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a
y2x(1−ylogx)
b
y2logyx(1−ylogx)
c
y2logyx(1−ylogxlogy)
d
y2logyx(1+ylogxlogy)
answer is C.
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Detailed Solution
y=ax axax.....…⇒y=axylogy=xylogalog(logy)=ylogx+log(loga)∣Diff w.r. to ‘x’1logy⋅1ydydx=yx+logxdydx+odydx1ylogy−logx=yxdydx1−ylogylogxylogy=yx⇒dydx=y2logyx[1−ylogylogx]