Question

If $x^y=e^{x-y}$ then $\frac{dy}{dx}=$

• A. $(1+\ln x{)}^{-1}$
• B. $(1+\ln x{)}^{-2}$
• C. $\ln x(1+\ln x{)}^{-2}$
• D. None of these

Answer 1

The correct option is C $\ln x(1+\ln x{)}^{-2}$

$Since,x^y=e^{x-y}\Rightarrow y\ln x=x-y\therefore y=\frac{x}{1+\ln x}\therefore \frac{dy}{dx}=\frac{\ln x}{(1+\ln x{)}^2}$