How would I go about finding the derivative of this?

Question

I have messed around a bit but I was unable to find: $$\frac{\mathrm{d}}{\mathrm{d}x}\,x^{\ln(x)}$$ I tried changing the $\ln(x)$ bit but was unable to come up with a nice solution. I am having trouble finding a good approach to this.

Answer 1

Answer: Try changing the base to $e$: $$\frac{\mathrm{d}}{\mathrm{d}x}\,x^{\ln(x)}=\frac{\mathrm{d}}{\mathrm{d}x}\,e^{\ln^2(x)}$$ Now apply $e$ derivative chain rule: $$\frac{\mathrm{d}}{\mathrm{d}x}\,e^{\ln^2(x)}=e^{\ln^2(x)}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\,\ln^2(x)=x^{\ln(x)}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\,\ln^2(x)$$ Now solve for the derivative of $\ln^2(x)$ by using chain rule: $$=x^{\ln(x)}\cdot\left(2\ln(x)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\,\ln(x)\right)$$ $$\boxed{\frac{\mathrm{d}}{\mathrm{d}x}\,x^{\ln(x)}=2\ln(x)\cdot x^{\ln(x)-1}}$$

Answer 2

We have $$y=x^{\ln(x)},$$ take natural logs on both sides $$\ln(y)=\ln(x^{\ln(x)}),$$ that is $$\ln(y)=(\ln(x))^2.$$ Now, implicit differentiation and the chain rule gives $$\frac{1}{y}\frac{dy}{dx}=2\ln(x)\frac{1}{x},$$ multiplying through by $y$ $$\frac{dy}{dx}=2y\ln(x)\frac{1}{x}.$$ Finally, substituting $y$ gives $$\frac{dy}{dx}=2x^{\ln(x)}\ln(x)\frac{1}{x}=2x^{\ln(x)-1}\ln(x).$$