EXAMPLES IN DIFFERENTIATION 95 



(6) Treating it as a product and working with the rule 



1 dy I du I dv 

 y dx u dx v dx 



Then u = 1 - x 



du . 1 du 1 



-T- = 1 and y- = 



dx u dx 1 x 



Also v = (1 + xrf 



dv x . 1 dv a: 



and - -T- = 



J 



Then !__ * + 

 * fte 1 x 1 + 



- x) 



(1 + a)(l - x) 



- 2x z + x - 1 



(1 + X 2 )(l x) 

 Multiplying across by y 



du &x -f- x 1 / 



^(i + ^d-*)"' 1 -*'^^ 



- 2,r 2 + z - 1 



(c) Working logarithmically 

 gy = lo 



dy_ - 



= log e (l -x) + - log e (l + a; 2 ) 



___ 

 I x 2 1 + x z 



1 a? 



l-x 1 + 



-2x*+x-I 



Multiplying across by y 



dy 2x 2 + x 1 



a? - 1 



