Question

8. log (log x), x >1

Answer 1

Given expression is log( logx ), x>1.

Let the expression be y=log( logx ).

Now, differentiate the expression on both sides with respect to x.

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

Further simplify equation (1) using d( logx ) dx = 1 x .

dy dx = 1 logx × 1 x (2)