If xy=xex-y+6, then dydx=
x-yx1+logex
1+x-yx1+logex
yx1+logex
None of these
Explanation to the correct option.
xy=xex-y+6⇒xyx=ex-y+6⇒xy-1=ex-y+6
By taking log on both sides we get,
y-1logx=x-y+6loge⇒y-1logx=x-y+6
By differentiating w.r.t x, we get
y-11x+logxdydx=1-dydx⇒logxdydx+dydx=1-y-1x⇒1+logxdydx=x-y+1x⇒dydx=x-y+1x1+logex
Hence, option B is correct