Question

9. log (logx)

Answer 1

Let, y=log( logx ).

The first order derivative is obtained by differentiating the function with respect to x.

dy dx = d( log( logx ) ) dx = 1 logx × d( logx ) dx = 1 logx × 1 x = 1 xlogx

Again differentiate the above function with respect to x.

d dx ( dy dx )= d dx ( 1 xlogx ) d 2 y d x 2 = d( 1 ) dx ×( xlogx )− d( xlogx ) dx ×1 ( xlogx ) 2 = 0×( xlogx )−( d( xlogx ) dx )×1 ( xlogx ) 2 = 0−( d( xlogx ) dx ) ( xlogx ) 2

Further simplify the above function.

d 2 y d x 2 = −{ d( x ) dx ×logx+ d( logx ) dx ×x } ( xlogx ) 2 = −{ 1×logx+ 1 x ×x } ( xlogx ) 2 = −( logx+1 ) ( xlogx ) 2

Therefore, the second order derivative of y=log( logx )is −( logx+1 ) ( xlogx ) 2 .