2
y2
( )
 dy 
y 2 (2 x) − x 2  2 y 
 dx 
=
4
y
2
2
2  x 
2 xy − 2 x y  2 
y 
dy
x2

=
∵
= 2
y4
dx
y
4
3
4
2x
2 xy − 2 x
2 xy 2 −
y
y
=
=
4
y
y4
=
OR
y2 =
(
− 2x x3 − y 3
)
Ans.
y5
( )
− 2x a 3
∵ x3 − y 3 = a3
5
y
2a 3 x
⇒ y2 = − 5
y
x = a cosθ ,
(iii)
Diff. x w.r.t θ
dx
d
= a
cosθ
dθ
dθ
= − a sin θ
dθ
1
⇒
= −
dx
a sin θ
Now by chain rule
dy
dy dθ
=
⋅
dx
dθ dx
= a cosθ ⋅
y = a sin θ
Diff y w.r.t θ
dy
d
= a
sin θ
dθ
dθ
= a cosθ
−1
a sin θ
Now diff. y1 w.r.t θ
dy1
d
= − cot θ
dx
dx
⇒ y2 = + cosec2 θ
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dθ
dx
⇒ y1 = − cot θ
FSc-II / Ex- 2.7 - 6
1 

= cosec 2 θ . −

 a sin θ 
−1
⇒ y2 =
a sin 3 θ
x = at 2 , y = bt 4
(iv)
Diff. x w.r.t t
Diff. y w.r.t t
dx
d 2
= a t
dy
d
= b (t 4 )
dt
dt
dt
dt
= 2at
= 4bt 3
dt
1
⇒
=
dx
2at
Now by chain rule
dy
dy dt
=
⋅
dx
dt dx
1
2b 2
= 4bt 3 ⋅
⇒ y1 =
t
2at
a
Now diff. y1 w.r.t x
dy1
2b d 2
2b d 2 dt
=
t =
t ⋅
dx
a dx
a dt
dx
2b
1
2b
⇒ y2 =
⇒ y2 = 2
( 2t ) ⋅
a
2at
a
( )
( )
x 2 + y 2 + 2 gx + 2 fy + c = 0
d 2
d
⇒
x + y 2 + 2 gx + 2 fy + c = (0)
dx
dx
dy
dy
⇒ 2 x + 2 y + 2 g (1) + 2 f
+0=0
dx
dx
dy
⇒ (2 y + 2 f ) + (2x + 2g ) = 0
dx
dy
⇒ (2 y + 2 f )
= − ( 2x + 2g )
dx
dy
( 2x + 2g ) ⇒ y = − x + g
⇒
= −
1
y+ f
dx
(2 y + 2 f )
Again diff. w.r.t x
dy1
d  x+g 
= − 

dx
dx  y + f 
(v)
(
)
d
d


 ( y + f ) dx ( x + g ) − ( x + g ) dx ( y + f ) 
⇒ y2 = − 

2
(y+ f )




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FSc-II / Ex- 2.7 - 7
( y + f )(1) − ( x + g )
= −
2
(y+
y2 = −
=
⇒ y2 =
( y + f ) + (x + g)
−
3
(y+ f )
=
2
Ans.
y 2 + 2 yf + f 2 + x 2 + 2 xg + g 2
(y+ f )
(x
−
= −
2
y+ f
2
(y+ f )
= −
OR
(y+ f )
2
2
f ) +(x + g)
 x+g 
dy
( y + f ) − ( x + g ) −

