Question 6 Share with your friends Share 0 Varun.Rawat answered this We have, xy = ex-y⇒logxy = log ex-y⇒y log x = x-y log e⇒y log x = x - y⇒y log x + y = x⇒y =x1 + log xDifferentiating both sides with respect to x, we get dydx = 1 + log x×ddxx - x × ddx1+ log x1 + log x2⇒dydx = 1 + log x - x ×1x1 + log x2⇒dydx = log x1 + log x2 1 View Full Answer