Complex derivative involving exponents and natural log Find: $\frac{d}{dx} a^{x\ln x}$
I have tried several methods involving u-substitution etc, but can't figure it out.

$$\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} a^{x \ln x} &= \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^{x \ln x \ln a} \\
&= \left( \frac{\mathrm{d}}{\mathrm{d}x} x \ln x \ln a \right)\mathrm{e}^{x \ln x \ln a} \\
&= \ln a \left( \frac{\mathrm{d}}{\mathrm{d}x} x \ln x  \right)\mathrm{e}^{x \ln x \ln a} \\
&= \ln a \left( 1+ \ln x  \right)\mathrm{e}^{x \ln x \ln a} \\
&= \ln a \left( 1+ \ln x  \right)a^{x \ln x}
\end{align*}$$