 y+ f 
dx = −
2
(y+ f )
2
3
)
+ y 2 + 2 gx + 2 fy + c − c + f 2 + g 2
(y+ f )
0 − c + f 2 + g2
(y+ f )
c − f 2 − g2
(y+ f )
3
3
3
∵ x 2 + y 2 + 2 gx + 2 fy + c = 0
Ans.
Question # 4
Find y4 if
(i) y = sin 3 x
(ii) y = cos3 x
(
(iii) y = ln x 2 − 9
)
Solution
(i)
y = sin 3 x
Diff. w.r.t x
dy
d
=
( sin 3x )
dx
dx
⇒ y1 = cos3 x (3) ⇒ y1 = 3cos3 x
Again diff. w.r.t x
dy1
d
= 3 cos3 x
⇒ y2 = 3 ( − sin 3 x (3) ) ⇒ y2 = − 9sin 3 x
dx
dx
Again diff. w.r.t x
dy2
d
= − 9 sin 3 x
dx
dx
⇒ y3 = − 9 cos3x (3) ⇒ y3 = − 27 cos3 x
Again diff. w.r.t x
dy3
d
= − 27 cos3 x ⇒ y4 = − 27 ( − sin 3 x (3) )
dx
dx
⇒ y4 = 81sin 3 x
y = cos3 x
(ii)
Diff w.r.t x
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FSc-II / Ex- 2.7 - 8
dy
d
=
cos3 x
dx
dx
d
⇒ y1 = 3 cos 2 x
cos x
dx
⇒ y1 = 3 cos 2 x ( − sin x )
(
⇒ y1 =
)
(
)
(
)
3 (1 − sin x ) ( − sin x )
⇒ y1 = − 3sin x + 3sin 3 x
2
Again diff. w.r.t x
dy1
d
d
= − 3 sin x + 3 sin 3 x
dx
dx
dx
d
⇒ y2 = − 3cos x + 9sin 2 x sin x
dx
⇒ y2 = − 3cos x + 9 1 − cos 2 x cos x
(
)
= − 3cos x + 9cos x − 9cos3 x
Again diff. w.r.t x
dy2
d
d
= 6 cos x − 9 cos3 x
dx
dx
dx
= 6cos x − 9cos3 x
d
(cos3 x) = −3sin x + 3sin 3 x
dx
= 21sin x − 27sin 3 x
(
⇒ y3 = 6 ( − sin x ) − 9 −3sin x + 3sin 3 x
)
∵
= − 6sin x + 27sin x − 27sin 3 x
Again diff. w.r.t x
dy3
d
d
= 21 sin x − 27 sin 3 x
dx
dx
dx
d
⇒ y4 = 21( cos x ) − 27 3sin 2 x
sin x
dx
= 21cos x − 81sin 2 x ( cos x ) = 21cos x − 81 1 − cos 2 x ( cos x )
(
)
(
)
= 21cos x − 81cos x + 81cos3 x = − 60cos x + 54cos3 x
Alternative:
y = cos3 x
Since cos3 x = 4cos3 x − 3cos x
⇒ cos3 x − 3cos x = 4cos3 x
⇒ cos3 x =
Therefore
1
( cos3x − 3cos x )
4
Now diff. w.r.t x
dy
1 d
d

