Giải câu 2 bài 5: Đạo hàm cấp hai.
a) y=11−x
y′=1′.(1−x)−1.(1−x)′(1−x)2=−(1−x)′(1−x)2=1(1−x)2
y"=−[(1−x)2]′(1−x)4=−2.(−1)(1−x)(1−x)4=2(1−x)3.
b) y=11−x
y′=−(1−x)′1−x=121−x(1−x)
y"=−12[(1−x)1−x]′(1−x)3
=−12−1−x+(1−x)−121−x(1−x)3
=34(1−x)21−x.
c) y=tanx
y′=1cos2x
y"=−(cos2x)′cos4x=2cosx.sinxcos4x=2sinxcos3x.
d) y=cos2x
y′=2cosx.(cosx)′=2cosx.(−sinx)=−2sinx.cosx=−sin2x,
y"=−(2x)′.cos2x=−2cos2x.
