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.

$$\dfrac{d(a^{x\ln(x)})}{dx}=\dfrac{d(a^{x\ln(x)})}{d(x\ln(x))}\left(\ln(x)+1\right)=\ln(a)a^{x\ln(x)}\left(\ln(x)+1\right)$$
The last result is obtained by the logarithm laws and using the fact that $a^x=e^{x\ln(a)}$, and this holds $\forall a\neq0,a\neq a(x)$.