⇒
=  cos3 x − 3 cos x 
dx
4  dx
dx

Do yourself
y =
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1
( cos3x − 3cos x )
4
FSc-II / Ex- 2.7 - 9
(
y = ln x 2 − 9
(iii)
)
= ln ( x + 3)( x − 3)  = ln ( x + 3) + ln ( x − 3)
Diff. w.r.t x
dy
d
d
=
ln( x + 3) + ln( x − 3)
dx
dx
dx
1
1
⇒ y1 =
+
x+3 x −3
−1
−1
= ( x + 3) + ( x − 3)
Again diff w.r.t x
dy1
d
d
−1
−1
=
( x + 3) + ( x − 3)
dx
dx
dx
−2
−2
⇒ y2 = − ( x + 3 ) − ( x − 3 )
Again diff. w.r.t x
dy2
d
d
−3
−3
−2
−2
= − ( x + 3 ) − ( x − 3)
⇒ y3 = 2 ( x + 3) + 2 ( x − 3)
dx
dx
dx
Again diff. w.r.t x
dy3
d
d
−3
−3
= 2 ( x + 3) + 2 ( x − 3)
dx
dx
dx
(
⇒ y4 = 2 −3 ( x + 3)
=
−6
( x + 3)
4
+
−4
) + 2 ( −3( x − 3) )
−4
 1
1 
= − 6
+
4
4
 ( x + 3) ( x + 3) 
−6
( x + 3)
4
Ans.
Question # 5
If x = sin θ , y = sin mθ , Show that 1 − x 2 y2 − xy1 + m 2 y = 0
(
)
Solution
x = sin θ …… (i) , y = sin mθ ……..(ii)
From (i) θ = sin −1 x , putting in (ii)
y = sin m sin −1 x
(
)
Diff. w.r.t x
dy
d
=
sin m sin −1 x
dx
dx
d
⇒ y1 = cos m sin −1 x
m sin −1 x
dx
1
= cos m sin −1 x ⋅ m
1 − x2
(
)
(
)
(
)
(
⇒ y1 1 − x 2 = m cos m sin −1 x
)
Taking square on both sides.
y12 1 − x 2 = m 2 cos 2 m sin −1 x
⇒
( )
(
)
y (1 − x ) = m (1 − sin (m sin x) )
2
1
2
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2
2
−1
∵ cos 2 x = 1 − sin 2 x
FSc-II / Ex- 2.7 - 10
(
)
(
⇒ y12 1 − x 2 = m 2 1 − y 2
)
From (ii)
Now again diff. w.r.t x
d 2
d
y1 1 − x 2 = m 2
1 − y2
dx
dx
d
d 2
dy 

