No links please Q.10. If y x = e y - x prove that, d y d x = log e y 2 log y Share with your friends Share 0 Varun Rawat answered this We have, yx = ey-x⇒log yx = log ey-x⇒x log y = y-x log e⇒x log y = y - x. 1⇒x log y = y - x⇒y = x log y + x ....1Differentiating 1 with respect to x, we get dydx = x×1ydydx + log y × 1 + 1⇒dydx - xydydx = 1 + log y⇒dydx1 - xy = 1 + log y⇒dydxy-xy = 1 + log y⇒dydx = y1 + log yy - x⇒dydx = x log y + x1 + log yx log y + x - x⇒dydx = x1 + log y1 + log yx log y⇒dydx = 1 + log y2log y⇒dydx = log e + log y2log y⇒dydx = log ey2log y 0 View Full Answer