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a
y2sinx1−logy
b
y2sinx1−logy
c
y2cotx1−logy
d
y2tanx1−logy
answer is C.
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Detailed Solution
y=sinxylogy=y⋅logsinxDifferentiate both sides with respect to x1y⋅dydx=y⋅cotx+logsinxdydxdydx1y−logsinx=ycotxdydx1−y⋅logsinxy=ycotxdydx=y2cotx1−y⋅logsinx=y2⋅cotx1−logy