Question

求大神帮忙解一下这道导数题！非常感谢！

Answer 1

复合函数求导，由外向内，逐步求导。
y'=e^[-sin²(1/x)]·[-sin²(1/x)]'
=e^[-sin²(1/x)]·[-2sin(1/x)]·[sin(1/x)]'
=e^[-sin²(1/x)]·[-2sin(1/x)]·cos(1/x)·(1/x)'
=e^[-sin²(1/x)]·[-2sin(1/x)]·cos(1/x)·(-1/x²)
=e^[-sin²(1/x)]·[2sin(1/x)]·cos(1/x)·(1/x²)
=e^[-sin²(1/x)]·sin(2/x) /x²

追问

谢谢

Answer 2

y = e^[-sin²(1/x)]

y ′ = e^[-sin²(1/x)] * [-sin²(1/x)] ′
= e^[-sin²(1/x)] * [-2sin(1/x)] * [sin(1/x)] ′
= e^[-sin²(1/x)] * [-2sin(1/x)] * [cos(1/x)] * (1/x)′
= e^[-sin²(1/x)] * [-2sin(1/x)] * [cos(1/x)] * (-1/x²)
= sin(2/x) * e^[-sin²(1/x)] / x²