MUKUND SIR’S MATHS FOR SSC

STD-12

Ex 5.4

Properties of logarithms
Let m and n be arbitrary positive numbers such that a  0, a  1, b  0, b  1 then

(1) loga a  1, loga 1  0

1
(2) loga b. logb a  1  loga b 
logb a
logb a
(3) logc a  logb a. logc b or logc a 
logb c
(4) loga (mn)  loga m  loga n

m 
(5) loga    loga m  loga n
n
(6) loga m n  n loga m (7) a loga m  m

1 1
(8) loga     loga n (9) loga  n  loga n
n 

(10) loga  n  loga n , (  0)

(11) a logc b  b logc a , (a, b, c  0 and c  1)

Laws of indices
(1) a 0  1 , (a  0)
1
(2) am  , (a  0)
am
(3) a m  n  a m .a n , where m and n are rational numbers

am
(4) a m n  , where m and n are rational numbers, a  0
an

(5) (a m )n  a mn
q
(6) a p / q  a p

(7) If x  y , then a x  ay , but the converse may not be true.

(1) Differentiation of logarithmic and exponential functions :

d 1 d x
(i) log x  , for x > 0 (ii) e  ex
dx x dx
d x
(iii) a  a x log a , for a > 0
dx
d 1
(iv) loga x  , for x > 0, a> 0, a  1
dx x log a

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 ex
(1) Differentiate the following w.r.t. x
sin x
ex
Ans :  Let y 
sin x
Differentiating both sides w.r.t. x, we have

dy d  ex 
  
dx dx  sin x 
d d
sin (e x )  ex (sin x)
dx dx sin x  ex  ex cos x
 
sin 2 x sin 2 x
ex ( s i n 
x cos x )
 2
sin x
1
(2) Differentiate the following w.r.t. x esin x
1
Ans :  Let y  esin x
Differentiating both sides w.r.t. x, we have

dy d sin 1 x
 e
dx dx
d 1 1 1
 esin (sin 1 x)  esin x 
x

dx 1 x2
3
(3) Differentiate the following w.r.t. x e x
3
Ans :  Let y  e x
Differentiating both sides w.r.t. x, we have

dy d x3
 e
dx dx
3 d 3 3
 ex  ( x3 ) ex  3 x2 32x x
e
dx

(4) Differentiate the following w.r.t. x sin (tan-1 e-x)

Ans :  Let y = sin (tan-1 e-x)
Differentiating both sides w.r.t. x, we have

dy d
 [sin(tan 1 e x )]
dx dx
d
 cos(tan 1 e x )  (tan 1 e x )
dx
1 d
 cos(tan 1 e x )  x 2
 (e x )
1  (e ) dx

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 1 d
 cos(tan 1 e x )  2x
e x  (x)
1 e dx
1
 cos(tan 1 e x )  2x
e x  (1)
1 e
x 1  x
e cos(tan e )

1  e2x
(5) Differentiate the following w.r.t. x log(cos ex)
Ans :  Let y = log(cos ex)
Differentiating both sides w.r.t. x, we have

dy d
 [ l o g ( c o xs e ) ]
dx dx
1 d
 x
 (cos e x )
cos e dx
1 d
 x
 ( sin ex )  (ex )
cos e dx
1
  ( sin e x )  e x
cos e x
e x sin e x
  e x tan e x
cos e x
2 3 5
(6) Differentiate the following w.r.t. x ex  ex  ex  ...  ex
2 3 5
Ans :  Let y  ex  ex  ex  ...  ex
Differentiating both sides w.r.t. x, we have

dy d 2 3 5
 [ ex  ex  ex  . . . xe ]
dx dx
2 d 3 d 4 d 5 d
 ex  ex  (x 2 )  e x  (x 3 )  e x  (x 4 )  e x  (x 5 )
dx dx dx dx
x2 x3 x4 x5
 e  e  2x  e  3x  e  4x  e  5x
x 2 3 4

2 3 4 5
 ex  2xex 3x3ex  4x 3ex  5x 4ex

(7) Differentiate the following w.r.t. x e x ,x>0

Ans :  Let y  e x
Differentiating both sides w.r.t. x, we have

dy d 1 d
 ( e x )  (e x )
dx dx 2 e x dx
1 d 1 1 e x e x
  (e x )  ( x )  e x   
2 e x dx 2 e x 2 x 4 x e x
4 xe x

(8) Differentiate the following w.r.t. x log (log x), x > 1

Ans :  Let y = log (log x)
Differentiating both sides w.r.t. x, we have

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 dy d
 [log(log x )]
dx dx
1 d 1 1 1
  (log x)   
log x dx log x x x log x
cos x
(9) Differentiate the following w.r.t. x ,x>0
log x
cos x
Ans :  Let y 
log x
Differentiating both sides w.r.t. x, we have

dy d  c o s x

d x d x l o gx
d d
log x (cos x)  cos x (log x)
 dx dx
2
(log x)
1 cos x
log x( sin x)  cos  sin x log x 
 x  x
(log x) 2 (log x) 2
(x sin x log x  cos x)

x(log x) 2
(10) Differentiate the following w.r.t. x cos (log x + ex), x > 0
Ans :  Let y = cos (log x + ex)
Differentiating both sides w.r.t. x, we have

dy d
 [ c o s ( l o gx x e ) ]
dx dx
d
  sin(log x  ex )  (log x  ex )
dx
1 
  sin(log x  ex )    ex 
x 
1 
    ex  sin(log x  e x )
x 

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