1. Given information

We have $f(x)=\int_a^x g(t)\,dt,$ where $g$ is differentiable.

Therefore, by the Fundamental Theorem of Calculus, $f'(x)=g(x), \qquad f''(x)=g'(x).$

We are told that the equation $f(x)=0$ has exactly five distinct roots in $(a,b)$.

We need the minimum number of roots of $g(x)g'(x)=0,$ i.e. the minimum number of points where either $g(x)=0 \quad \text{or} \quad g'(x)=0.$

2. Use Rolle's theorem on the roots of $f$

Let the five distinct roots of $f$ in $(a,b)$ be $x_1<x_2<x_3<x_4<x_5.$ Since $f(x_i)=0$ for all $i$, applying Rolle's theorem on each interval $[x_1,x_2], [x_2,x_3], [x_3,x_4], [x_4,x_5],$ there exists at least one point in each open interval where $f'(x)=0.$ But $f'(x)=g(x)$, so $g(x)=0$ has at least one root in each of the four intervals.

Hence, $g(x)=0$ has at least $4$ distinct roots in $(a,b)$.

3. Now apply Rolle's theorem to $g$

Let these four distinct roots of $g$ be $y_1<y_2<y_3<y_4.$ Applying Rolle's theorem to $g$ on each interval $[y_1,y_2], [y_2,y_3], [y_3,y_4],$ there exists at least one point in each open interval where $g'(x)=0.$ So $g'(x)=0$ has at least $3$ distinct roots in $(a,b)$.

4. Count roots of $g(x)g'(x)=0$

The equation $g(x)g'(x)=0$ is satisfied at all roots of $g$ and all roots of $g'$.

So, from above, we get at least $4+3=7$ roots.

These are distinct because:

• the roots of $g'$ obtained from Rolle's theorem lie in the open intervals between consecutive roots of $g$,
• so they cannot coincide with the roots of $g$.

Thus, $g(x)g'(x)=0$ has at least $7$ distinct roots in $(a,b)$.

5. Check options

• A: twelve roots — not guaranteed
• B: five roots — too small
• C: seven roots — guaranteed minimum
• D: three roots — too small

Hence the correct option is $\boxed{\text{C}}.$