Answer
$-\frac{1}{4x \sqrt{x}}+(e^x)(ln x + \frac{2x-1}{x^2})$
Work Step by Step
$g(x)=\sqrt{x}+e^x ln x$
$u=e^x$
$u’=e^x$
$v=ln x$
$v’=\frac{1}{x}$
$g’(x)=\frac{1}{2}x^{-\frac{1}{2}}+(e^x)(ln x)+\frac{e^x}{x}$
$=\frac{1}{2\sqrt{x}}+e^{x}(ln x+\frac{1}{x})$
$u=e^x$
$u’=e^x$
$v=ln x+ \frac{1}{x}$
$v’=\frac{1}{x}-\frac{1}{x^2}$
$g’’(x)=-\frac{1}{4}x^{-\frac{3}{2}}+(e^x)(ln x+\frac{1}{x})+(e^x)(\frac{x-1}{x^2})$
$=-\frac{1}{4x \sqrt{x}}+(e^x)(ln x + \frac{2x-1}{x^2})$
