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Question

If y=12(sin−1x)2, then find (1−x2)y2−xy1.
Where y1 and y2 denote first and second derivatives of y respectively.

A
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Solution

The correct option is A 1
Given y=12(sin1x)2.

We have to find (1x2)y2xy1 where y1,y2 denote the first and second derivatives y respectively.

y1=ddx(12(sin1x)2)

=12ddx(sin1x)2

=122sin1x11x2

y1=sin1x1x2

Now let us find y2

y2=ddxy1

=ddx(sin1x1x2)

=1x2ddx(sin1x)sin1xddx(1x2)(1x2)2

=1x211x2sin1x(x1x2)(1x2)2

=1+xsin1x1x21x2

y2=1x2+xsin1x(1x2)(1x2)

Now we have to find (1x2)y2xy1.

(1x2)y2xy1=(1x2)1x2+xsin1x(1x2)(1x2)xsin1x1x2

=1x2+xsin1x1x2xsin1x1x2

=1+xsin1x1x2xsin1x1x2

(1x2)y2xy1=1


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