If y=axax..... then the value of dydx is
y2x(1−ylogx)
y2logyx(1−ylogx)
y2logyx(1−ylogxlogy)
y2logyx(1+ylogxlogy)
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]