Derivative of $x^x$ with respect to $x\log x$ is

$x^x$

The correct answer is Option (3) → $x^x$

Let $y = x^x$

Take natural logarithm: $\ln y = x \ln x$

Differentiate both sides w.r.t $x \ln x$:

$\frac{d(\ln y)}{d(x \ln x)} = \frac{d(\ln y)/dx}{d(x \ln x)/dx} = \frac{\frac{1}{y} \frac{dy}{dx}}{\ln x + 1}$

Also, $\frac{dy}{dx} = x^x (\ln x + 1)$

So, $\frac{dy}{d(x \ln x)} = \frac{x^x (\ln x + 1)}{\ln x + 1} = x^x$