If $$y = {\log _{10}}x + {\log _x}10 + {\log _x}x + {\log _{10}}10$$        then what is $${\left( {\frac{{dy}}{{dx}}} \right)_{x = 10}}$$   equal to ?

Solution :
$$\eqalign{ & y = {\log _{10}}x + {\log _x}10 + {\log _x}x + {\log _{10}}10 \cr & y = {\log _{10}}x + {\log _x}10 + 1 + 1 \cr} $$
Differentiating equation w.r.t. $$x$$
$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{1}{{x\,{{\log }_e}10}} - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}}}.\frac{1}{{\left( {x\,\log \,10} \right)}} \cr & = \frac{1}{{x\,{{\log }_e}10}}\left[ {1 - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}}}} \right] \cr & {\left( {\frac{{dy}}{{dx}}} \right)_{x = 10}} = \frac{1}{{10\,{{\log }_e}10}}\left[ {1 - 1} \right] = 0 \cr} $$

\[\left[ \begin{array}{l} {\bf{Note :}}\\ {\log _x}10 = \frac{{{{\log }_{10}}10}}{{{{\log }_{10}}x}} = \frac{1}{{{{\log }_{10}}x}}\\ \frac{d}{{dx}}\left[ {\frac{1}{{{{\log }_{10}}x}}} \right] = - {\left( {{{\log }_{10}}x} \right)^{ - 2}} \times \frac{1}{{x\,{{\log }_e}10}}\\ = - \frac{1}{{{{\left( {{{\log }_{10}}x} \right)}^2}x\,{{\log }_e}10}} \end{array} \right]\]