Solution:f(x)=cos-1x2n-1x2n+1f'(x)=11-x2n-1x2n+12ddxx2n-1x2n+1=-x2n+1x2n+12- x2n-12×2nx2n-1(x2n+1)-2nx2n-1(x2n-1)(x2n+1)2=-14x2n4nx2n-1(x2n-1)=-2nx2n-1|xn|(x2n+1)=-2n(|x|2)nx|x|n(x2n+1)=-2x|x|nx(x2n+1)Thus we get,-2nxn-1(x2n+1) if n is even and 2nxn|x|(x2n+1) if n is odd Therefore Option A is correct

if fx is equal to

S o l u t i o n : f (x) = \cos^{- 1} (\frac{x^{2 n} - 1}{x^{2 n} + 1}) f' (x) = \frac{1}{\sqrt{1 - {(\frac{x^{2 n} - 1}{x^{2 n} + 1})}^{2}}} \frac{d}{d x} (\frac{x^{2 n} - 1}{x^{2 n} + 1}) = - \frac{x^{2 n} + 1}{\sqrt{{(x^{2 n} + 1)}^{2}} - {(x^{2 n} - 1)}^{2}} \times \frac{2 n x^{2 n - 1} (x^{2 n} + 1) - 2 n x^{2 n - 1} (x^{2 n} - 1)}{{(x^{2 n} + 1)}^{2}} = - \frac{1}{\sqrt{4 x^{2 n}}} \frac{4 n x^{2 n - 1}}{(x^{2 n} - 1)} = - \frac{2 n x^{2 n - 1}}{| x^{n} | (x^{2 n} + 1)} = - \frac{2 n {(| x |^{2})}^{n}}{x | x |^{n} (x^{2 n} + 1)} = - \frac{2 x | x |^{n}}{x (x^{2 n} + 1)} T h u s w e g e t, - \frac{2 n x^{n - 1}}{(x^{2 n} + 1)} i f n i s e v e n a n d \frac{2 n x^{n}}{| x | (x^{2 n} + 1)} i f n i s o d d .

Therefore Option A is correct

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