​ If x^y = e^x.y, show that dy/dx = ​logx/(log(xe))^2

Question

If x^y = e^x.y, show that \(\frac{dy}{dx}\) = \(\frac{logx}{(log(xe))^2}\)

Answer 1

x^y = e^{x - y}

ylogx = (x - y)loge

ylogx + y = x

y(1 + logx) = x