Bài 2. Tìm đạo hàm cấp hai của các hàm số sau:
a) y=11−xy=11−x;
b) y=1√1−xy=11−x;
c) y=tanxy=tanx;
d) y=cos2xy=cos2x .
Lời giải:
a) y′=−(1−x)′(1−x)2y′=−(1−x)′(1−x)2 = 1(1−x)21(1−x)2, y"=−[(1−x)2]′(1−x)4=−2.(−1)(1−x)(1−x)4y"=−[(1−x)2]′(1−x)4=−2.(−1)(1−x)(1−x)4 = 2(1−x)32(1−x)3.
b) y′=−(√1−x)′1−xy′=−(1−x)′1−x = 12(1−x)√1−x12(1−x)1−x;
y"=−12[(1−x)√1−x]′(1−x)3y"=−12[(1−x)1−x]′(1−x)3 = −12−√1−x+(1−x)−12√1−x(1−x)3−12−1−x+(1−x)−121−x(1−x)3 = 34(1−x)2√1−x34(1−x)21−x.
c) y′=1cos2xy′=1cos2x; y"=−(cos2x)′cos4x=2cosx.sinxcos4xy"=−(cos2x)′cos4x=2cosx.sinxcos4x = 2sinxcos3x2sinxcos3x.
d) y′=2cosx.(cosx)′=2cosx.(−sinx)y′=2cosx.(cosx)′=2cosx.(−sinx)
=−2sinx.cosx=−sin2x=−2sinx.cosx=−sin2x,
y"=−(2x)′.cos2x=−2cos2xy"=−(2x)′.cos2x=−2cos2x.
