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B
1√(1−x4)
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C
1√(1−x3)
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D
1√(1+x2)
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Solution
The correct option is A1√(1−x2) y=sin−1x, y+δy=sin−1(x+δx) ∴(x+δx)=sin(y+δy) dydx=limx→01δx(sin−1(x+δx)−sin−1x) =limx→0δysin(y+δy)−siny=1cosy =1√(1−sin2y)=1√(1−x2) Note that when :δx→0,δy also tends to zero.