Given $y=e^{(xy)^y}$, find $\frac {dy}{dx}$.
This problem originated from a misunderstanding of this question. I solved the problem in this post first, before I realized my solution was incorrect. However, I've decided to upload the problem and solution anyway as I imagine someone can find some use from it.
We can solve this problem using a combination of implicit and logarithmic differentiation: $$y=4e^{(xy)^y}$$ $$\ln{\frac y4}=(xy)^y$$ $$\ln{\left(\ln \frac y4\right)}=y\ln{(xy)}$$ Taking the derivative of both sides: $$\frac 1{\ln \frac y4}\left(\frac{\frac{y'}4}{\frac y4}\right)=y\left(\frac{y+xy'}{xy}\right)+y'\ln{(xy)}$$ $$\frac{y'}{y\ln{\left(\frac y4\right)}}=\frac yx+y'+y'\ln{(xy)}$$ $$xy'=y^2\ln \frac y4+xyy'\ln \frac y4+xyy'\left(\ln \frac y4\right)\ln{(xy)}$$ $$y'=\frac{y^2\ln \frac y4}{x-xy\ln \frac y4-xy\left(\ln \frac y4\right)\ln {(xy)}}$$ Not beautiful, but correct.
I hope someone can learn something from this interesting use of repeated logarithms to differentiate functions, or just enjoy an unusual but cool derivative.