Convert the base to base e {\displaystyle e} . Since a b = e log ( a b ) = e b log ( a ) {\displaystyle a^{b}=e^{\log(a^{b})}=e^{b\log(a)}} we obtain
y = x log x = e log ( x ) log ( x ) = e ( ( log ( x ) ) 2 ) . {\displaystyle y=x^{\log x}=e^{\log(x)\log(x)}=e^{\left((\log(x))^{2}\right)}.}
To find the derivative we use the chain rule ( f ( g ( h ( x ) ) ) ) ′ = f ′ ( g ( h ( x ) ) ) g ′ ( h ( x ) ) h ′ ( x ) {\displaystyle (f(g(h(x))))'=f'(g(h(x)))g'(h(x))h'(x)} with f ( x ) = e x {\displaystyle f(x)=e^{x}} , g ( x ) = x 2 {\displaystyle g(x)=x^{2}} and h ( x ) = log ( x ) {\displaystyle h(x)=\log(x)} . With this we calculate
y = e ( ( log ( x ) ) 2 ) d y d x = ( f ( g ( h ( x ) ) ) ) ′ = f ′ ( g ( h ( x ) ) ) g ′ ( h ( x ) ) h ′ ( x ) = e ( ( log ( x ) ) 2 ) ( 2 log ( x ) ) ( 1 / x ) = 2 log ( x ) ⋅ 1 x ⋅ x log ( x ) = 2 log ( x ) ⋅ x log ( x ) − 1 {\displaystyle {\begin{aligned}y&=e^{\left((\log(x))^{2}\right)}\\{\frac {dy}{dx}}&=(f(g(h(x))))'\\&=f'(g(h(x)))g'(h(x))h'(x)\\&=e^{\left((\log(x))^{2}\right)}(2\log(x))(1/x)\\&=2\log(x)\cdot {\frac {1}{x}}\cdot x^{\log(x)}\\&=2\log(x)\cdot x^{\log(x)-1}\end{aligned}}}
