Let #f(x)=x^2(x^2-1)(x^2-2)(x^2-3)...(x^2-n)# What is the value of #f'(1)#?

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Answer choices:
#f'(1)=2(n-1)!#

#f'(1)=(-1)^(n-1)2(n-1)!#

#f'(1)=(-1)^(n)(2n)!#

#f'(1)=(-1)^(n-1)*(2n-2)!#

None of the above

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Answer 1 · 2016-12-10T18:15:51

#f'(1)=2(-1)^(n-1)(n-1)!#

#f(x) = prod_(k=0)^n(x^2-k)# but

#(df)/(dx) = sum_(j=0)^n 2xprod_(knej)^n(x^2-k)# then

#f'(1)=2(1)prod_(kne1)^n(1-k)# because all the others terms contain the factor #(x-1)#

Finally

#f'(1)=2(-1)^(n-1)(n-1)!#