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)

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