Question 550906
<pre>
y = {{{1/log((x))}}}

Use the change of base formula:

Y = {{{1/(ln(x)/ln(10))}}}

Write as a division:

y = 1÷{{{ln(x)/ln(10)}}}

Invert and multiply:

y = 1·{{{ln(10)/ln(x)}}}

y = {{{ln(10)/ln(x)}}}

Bring the denominator to the numerator with a -1 power 

y = ln(10)·[ln(x)]<sup>-1</sup>

                       Now we need two formulas 

                       1) {{{expr(d/dx)}}}{{{(u^n)}}} = {{{n*u^(n-1)expr((du)/(dx))}}}

                       and

                       2) {{{expr(d/dx)}}}{{{ln(u)}}} = {{{1/u}}}·{{{(du)/(dx)}}}

y = ln(10)·[ln(x)]<sup>-1</sup>

{{{dy/dx}}} = -1·ln(10)·[ln(x)]<sup>-2</sup>{{{(1/x)}}}
{{{dy/dx}}} = -ln(10)·{{{1/("ln_x")^2}}}{{{(1/x)}}}
{{{dy/dx}}} = -{{{ln(10)/(x*("ln_x")^2)}}}

Edwin</pre>