Science:Math Exam Resources/Courses/MATH100/December 2014/Question 03 (c)/Solution 1

Convert the base to base $e$. Since $a^{b}=e^{\log(a^{b})}=e^{b\log(a)}$ we obtain

$y=x^{\log x}=e^{\log(x)\log(x)}=e^{\left((\log(x))^{2}\right)}.$

To find the derivative we use the chain rule $(f(g(h(x))))'=f'(g(h(x)))g'(h(x))h'(x)$ with $f(x)=e^{x}$, $g(x)=x^{2}$ and $h(x)=\log(x)$. With this we calculate

${\begin{aligned}y&=e^{\left((\log(x))^{2}\right)}\\{\frac {dy}{dx}}&=(f(g(h(x))))'\\&=f'(g(h(x)))g'(h(x))h'(x)\\&=e^{\left((\log(x))^{2}\right)}(2\log(x))(1/x)\\&=2\log(x)\cdot {\frac {1}{x}}\cdot x^{\log(x)}\\&=2\log(x)\cdot x^{\log(x)-1}\end{aligned}}$