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

Question

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

Isabella Ackerman · Accepted Answer

#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)!#

Emilia Aguilar · Answer

undefined The value of $ f'(1) $ can be found by taking the derivative of the function $ f(x) $ with respect to $ x $ and then evaluating it at $ x = 1 $. Using the product rule for differentiation, the derivative of $ f(x) $ is:

$$ f'(x) = 2x(x^2-1)(x^2-2)(x^2-3)...(x^2-n) + x^2(x^2-1)(2x)(x^2-2)(x^2-3)...(x^2-n) + \cdots + x^2(x^2-1)(x^2-2)(x^2-3)...(2x)(x^2-n) + x^2(x^2-1)(x^2-2)(x^2-3)...(x^2-n)(2x) $$

Simplifying and evaluating at $ x = 1 $:

$$ f'(1) = 2(1)(1^2-1)(1^2-2)(1^2-3)...(1^2-n) + 1^2(1^2-1)(2 \cdot 1)(1^2-2)(1^2-3)...(1^2-n) + \cdots + 1^2(1^2-1)(1^2-2)(1^2-3)...(2 \cdot 1)(1^2-n) + 1^2(1^2-1)(1^2-2)(1^2-3)...(1^2-n)(2 \cdot 1) $$

$$ f'(1) = 2(0)(-1)(-2)(-3)...(-n) + 1^2(0)(2)(-1)(-2)(-3)...(-n) + \cdots + 1^2(0)(-1)(-2)(-3)...(2)(-n) + 1^2(0)(-1)(-2)(-3)...(-n)(2) $$

$$ f'(1) = 0 + 0 + \cdots + 0 = 0 $$

Therefore, $ f'(1) = 0 $.

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

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

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