[tan(2x+ 1)]' = sec & amp； sup 2(2x+ 1)*(2x+ 1)' = 2 sec & amp； sup 2(2x+ 1)；

2.

Because 2 sec &；; Sup2(2x+ 1) constant >: 0, so tan(2x+ 1) has no stagnation point. (stagnation point is the point whose derivative is 0)

3.

For a very small p, find the approximate variable of the function tan (2x+1) when x increases from1to1+p.

In fact, it is for differentiation: when x = 1 and dx = p, dtan (2x+1) = 2sec &; sup 2(2x+ 1)dx = 2psec & amp； sup23。