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Mathematics Question on limits and derivatives

If f(x)=(x−2)(x−4)(x−6)....(x−2n),f(x)=(x-2)(x-4)(x-6)....(x-2n), then f′(2)f'(2) is

A

(−1)n2n−1(n−1)!{{(-1)}^{n}}{{2}^{n-1}}(n-1)!

B

(−2)n−12n(n−1)!{{(-2)}^{n-1}}{{2}^{n}}(n-1)!

C

(−2)nn!{{(-2)}^{n}}n!

D

(−1)n−12n(n−1)!{{(-1)}^{n-1}}{{2}^{n}}(n-1)!

Answer

(−2)n−12n(n−1)!{{(-2)}^{n-1}}{{2}^{n}}(n-1)!

Explanation

Solution

∵\because f(x)=(x−2)(x−4)(x−6)....(x−2n)f(x)=(x-2)(x-4)(x-6)....(x-2n) Taking log on both sides, we get log⁡f(x)=log⁡(x−2)+log⁡(x−4)\log f(x)=\log (x-2)+\log (x-4) +....+log⁡(x−2n)+....+\log (x-2n)
aOn differentiating w.r.t. x,x, we get
1f(x)f(x)=1(x−2)+1(x−4)\frac{1}{f(x)}f(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)} +...+1(x−2n)+...+\frac{1}{(x-2n)} f(x)=(x−4)(x−6)...(x−2n)f(x)=(x-4)(x-6)...(x-2n) +(x−2)(x−6)....(x−2n)+(x-2)(x-6)....(x-2n) +.....+(x−2)(x−6)...(x−2(n−1))+.....+(x-2)(x-6)...(x-2(n-1))
∴\therefore f(2)=(−2)(−4)....(2−2n)f(2)=(-2)(-4)....(2-2n)
=(−2)n−1(1.2....(n−1))=(−2)n−1(n−1)!={{(-2)}^{n-1}}(1.2....(n-1))={{(-2)}^{n-1}}(n-1)!