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

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

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

$\begin{array}{rl}y& =e^{((\log (x){)}^2)}\\ \frac{dy}{dx}& =(f(g(h(x))){)}^{\prime }\\ & =f^{\prime }(g(h(x)))g^{\prime }(h(x))h^{\prime }(x)\\ & =e^{((\log (x){)}^2)}(2\log (x))(1/x)\\ & =2\log (x)\cdot \frac{1}{x}\cdot x^{\log (x)}\\ & =2\log (x)\cdot x^{\log (x)-1}\end{array}$