Question

If $y = x^x , x > 0$ , then $\frac{dy}{dx}$ is

• A. log x
• B. 2 + log x
• C. $x^x \log x$
• D. $x^x ( 1 + \log x )$

Answer 1

Answer: (D)

$x^x ( 1 + \log x )$

Answer 2

$y =x^{x} \Rightarrow \log y =x \log x$ $\Rightarrow \frac{1}{y} \frac{dy}{dx} = x. \frac{1}{x} + \log x.1$ $\Rightarrow \frac{dy}{dx} = y\left(1+\log x\right) =x^{x} \left(1+\log x\right)$