The derivative of (log x)^x with respect to log x is

The correct answer is Option (3) → \(x(\log x)^{x-1} [1 + \log x.\log(\log x)]\)

Let \(y = (\log x)^x\) and \(t = \log x\). The required derivative is \(\displaystyle\frac{dy}{dt}\), found by dividing \(\displaystyle\frac{dy}{dx}\) by \(\displaystyle\frac{dt}{dx}\).

By logarithmic differentiation:

    \(\log y = x \log(\log x)\)

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

    \( \displaystyle \frac{dy}{dx} = (\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right] \)

Also \(\displaystyle\frac{dt}{dx} = \frac{1}{x}\), so

    \( \displaystyle \frac{dy}{dt} = x(\log x)^x\left[\log(\log x) + \frac{1}{\log x}\right] \)

Taking \(\displaystyle\frac{1}{\log x}\) out of the bracket:

    \( \displaystyle \frac{dy}{dt} = x(\log x)^{x-1}\left[1 + \log x \cdot \log(\log x)\right] \)