Complex derivative involving exponents and natural log

Question

Find: $\frac{d}{dx} a^{x\ln x}$

I have tried several methods involving u-substitution etc, but can't figure it out.

Answer 1

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

Answer 2

Let $y = a^{x \ln x}$. Then $\ln|y| = x \ln(x) \ln|a|$. Try using implicit differentiation from here.

Answer 3

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