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.)
To find the derivative of g(x)=(1-x)(1-2x)...(1-nx), we can use the product rule.
The product rule states that if you have a function h(x) = f(x) * g(x), then the derivative of h(x) with respect to x, denoted as h'(x), is given by:
h'(x) = f'(x) * g(x) + f(x) * g'(x)
Applying the product rule to g(x), we can set f(x) = (1-x)(1-2x)...(1-(n-1)x) and g(x) = (1-nx).
Let's start by finding f'(x). We have:
f(x) = (1-x)(1-2x)...(1-(n-1)x)
Taking the derivative of f(x) with respect to x, we get:
f'(x) = [(1-2x)(1-3x)...(1-(n-1)x)] + [(1-x)(1-3x)...(1-(n-1)x)(-2)] + ... + [(1-x)(1-2x)...(1-(n-2)x)(-1)]
Notice that for each term, we are essentially taking one factor out at a time and multiplying the remaining factors together. The factor taken out contributes a derivative term (-k), where k is the number corresponding to the factor being taken out.
Now, let's find g'(x):
g(x) = (1-nx)
Taking the derivative of g(x) with respect to x, we get:
g'(x) = -n
Now that we have f'(x) and g'(x), we can find g'(0) by substituting x = 0 into the expression for g'(x):
g'(0) = -n
Therefore, the derivative g'(0) is -n.