The derivative of e^xlog_e x , x gt 0 , with respect to x is

The correct answer is Option (2) → \(e^{x}\left(\log_e x + \frac{1}{x}\right)\)

By the product rule, \((uv)' = u'v + uv'\), with \(u = e^{x}\) and \(v = \log_e x\):

    \( \displaystyle \frac{d}{dx}\left(e^{x}\log_e x\right) = e^{x}\log_e x + e^{x} \cdot \frac{1}{x} \)

    \( \displaystyle = e^{x}\left(\log_e x + \frac{1}{x}\right) \)