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!
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
Sorry, I don't understand how to use the product rule to make g(x) become g'(x)
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 = dy/dx (1-x) (1-2x)...(1-nx) + (1-x)dy/dx (1-2x)...(1-nx) + ...
= (-1)(1-2x)(1-3x)...(1-nx) + (1-x)(-2)(1-3x) + (1-x)(1-2x)(-3)(1-4x) ....+ (1-x)(1-2x)..(-n)
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
To find g'(0), we need to take the derivative of g(x) with respect to x and then evaluate it at x = 0.
To start, let's expand g(x) using the product rule:
g(x) = (1-x)(1-2x)...(1-nx)
= (1-x)(1-2x)...(1-(n-1)x)(1-nx)
Now, we can differentiate each term in the product using the chain rule:
g'(x) = -1(1-2x)...(1-(n-1)x)(1-nx) + (1-x)(-2)(1-2x)...(1-(n-1)x)(1-nx) + ... + (1-x)(1-2x)...(1-(n-1)x)(-nx)
Notice that each term in the sum has a factor of (1-x)(1-2x)...(1-(n-1)x), except for the last term, which has a factor of (-nx) instead. We can factor out this common factor:
g'(x) = (1-x)(1-2x)...(1-(n-1)x) * (-n) + (1-x)(1-2x)...(1-(n-1)x)(-2) + ... + (1-x)(1-2x)...(1-(n-1)x)(-n)
Simplifying further, we get:
g'(x) = (1-x)(1-2x)...(1-(n-1)x)(-n - 2 - ... - n)
= (-1)^(n-1) (1-x)(1-2x)...(1-(n-1)x)(1 + 2 + ... + n)
Now, let's evaluate g'(x) at x = 0:
g'(0) = (-1)^(n-1) (1-0)(1-0)...(1-0)(1 + 2 + ... + n)
= (-1)^(n-1) (1)(1)...(1)(1 + 2 + ... + n)
= (-1)^(n-1) (1 + 2 + ... + n)
Therefore, g'(0) is equal to (-1)^(n-1) multiplied by the sum of the integers from 1 to n, which can be represented as:
g'(0) = (-1)^(n-1) * (n(n+1))/2
So, g'(0) is equal to (-1)^(n-1) multiplied by n multiplied by (n+1) divided by 2.