1. We are given $f'(x)=2010(x-2009)(x-2010)^2(x-2011)^3(x-2012)^4.$ Also, $f(x)=\ln(g(x)),$ with $g(x)>0$ for all $x\in \mathbb R$.

2. Since $f(x)=\ln(g(x)),$ and $\ln x$ is a strictly increasing function on $(0,\infty)$, the local maxima/minima of $f$ and $g$ occur at the same points.

Indeed, $f'(x)=\frac{g'(x)}{g(x)}.$ Since $g(x)>0$, $f'(x)$ and $g'(x)$ always have the same sign. Hence the intervals of increase/decrease of $f$ and $g$ are identical.

So it is enough to count the number of local maxima of $f$.

3. Critical points come from $f'(x)=0$: $x=2009,\;2010,\;2011,\;2012.$ Now analyze sign changes of $f'(x)$.

Because the constant $2010>0$, the sign depends on $(x-2009)(x-2010)^2(x-2011)^3(x-2012)^4.$

Notice:

• $(x-2010)^2$ is always nonnegative and has even multiplicity, so it does not change sign across $x=2010$.
• $(x-2012)^4$ is always nonnegative and has even multiplicity, so it does not change sign across $x=2012$.
• $(x-2009)$ changes sign across $x=2009$.
• $(x-2011)^3$ changes sign across $x=2011$.

Thus the sign of $f'(x)$ is essentially the sign of $(x-2009)(x-2011).$

4. Check intervals:

• For $x<2009$: $x-2009<0,\quad x-2011<0 \implies f'(x)>0.$

• For $2009<x<2010$: $x-2009>0,\quad x-2011<0 \implies f'(x)<0.$

• For $2010<x<2011$: again $x-2009>0,\quad x-2011<0 \implies f'(x)<0.$

• For $2011<x<2012$: $x-2009>0,\quad x-2011>0 \implies f'(x)>0.$

• For $x>2012$: $x-2009>0,\quad x-2011>0 \implies f'(x)>0.$

So the sign chart is: $+ o - \text{ at } x=2009,$ no sign change at $x=2010$, $-\to + \text{ at } x=2011,$ no sign change at $x=2012$.

5. Therefore:

• At $x=2009$, $f$ changes from increasing to decreasing, so $f$ has a local maximum.
• At $x=2011$, $f$ has a local minimum.
• At $x=2010$ and $x=2012$, there is no local extremum.

Hence $g$ also has exactly one local maximum.

6. Final answer: $\boxed{1}$