If y=x log x, then which of the following is correct ?

$y=x \log x-1$     ......(1)

differentiating (1) w.r.t (x)

$\frac{d y}{d x} =\frac{d}{d x}(x \log x) \Rightarrow \frac{d y}{d x}=\log x \frac{d x}{d x}+x \frac{d \log x}{d x}$

$\Rightarrow \frac{d y}{d x} =\log x . 1+x \times \frac{1}{x}$

$\frac{d y}{d x} =\log x+1$      ......(2)

now multiplying (2) with (x)

$\frac{x d y}{d x}=x \log x+x$

Substituting x log x with y from eq (1)

$\frac{x d y}{d x}=x+y \Rightarrow x\frac{d y}{d x}-x=y$