Question

If u , v and w are functions of x , then show that in two ways-first by repeated application of product rule, second by logarithmic differentiation.

Answer 1

Let the given function be,

y=( x 2 −5x+8 )( x 3 +7x+9 )

(i)

Differentiating the given function by product rule,

dy dx =( x 2 −5x+8 ) d dx ( x 3 +7x+9 )+( x 3 +7x+9 ) d dx ( x 2 −5x+8 ) =( x 2 −5x+8 )( 3 x 2 +7 )+( x 3 +7x+9 )( 2x−5 ) =3 x 4 +7 x 2 −15 x 3 −35x+24 x 2 +56+2 x 4 −5 x 3 +14 x 2 −35x+18x−45 =5 x 4 −20 x 3 +45 x 2 −52x+11

(ii)

Simplify the given function,

y= x 2 ( x 3 +7x+9 )−5x( x 3 +7x+9 )+8( x 3 +7x+9 ) y= x 5 +7 x 3 +9 x 2 −5 x 4 −35 x 2 −45x+8 x 3 +56x+72 y= x 5 −5 x 4 +15 x 3 −26 x 2 +11x+72

Differentiate both sides with respect to x,

dy dx = d dx x 5 −5 d dx ( x 4 )+15 d dx ( x 3 )−26 d dx ( x 2 )+11 d dx ( x )+ d dx ( 72 ) =5 x 4 −5( 4 x 3 )+15( 3 x 2 )−26( 2x )+11( 1 )+0 =5 x 4 −20 x 3 +45 x 2 −52x+11

(iii)

Take log on both the sides of the given function,

logy=log( x 2 −5x+8 )+log( x 3 +7x+9 )

Differentiate both sides with respect to x,

1 y dy dx = d dx log( x 2 −5x+8 )+ d dx log( x 3 +7x+9 ) 1 y dy dx = 1 x 2 −5x+8 ( 2x−5 )+ 1 x 3 +7x+9 ( 3 x 2 +7 ) dy dx =y[ ( 2x−5 )( x 3 +7x+9 )+( 3 x 2 +7 )( x 2 −5x+8 ) ( x 2 −5x+8 )( x 3 +7x+9 ) ] dy dx =( x 2 −5x+8 )( x 3 +7x+9 )[ ( 2x−5 )( x 3 +7x+9 )+( 3 x 2 +7 )( x 2 −5x+8 ) ( x 2 −5x+8 )( x 3 +7x+9 ) ]

Simplify further,

dy dx =3 x 4 +7 x 2 −15 x 3 −35x+24 x 2 +56+2 x 4 −5 x 3 +14 x 2 −35x+18x−45 =5 x 4 −20 x 3 +45 x 2 −52x+11

It can be observed that the value of dy dx is same in all the three cases.