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Question

Let y=sin1x, then find (1x2)y2xy1.
Where y1 and y2 denote the first and second order derivatives respectively.

A
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B
1
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C
0
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D
2
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Solution

The correct option is D 0

Given, y=sin1x

y1=dydx=11x2

y2=d2ydx2=121(1x2)32(2x)=x(1x2)1x2

Thus the value of (1x2)y2xy1 is

(1x2)x(1x2)1x2x11x2

=x1x2x1x2=0

So, option C is correct

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