Answer
$$\frac{{dy}}{{dx}} = {x^{9 + {x^{10}}}} + 10{x^{9 + {x^{10}}}}\ln x$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left( {{x^{\left( {{x^{10}}} \right)}}} \right) \cr
& {\text{let }}y = {x^{\left( {{x^{10}}} \right)}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \,\,\,\ln y = \ln {x^{\left( {{x^{10}}} \right)}} \cr
& {\text{using logarithmic properties}} \cr
& \,\,\,\ln y = {x^{10}}\ln x \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {{x^{10}}\ln x} \right] \cr
& {\text{product rule}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = {x^{10}}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {{x^{10}}} \right] \cr
& {\text{compute derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = {x^{10}}\left( {\frac{1}{x}} \right) + \ln x\left( {10{x^9}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = {x^9} + 10{x^9}\ln x \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left( {{x^9} + 10{x^9}\ln x} \right) \cr
& {\text{replace }}y{\text{ with }}{x^{\left( {{x^{10}}} \right)}} \cr
& \frac{{dy}}{{dx}} = {x^{\left( {{x^{10}}} \right)}}\left( {{x^9} + 10{x^9}\ln x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = {x^{9 + {x^{10}}}} + 10{x^{9 + {x^{10}}}}\ln x \cr} $$
