If x^y =e^x-y show that

dy/dx=(logx)/{log(xe)}²

 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 prove

dy/dx = (logx)(log e)/(log(xe))²

i hope you can understand what i meant to say :P

ylnx=x-y    ---1

lnxdy/dx+y/x=1-dy/dx

dy/dx(lnx+1)=1-x/y        ----3

we can see from 1 that

y=x/1+lnx

putting in 3

dy/dx=y-x/x

put the value of y

dy/dx=1/1+lnx

now 1-= lne

now u can see furthur to conversion