Derivative of x^x with respect to xlog x is

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

Let \(u = x^x\) and \(v = x\log x\). The derivative of u with respect to v is found by the chain rule:

    \( \displaystyle \frac{du}{dv} = \frac{du/dx}{dv/dx} \)

By logarithmic differentiation of \(u = x^x\):

    \(\log u = x\log x\)

    \( \displaystyle \frac{1}{u}\,\frac{du}{dx} = \log x + x \cdot \frac{1}{x} = 1 + \log x \)

    \( \displaystyle \frac{du}{dx} = x^x(1 + \log x) \)

By the product rule:

    \( \displaystyle \frac{dv}{dx} = \log x + x \cdot \frac{1}{x} = 1 + \log x \)

Hence

    \( \displaystyle \frac{du}{dv} = \frac{x^x(1 + \log x)}{1 + \log x} = x^x \)

Shortcut: Since \(\log u = x\log x = v\), we have \(u = e^{v}\), so \(\displaystyle\frac{du}{dv} = e^{v} = u = x^x\).