Deriving f(x) = e[x] * ln[x]

I thought that if you derive

e[x] * ln[x]

you will get

x * e[x] * ln[x] + (e[x]/x)

but my mathbook says you'll get

e[x] * ln[x] + (e[x]/x)

Who is retarded, me or it?

you...

(%i2) derivative(exp(x)*log(x), x);

                                              x
                                 x          %E
(%o2)                          %E  LOG(x) + ---
                                             x

fucking fonts...

you get e(x)*log(x)+e(x)/x

I can't see why, shouldn't:

y = f(x) * g(x)
y'= f'(x) * g(x)  +  g'(x) * f(x)

?

yes... what's the problem?
d/dx(e(x) = e(x)
d/dx(log(e)) = 1/x

d/dx(e(x)*log(x)) = e(x)*log(x) + e(x) * 1/x = e(x) * log(x) + e(x)/x

I thought that the rules for deriving e would be:
f(x)= e[a]
f'(x)= a*e[a]

You are not taking the derivative of e[a], you're taking the derivative of e[x]. The rule I think you're referencing is
f(x)  = e[u(x)]
f' = e[u(x)] * d/dx(u).
Therefore:
f(x) = e[x]
f'(x) = e[x] * d/dx(x) = e[x] * 1 = e[x]

Oh yes, now I remember. Many thanks, it's greatly appreciated.