# 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

Answer choices:

None of the above

Finally

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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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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