Question

Find the derivative of the function given by and hence find .

Answer 1

The function is given as,

f( x )=( 1+x )( 1+ x 2 )( 1+ x 4 )( 1+ x 8 )

Take log on both the sides of the given function,

logf( x )=log( 1+x )+log( 1+ x 2 )+log( 1+ x 4 )+log( 1+ x 8 )

Differentiate both the sides,

1 f( x ) d dx ( f( x ) )= d dx ( log( 1+x ) )+ d dx ( log( 1+ x 2 ) )+ d dx ( log( 1+ x 4 ) )+ d dx ( log( 1+ x 8 ) ) 1 f( x ) f ′ ( x )= 1 1+x + 1 1+ x 2 ( 2x )+ 1 1+ x 4 ( 4 x 3 )+ 1 1+ x 8 ( 8 x 7 ) f ′ ( x )=f( x )[ 1 1+x + 1 1+ x 2 ( 2x )+ 1 1+ x 4 ( 4 x 3 )+ 1 1+ x 8 ( 8 x 7 ) ] f ′ ( x )=( 1+x )( 1+ x 2 )( 1+ x 4 )( 1+ x 8 )[ 1 1+x + 1 1+ x 2 ( 2x )+ 1 1+ x 4 ( 4 x 3 )+ 1 1+ x 8 ( 8 x 7 ) ]

Substitute x=1 in the above function,

f ′ ( 1 )=( 1+1 )( 1+ 1 2 )( 1+ 1 4 )( 1+ 1 8 )[ 1 1+1 + 1 1+ 1 2 ( 2( 1 ) )+ 1 1+ 1 4 ( 4 ( 1 ) 3 )+ 1 1+ 1 8 ( 8 ( 1 ) 7 ) ] =( 2 )( 2 )( 2 )( 2 )[ 1 2 + 2 2 + 4 2 + 8 2 ] =16( 1+2+4+8 2 ) =120

Therefore, the value of f ′ ( 1 ) is 120.