Question

Let n be a positive integer. Find g'(0), where g(x)=(1-x)(1-2x)...(1-nx). (Note: The answer is in terms of n.)
I hope someone can help me, please.
Thank you so much!

Answer 1

using the product rule,
g'(x) = -1(1-2x)... -2(1-x)(1-3x) ... -3(1-x)(1-2x)(1-4x) ...
so, since at x=0 all those linear factors are just 1,
g'(0) = -1-2-3-...-n = -n(n+1)/2

Answer 2

Sorry, I don't understand how to use the product rule to make g(x) become g'(x)

Answer 3

Don't know if oobleck is on line right now ....

if you have a product of several factors, the derivative is as follows

y = (a)(b)(c)(d) ....
then y' = a'(b)(c)(d)... + (a)b'(c)(d)... + (a)(b)c'(d)... +
in other words, you differentiate the first one of the factors times all the the others + the derivative of the 2nd times all the others + ...
e.g. in your case: y = (1-x)(1-2x)...(1-nx)
dy/dx = <b>dy/dx </b>(1-x) (1-2x)...(1-nx) + (1-x)<b>dy/dx </b>(1-2x)...(1-nx) + ...

= <b>(-1)</b>(1-2x)(1-3x)...(1-nx) + (1-x)<b>(-2)</b>(1-3x) + (1-x)(1-2x)<b>(-3)</b>(1-4x) ....+ (1-x)(1-2x)..<b>(-n)</b>

so when x = 0 this becomes
dy/dx = -1(1)(1)... + (1)(-2)(1)... + .... (1)(1)...(-n)
= -1 -2 -3 - ... -n
= -1(1+2+3+...+n) < we have a formula for this, as shown by oobleck