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Question

# If y=12(sinâˆ’1x)2, then find (1âˆ’x2)y2âˆ’xy1. Where y1 and y2 denote first and second derivatives of y respectively.

A
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Solution

## The correct option is A 1Given y=12(sin−1x)2.We have to find (1−x2)y2−xy1 where y1,y2 denote the first and second derivatives y respectively.y1=ddx(12(sin−1x)2) =12⋅ddx(sin−1x)2 =12⋅2sin−1x⋅1√1−x2y1=sin−1x√1−x2Now let us find y2y2=ddxy1 =ddx(sin−1x√1−x2) =√1−x2⋅ddx(sin−1x)−sin−1x⋅ddx(√1−x2)(√1−x2)2 =√1−x2⋅1√1−x2−sin−1x⋅(−x√1−x2)(√1−x2)2 =1+xsin−1x√1−x21−x2∴y2=√1−x2+xsin−1x(1−x2)(√1−x2)Now we have to find (1−x2)y2−xy1.(1−x2)y2−xy1=(1−x2)⋅√1−x2+xsin−1x(1−x2)(√1−x2)−x⋅sin−1x√1−x2 =√1−x2+xsin−1x√1−x2−xsin−1x√1−x2 =1+xsin−1x√1−x2−xsin−1x√1−x2∴(1−x2)y2−xy1=1

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