⇒ y12
1 − x2 + 1 − x2
y1 = m 2  0 − 2 y 
dx
dx
dx 

dy
dy
⇒ y12 ( −2 x ) + 1 − x 2 2 y1 1 = − 2m 2 y
dx
dx
2
2
2
⇒ − 2 xy1 + 1 − x 2 y1 y2 = − 2m y y1
(
)
(
(
) (
(
)
)
(
(
)
)
) )
(
⇒ 2 y1 − xy1 + 1 − x 2 y2
(
(
= 2 y1 −m 2 y
)
)
⇒ − xy1 + 1 − x 2 y2 = − m 2 y
(
)
⇒ 1 − x 2 y2 − xy1 + m 2 y = 0
Proved
Question # 6
d2y
dy
If y = e sin x , show that
− 2 + 2y = 0
2
dx
dx
x
Solution
y = e sin x
Diff. w.r.t x
dy d x
= e sin x
dx dx
d
d
= e x sin x + sin x e x
dx
dx
x
x
= e cos x + sin x e = e x ( cos x + sin x )
Again diff. w.r.t x
d  dy 
d x
e ( cos x + sin x )
  =
dx  dx 
dx
d2y
d
d
⇒
= e x ( cos x + sin x ) + ( cos x + sin x ) e x
2
dx
dx
dx
x
x
= e ( − sin x + cos x ) + ( cos x + sin x ) e = e x ( − sin x + cos x + cos x + sin x )
x
= e x ( 2cos x ) = 2e x cos x
Now
d2y
dy
L.H.S =
−
2
+ 2y
dx 2
dx
= 2e x cos x − 2e x ( cos x + sin x ) + 2e x sin x
= 2e x ( cos x − cos x − sin x + sin x )
= 0
d2y
dy
−
2
+ 2y = 0
i.e.
dx 2
dx
Question # 7
[
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Proved
FSc-II / Ex- 2.7 - 11
If y = eax sin bx , show that
d2y
dy
− 2a + a 2 + b 2 y = 0
2
dx
dx
(
)
Solution
y = eax sin bx
Diff. w.r.t x
dy d ax
= e sin bx
dx dx
d
d
= e ax sin bx + sin bx e ax = e ax cos bx(b) + sin bx eax (a)
dx
dx
ax
= e ( b cos bx + a sin bx )
Again diff. w.r.t x
d  dy 
d ax
e ( b cos bx + a sin bx )
  =
dx  dx 
dx
d2y
d ax
ax d
⇒
=
e
b
cos
bx
+
a
sin
bx
+
b
cos
bx
+
a
sin
bx
e
(
)
(
)
dx 2
dx
dx
= e ax ( − b sin bx (b) + a cos bx (b) ) + ( b cos bx + a sin bx ) eax (a )
(
)
e ( 2ab cos bx + a sin bx − b sin bx )
e ( 2ab cos bx + 2a sin bx − a sin bx − b sin bx )
e  2a ( b cos bx + a sin bx ) − ( a + b ) sin bx 
2ae ( b cos bx + a sin bx ) − ( a + b ) e sin bx
dy
d y
dy
2a − ( a + b ) y ⇒
− 2a + ( a + b ) y
dx
dx
dx
= e ax − b 2 sin bx + ab cos bx + ab cos bx + a 2 sin bx
=
=
=
=
d2y
⇒
=
dx 2
Question # 8
(
ax
2
ax
2
2
2
ax
2
2
ax
2
2
ax
2
2
2
2
2
2
2
2
)
y = ( Cos x )
(
)
If y = Cos −1 x , prove that 1 − x 2 y2 − xy1 − 2 = 0
Solution
−1
2
Diff. w.r.t x
dy
d
=
Cos −1 x
dx
dx
(
(
)
⇒ y1 = 2 Cos −1 x ⋅
)
2
(
⇒ y1 = 2 Cos −1 x
−1
) dxd Cos
(
On squaring both sides
( ) = 4 ( Cos x )
(1 − x ) = 4 y
⇒ y12
−1
2
Again diff. w.r.t x
d 2
y1 1 − x 2
dx
(
http://www.mathcity.org
)
=4
dy
dx
x
⇒ y1 1 − x 2 = − 2 Cos −1 x
1 − x2
y12 1 − x 2
−1
2
(
∵ y = Cos −1 x
)
2
)
= 0
FSc-II / Ex- 2.7 - 12
(
⇒ 1 − x2
) dxd y
2
1
( )
(1 − x ) y
+ y12
d
1 − x2
dx
(
)
= 4 y1
dy1
+ y12 ( −2 x ) = 4 y1 ⇒ 2 y1  1 − x 2 y2 − xy1  = 4 y1
dx
2 − xy1 − 2 = 0
(
⇒ 1 − x 2 ⋅ 2 y1
⇒
2
)
]
Question # 9
If y = a cos ( ln x ) + b sin ( ln x ) , prove that x 2
d2y
dy
+x +y= 0
2
dx
dx
Solution
y = a cos ( ln x ) + b sin ( ln x )
Diff. w.r.t x
dy
d
d
= a cos ( ln x ) + b sin ( ln x )
dx
dx
dx
d
d
= a  − sin ( ln x )  (ln x) + b cos ( ln x ) (ln x)
dx
dx
1
1
= − a sin ( ln x ) + b cos ( ln x )
x
x
dy
⇒ x = − a sin ( ln x ) + b cos ( ln x )
dx
Again diff. w.r.t x
d  dy 
d
d
x  = − a sin ( ln x ) + b cos ( ln x )

dx  dx 
dx
dx
d  dy  dy  dx 
d
d
⇒ x   + ⋅   = − a cos ( ln x ) ( ln x ) + b ( − sin ( ln x ) ) ( ln x )
dx  dx  dx  dx 
dx
dx
1
1
d 2 y dy
⇒ x 2 + ⋅ (1) = − a cos ( ln x ) ⋅ − b sin ( ln x ) ⋅
dx
dx
x
x
2
2
d y dy
1
dy
2 d y
⇒ x 2 +
= − ( a cos ( ln x ) + b sin ( ln x ) ) ⇒ x
+x = −y
2
dx
dx
x
dx
dx
2
d y
dy
⇒ x 2 2 + x + y = 0 Proved
dx
dx
Please report us error at www.mathcity.org/error
Book:
Exercise 2.7, page 94
Text Book of Algebra and Trigonometry Class XII
Punjab Textbook Board, Lahore.
Available online at http://www.MathCity.org in PDF Format
(Picture format to view online).
Updated: September,14,2017.
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