If x^y = e^x - y, then show that dy/dx = (log x)/(1 + log x)^2

Question

If x^y = e^{x - y}, then show that dy/dx = \(\frac{log\,x}{(1 + log\,x)^2}\)

(log x)/(1 + log x)²

Answer 1

x^y = e^{x - y} 

log x^y = log e^{x - y} 

y log x = (x – y) log e 

y log x = (x – y) ….. [∵ log e = 1] 

y + y log x = x – y 

y + y log x = x 

y(1 + log x) = x