derivative $\frac{\ln{x}}{e^x}$ Im asked to solve find the derivative of: $$ \frac{\ln x}{e^x}$$
my attempt
$$D\frac{\ln x}{e^x} = \frac{\frac{1}{x}e^x + \ln (x) e^x}{e^x} = e^x \frac{\frac{1}{x}+\ln x}{e^{2x}} = \frac{\frac{1}{x}+\ln x}{e^x}$$
But this is apparently wrong and the correct answer is: 
$$\frac{\frac{1}{x} - \ln x}{e^x}$$
Where do I go wrong?

Note that:
$$ \frac{\ln x}{e^x}=e^{-x}\ln {x}$$
And
$$(uv)'=u'v+uv'$$
Thus
$$ (\frac{\ln x}{e^x})'=(e^{-x}\ln {x})'=(-e^{-x}\ln {x})+(\frac{e^{-x}}{x})=\frac{\frac{1}{x} - \ln x}{e^x}$$
