How would you differentiate ln(sin(3x))?
To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx)
To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x)
Firstly, using a standard result we have dy/du=1/u
Secondly, we must work out du/dx. Another standard result is that d/dx(sin(ax))=acos(ax) for any constant number a. This means du/dx=3cos(3x)
Putting the two parts together, we find that our answer, given by dy/dx, is equal to
3*cos(3x)1/u=(3cos(3x))/(ln(sin(3x)))
HD
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