Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
  y &= x{a^x}  \cr 
  \ln y &= \ln x + \ln {a^x}  \cr 
  \ln y &= \ln x + x\ln a  \cr 
  {1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right)  \cr 
  {1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\right)  \cr 
  {1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({{x + a\ln a} \over a}\right)  \cr 
  {1 \over y}{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}}  \cr 
  {{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \times x{a^x}  \cr 
  {{dy} \over {dx}} &= {{x{a^{x + 1}} + {x^3}{a^x} + {a^{x + 1}}{x^2}\ln a} \over {ax}}  \cr 
  {{dy} \over {dx}} &= {a^x} + {x^2}{a^{x - 1}} + {a^x}x\ln a  \cr 
  {{dy} \over {dx}} &= {a^x}\left(1 + {x^2}{a^{ - 1}} + x\ln a\right)  \cr 
  {{dy} \over {dx}} &= {a^x}\left({{a + {x^2} + x\ln a} \over a}\right) \cr} $$
However the answer in the back of the book is:
$${dy \over dx} = a^x (1 + x\ln a)$$
What have I done wrong?

Proceeding with your method in approaching the problem, but recalling that we can take $a$ as representing some constant: that is, $\;$ '$a$' $\;$ is a value that is NOT a function of $x$, and so does not get differentiated with respect to $x$. Similarly, so we can take $\ln a$ to be a constant. 
So in your third line, when we see $\,x\ln a,\,$ we will treat it no differently than we would treat, say, $x\ln 3$; that is, $\;\dfrac{d}{dx}\left(x\ln a\right) = \dfrac{d}{dx}\Big((\ln a)\cdot x\Big) = \ln a.\;\;$ (And we will use this when moving from the third line to the fourth line below). 
In contrast, $\,y\,$ IS a function of $x$, so you're correct that $\;\dfrac{d}{dx}(\ln y) = \dfrac 1y \cdot \dfrac{dy}{dx}$:
$$\eqalign{
  & y = x{a^x} \\ 
  & \ln y = \ln x + \ln {a^x}\\ 
  & \ln y = \ln x + x\ln a  \\
  & {1 \over y}{{dy} \over {dx}} = {1 \over x} + \ln a \\
  & {{dy} \over {dx}} = \left({1 \over x} + \ln a\right)\times x{a^x} \\
  & {{dy} \over {dx}} = a^x + xa^x\cdot \ln a  \\
  & {{dy} \over {dx}} = a^x(1 + x\ln a)\cr}$$
