Question

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

Answer 1

Given,

x^y=e^(x-y)

Taking log of both sides​

=>log x^y = log e^{(x - y)} 

Answer 2

Given,

x^y = e^(x-y)

Taking log of both sides

\(\Rightarrow\) log x^y = log e^(x-y)

\(\Rightarrow\) y.log x = (x-y) log e [\(\because\) log e = 1]

\(\Rightarrow\) y.log x = (x-y) \(\Rightarrow\) y log x+y = x

\(\Rightarrow\) y = \(\frac{X}{1+logX}\)

\(\Rightarrow\) \(\frac{dy}{dX}\) = \(\frac{(1+logX).1-X.\big(0+\frac{1}{X}\big)}{(1+logX)^2}\)

\(\Rightarrow\) \(\frac{dy}{dX}\) = \(\frac{1+logX-1}{(1+logX)^2}\) = \(\frac{log.X}{(loge+logX)^2}\)  [\(\because\) 1 = log e]

\(\Rightarrow\) \(\frac{dy}{dX}\) = \(\frac{logX}{(log\,eX)^2}\) \(\Rightarrow\) \(\frac{dy}{dX}\) = \(\frac{logX}{\{log\,(eX)\}^2}\)