If x^{y }=e^{x-y }show that dy/dx=(logx)/{log(xe)}^{2 } Share with your friends Share 55 Koka Sri Lakshmi Divya Sai answered this 85 View Full Answer Exuberant.me answered this I would say "this question is wonderful" for the ones who take 'log x as ln x' and write d/dx(log x) = 1/x where it must be d/dx (ln x) = 1/x but only if in your question, one had to provedy/dx = (logx)(log e)/(log(xe))^{2 }i hope you can understand what i meant to say :P -36 pulkitagarwal389 answered this ylnx=x-y ---1lnxdy/dx+y/x=1-dy/dxdy/dx(lnx+1)=1-x/y ----3we can see from 1 thaty=x/1+lnxputting in 3dy/dx=y-x/xput the value of ydy/dx=1/1+lnxnow 1-= lnenow u can see furthur to conversion -21 Manan Chheda answered this first take log on both sidesylog(x)=(x-y)loge since loge=1ylogx=x-yylogx+y=xy(logx+1)=xy=x/(1+logx) now solve by basic rule of diffrentiation ie. apply d/dx(u/v) 21