Question n.o 7 Share with your friends Share 0 Manbar Singh answered this We have, xx-y = logax-y⇒xx-y = log a - logx - yDifferentiating both sides with respect to x, we getx-y×ddxx - x×ddxx-yx-y2 = 0 - 1x-y×1 - dydx⇒x - y - x1 - dydxx-y2 = 1x-ydydx - 1⇒x-y-x+xdydxx-y = dydx - 1⇒xdydx - yx-y = dydx - 1⇒xdydx - y = x-ydydx - x + y⇒xdydx - y = xdydx - ydydx - x + y⇒-y - y = -ydydx - x⇒-2y = -ydydx - x⇒ydydx = 2y - x⇒dydx = 2 - xy 0 View Full Answer