Q. If $x^{y} = e^{2(x - y)},\ then\frac{dy}{dx}\ \$is equal to
A) $\frac{2(1 + \log{x)}}{\left( 2 + \log x \right)^{2}}$
B) $\frac{1 + \log x}{(2 + \log{x)}^{2}}$
C) $\frac{2}{2 + log\ x}$
D) $\frac{2(1 - \log{x)}}{(2 + \log{x)}^{2}}$
Explanation
(a)
$x^{y} = e^{2(x - y)}$
$\therefore y\ \log{x = 2(x - y)}$
$\Rightarrow \ y\ (\log{x + 2) = 2x}$
$y = \frac{2x}{\log{x + 2}}$
$\frac{dy}{dx}\text{= }\frac{(\log{x + 2)(2)} - 2x.\frac{1}{x}}{{(\log{x + 2)}}^{2}}\$
$= \frac{2\log{x + 4 - 2}}{{(\log{x + 2)}}^{2}} = \frac{2(\log{x + 1)}}{{(\log{x + 2)}}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.