The derivative of x lnx is
$\boxed{\frac{d}{dx} (x\ln x) = \ln x + 1}$.
To show this, we will use the product rule, which states that for two functions $f(x)$ and $g(x)$
\begin{align*} &\dfrac{d}{dx} \bigg(f(x)g(x)\bigg) = f'(x)g(x) + f(x)g'(x) \end{align*}
In our case, we will let \(f(x)=x\) and \(g(x)=\ln x\). Then the product rule states that
\[ \begin{align*} \frac{d}{dx} (x\ln x) &= \dfrac{d}{dx} (x)\ln x+x\dfrac{d}{dx}(\ln x)\\ &= (1)\ln x + x\left(\dfrac1x\right) \\ &= \ln x + 1, \end{align*} \]
which shows that, indeed, $\boxed{\frac{d}{dx} (x\ln x) = \ln x + 1}$.
Similar Example: Derivative of $x \ln(2x)$
We will now find the derivative of $x \ln(2x)$.
We will use the product rule again:
\begin{align*} &\dfrac{d}{dx} \bigg(f(x)g(x)\bigg) = f'(x)g(x) + f(x)g'(x) \end{align*}
We now let $f(x)=x$ and $g(x)=\ln(2x)$.
\[ \begin{align*} \frac{d}{dx} (x\ln (2x)) &= \dfrac{d}{dx} (x)\ln (2x)+x\dfrac{d}{dx}(\ln (2x))\\ &= (1)\ln (2x) + x\left(2\cdot\dfrac1{2x}\right) \\ &= \ln (2x) + 1, \end{align*} \]
Thus the final result is $\boxed{\frac{d}{dx} (x\ln(2x)) = \ln(2x) + 1}$.
Generalization: Derivative of $x \ln(kx)$
If $k$ is any non-zero constant, we may apply the same technique as above. Also note that using the properties of logarithms we may write
\begin{align*}\dfrac d{dx} (\ln (kx)) = \dfrac d{dx} (\ln k+ \ln x) = \dfrac d{dx} (\ln k)+ \dfrac d{dx}(\ln x) = 0+\frac1x =\frac1x\end{align*}
Using the product rule just like before with $f(x) = x$ and $g(x) = \ln (kx)$, we get
\[ \begin{align*} \frac{d}{dx} (x\ln (kx)) &= \dfrac{d}{dx} (x)\ln (kx)+x\dfrac{d}{dx}(\ln (kx))\\ &= (1)\ln (2x) + x\left(\dfrac1{x}\right) \\ &= \ln (kx) + 1, \end{align*} \]
which shows that $\boxed{\frac{d}{dx} (x\ln(kx)) = \ln(kx) + 1}$ for all non-zero $k$. Note that we must stipulate that $k$ is non-zero, since otherwise the logarithm would not be defined.