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