How to find root of derivative of any polynomial/equation? 
Let
  $$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5),\quad -\infty< x<\infty.$$
  The number of distinct roots of equation
  $$\frac{d}{dx}f(x) = 0$$
  is exactly ?

Source.
The only method that I know is to multiply and then find the derivative of function then apply Sturm's theorem but it seems vague when you have to solve the question in 3 to 5 minutes . So you are requested to suggest a plausible alternative approach.
 A: $f(x)$ is a polynomial of degree 5, $f'(x)$ is one of degree 4. So $f'$ has 4 zeros. As the zeros of $f$ are distinct, it has no zeros in common with $f'$. By the mean value theorem, there has to be a zero of $f'$ between each consecutive pair of zeros of $f$, i.e., there are at least 4 different zeros of $f'$ (there might be several between consecutive zeros, or ones outside the range of zeros of $f$). But as $f'$ is a quartic, there are exactly 4.
A: Here's an alternative to using Rolle's Theorem, which can also be done in a couple of minutes.
Let $x-3=u$.  Then the polynomial becomes
$$u(u^2-1)(u^2-4)=u(u^4-5u^2+4)=u^5-5u^3+4u$$
so the derivative is
$$5u^4-15u^2+4$$
which is quadratic in $u^2$.  You can do is either using the quadratic formula,
$$U=u^2={15\pm\sqrt{225-80}\over10}$$
so simply by noting that $5U^2-15U+4$ is clearly positive for $U\le0$ and negative at $U=1$, so it has two distinct positive roots, hence the quartic derivative has $4$ real roots, all distinct.
It took me more than five minutes to write this up, but the scratchwork and thinking took less than three.  The key was to notice that a simple shift makes it easy to expand the factored quintic.
Remark:  This approach clearly only works because the roots of the quintic are equally spaced.  If the problem started, say with $(x-1)(x-2)(x-4)(x-8)(x-16)$, then Rolle's Theorem would be your best bet.  Still, it's a useful alternative to have on tap; suppose, for example, the question had asked for the number of solutions to $f'(x)=1$.  Try answering that armed only with Rolle!
A: Think about the graph of the function: it will have zeroes at $x = 1, 2, 3, 4, 5$, and will be nonzero in between (since $f(x) = 0$ requires that one of the factors of $(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)$ is zero). So $f$ will have at least one local extremum on each of the intervals $(1, 2), (2, 3), (3, 4)$ and $(4, 5)$ (one can see this intuitively and make it rigorous by using the extreme value theorem). One can make this rigorous by using the mean value theorem.
Since the degree of $f'(x)$ will be four (which can be seen by writing $f(x)$ as a polynomial and differentiating it), it won't have more than four zeroes. So $f'$ must have exactly four zeroes.
A: The multiplicity of each of the roots of $f(x)$ is $1$, which means that the graph of the function of $f(x)$ crosses the $x$-axis at those roots. By drawing a sketch, it becomes obvious that the derivative must have precisely four real roots.
