Question

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

• A. $-\frac {x} {y}$
• B. $-\frac {y} {x}$
• C. $\frac {1+ \log \; x} {1+ \log \; y}$
• D. $1+\log \left(\frac{x}{y}\right)$

Answer 1

Answer: (C)

Given, $x^x = y^y$
Taking log on both sides, we get
$x \,\log \,x = y\,\log \, y$
Differentiating w.r.t. y, we get
$y . \frac{1}{y} . \frac{dy}{dx} + \log y$
$ \frac{dy}{dx} =x \frac{1}{x} + \log x $
$ \Rightarrow \frac{dy}{dx} \left(1+ \log y\right)$
$ = 1 + \log x$
$ \Rightarrow \frac{dy}{dx} = \frac{1+\log x}{1+ \log y} $