Find the number of real roots of the derivative of $f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$ 
Find out the number of real roots of equation $f'(x) = 0$, where 
  $$f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)$$ 

How can I differentiate this function without expanding it to the polynomial form. Am I underestimating some theory of equation concept associated with it? (I know the product rule approach and solving by simplifying but I want to know is there any other way to solve it) 
 A: This is a question about Rolle's Theorem.
The derivative is a polynomial of degree 4, so it has at most four zeros.
Can you explain why it has exactly four zeros.
A: You can extend the idea of the product rule. You already know that if $y=f\times g$ then $y'=f'\times g+f\times g'$. If you have more functions multiplied then we get, if:
$$y=f_1\times f_2\times f_3\times...\times f_n$$
then
$$y'=f'_1\times f_2\times f_3\times...\times f_n+f_1\times f'_2\times f_3\times...\times f_n+f_1\times f_2\times f'_3\times...\times f_n+...+f_1\times f_2\times f_3\times...\times f'_n$$
Hence for your question you get:
$$y'=(x-2)(x-3)(x-4)(x-5)+(x-1)(x-3)(x-4)(x-5)+(x-1)(x-2)(x-4)(x-5)+(x-1)(x-2)(x-3)(x-5)+(x-1)(x-2)(x-3)(x-4)$$
Note that this now won't nicely factorize so you'd need to expand from here...
$$$$
However none of this is needed to answer your question. As $f(x)$ is of degree 5 then the derivative will be of degree 4. If you think about the graph of $f(x)$ there will not be any repeated roots of $f'(x)$ so there will be 4 solutions.
