LOGARITHMIC DIFFERENTIATION 325

G

(ii) Apply the laws of logarithms.

Thus lny=ln(1+x)^2 +ln(x−1)

1

2

−lnx−ln(x+2)

1

(^2) ,bylaws(i)
and (ii) of Section 31.2
i.e. lny=2ln(1+x)+^12 ln(x−1)
−lnx−^12 ln(x+2), by law (iii)
of Section 31.2
(iii) Differentiate each term in turn with respect tox
using equations (1) and (2).
Thus
1
y
dy
dx

2
(1+x)

• 1
  2
  (x−1)
  −
  1
  x
  −
  1
  2
  (x+2)
  (iv) Rearrange the equation to make
  dy
  dx
  the subject.
  Thus
  dy
  dx
  =y
  {
  2
  (1+x)


• 
  

  1
  2(x−1)
  −
  1
  x
  −
  1
  2(x+2)
  }
  (v) Substitute foryin terms ofx.
  Thus
  dy
  dx

  
  

  ( 1 +x)^2
  √
  (x− 1 )
  x
  √
  (x+ 2 )
  {
  2
  ( 1 +x)

  
  


• 
  

  1
  2 (x− 1 )
  −
  1
  x
  −
  1
  2 (x+ 2 )
  }
  Problem 1. Use logarithmic differentiation to
  differentiatey=
  (x+1)(x−2)^3
  (x−3)
  Following the above procedure:
  (i) Since y=
  (x+1)(x−2)^3
  (x−3)
  then lny=ln
  {
  (x+1)(x−2)^3
  (x−3)
  }
  (ii) lny=ln(x+1)+ln(x−2)^3 −ln(x−3),
  by laws (i) and (ii) of Section 31.2,
  i.e. lny=ln(x+1)+3ln(x−2)−ln(x−3),
  by law (iii) of Section 31.2.
  (iii) Differentiating with respect toxgives:
  1
  y
  dy
  dx

  
  

  1
  (x+1)

  
  


• 
  

  3
  (x−2)
  −
  1
  (x−3)
  ,
  by using equations (1) and (2)
  (iv) Rearranging gives:
  dy
  dx
  =y
  {
  1
  (x+1)

  
  


• 
  

  3
  (x−2)
  −
  1
  (x−3)
  }
  (v) Substituting forygives:
  dy
  dx

  
  

  (x+ 1 )(x− 2 )^3
  (x− 3 )
  {
  1
  (x+ 1 )

  
  


• 
  

  3
  (x− 2 )
  −
  1
  (x− 3 )
  }
  Problem 2. Differentiate
  y=
  √
  (x−2)^3
  (x+1)^2 (2x−1)
  with respect toxand eval-
  uate
  dy
  dx
  whenx=3.
  Using logarithmic differentiation and following the
  above procedure:
  (i) Since y=
  √
  (x−2)^3
  (x+1)^2 (2x−1)
  then lny=ln
  { √
  (x−2)^3
  (x+1)^2 (2x−1)
  }
  =ln
  {
  (x−2)
  3
  2
  (x+1)^2 (2x−1)
  }
  (ii) lny=ln(x−2)
  3
  (^2) −ln(x+1)^2 −ln(2x−1)
  i.e. lny=^32 ln(x−2)−2ln(x+1)
  −ln(2x−1)
  (iii)
  1
  y
  dy
  dx

  
  

  3
  2
  (x−2)
  −
  2
  (x+1)
  −
  2
  (2x−1)
  (iv)
  dy
  dx
  =y
  {
  3
  2(x−2)
  −
  2
  (x+1)
  −
  2
  (2x−1)
  }
  (v)
  dy
  dx

  
  

  √
  (x−2)^3
  (x+1)^2 (2x−1)
  {
  3
  2(x−2)
  −
  2
  (x+ 1 )
  −
  2
  ( 2 x− 1 )
  }