Originally Posted by

**ebaines** You don't need to find the general form of f'(x), you only need to find f'(0). The function f(x) is a big polynomial, of order x^97:

f(x) = x(x-1)(x-2)...(x-96) = x^(97) + Ax^(96) + Bx^(95) + .... + Wx^3 + Yx^2 + Zx

where the A, B, etc terms are constants. Its derivative is also a polynomial, of order one less than the original polynomial:

f'(x) = 97x^(96) + 96Ax^(95) + 95Bx^(94) + ... + 3Wx^2 + 2Yx + Z

Note that at x=0 all these terms equal zero except for the last, so you have f'(0) = Z. Now the only thing left to do is figure out the value of Z, which is the coefficient of the x^1 term in the original polynomial. I'll leave that to you!