Q. If $x^{y} = y^{x},\ then\ x(x - y\log{x)\frac{dy}{dx}\ }$is equal to
A) $y(y - x\log{y)}$
B) $y(y + x\log{y)}$
C) $x(x + y\log{x)}$
D) $x(y - x\log{y)}$
Explanation
(a)
Since, $x^{y} = y^{x} \Rightarrow y\log{x = x\log y}$
On differentiating w.r.t. $x,\$we get
$y.\frac{1}{x}$+log $x\ \frac{dy}{dx} = x.\frac{1}{y}\frac{dy}{dx} + \log y$
$\Rightarrow x(x - y\log{x)\frac{dy}{dx} = y( - x\log{y + y)}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.