Differentiation

As follows:


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2011-03-14 23:50:47 補充：
Q1:
=nsin^(n-1)x(sinnxcosx+sinxcosnx)
=n[sin^(n-1)x][sin(n+1)x]

Q2:
Since when 0≤ x<π/3, f is increasing and when π/3≤ x≤π, f is decreasing, so we know that the min. pt should be lying on the eding point and we check f(0) and f(π) to see which one is smaller, the smaller one is the min. value.

I wonder if the answer for Q1 can be further simplified...
Have you any ideas about it?

2011-03-14 20:48:15 補充：
Btw for Q2, why can you sub. the ending points to get the minimum value...?