My interpretation, though it is written badly, is to show that the derivative of f(x)= sin^n(x)cos(nx) is f'(x)= n sin^{n-1}(x)cos((n+1)x).
To start with use the product rule and chain rule:
f'(x)= n sin^{n-1}(x)cos(x)cos(nx)+ sin^n(x)(-n sin(nx))= n sin^{n-1}(x)[cos(x)cos(nx)- sin(x)sin(nx)]
Now use the trig identity cos(a+b)= cos(a)cos(b)- sin(a)sin(b) to rewrite the quantitiy inside the [ ].
Thank you so much! But I still need some help. Please bear with me. Also, topsquark, can you show me how you made the equation look like that on this board?
I get that (sin^n)x turns into n(sin^n-1) but can you explain in more detail what's going on here cos(x)cos(nx)+ sin^n(x)(-n sin(nx)). Like I don't quite see where the chain rule is applied.