In my textbook, there's something like:
$$\ln(\sin 2x)^{\ln x} = \ln x \cdot \sin 2x$$
I thought it should be
$$\ln x\cdot \ln(\sin 2x).$$
In my textbook, there's something like:
$$\ln(\sin 2x)^{\ln x} = \ln x \cdot \sin 2x$$
I thought it should be
$$\ln x\cdot \ln(\sin 2x).$$
If the textbook means to imply that $$\ln\left[(\sin 2x)^{\ln x}\right]=(\ln x) \cdot \sin 2x,\quad\;\;\text{ }$$ then it is in error, because in actuality it is as you say it is: $$\ln\left[(\sin 2x)^{\ln x}\right]=(\ln x)\cdot \ln(\sin2x).$$ If the rest of the derivation proceeds from this folly, it is completely wrong.