![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018071159-76007.gif)
![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018071159-60727.gif) (2)解: y=μ3,μ=ax-bsin2ωx,μ=av-by v=x,y=sinγ γ=ωx y′=(μ3)′=3μ2·μ′=3μ2(av-by)′ =3μ2(av′-by′)=3μ2(av′-by′γ′) =3(ax-bsin2ωx)2(a-bωsin2ωx) (3)解法一: 设y=f(μ),μ= ,v=x2+1,则 y′x=y′μμ′v·v′x=f′(μ)· v- ·2x =f′( )·![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018071200-68889.gif) ·2x =![](http://img.shitiku.com.cn/uploads/allimg/20191018/20191018071200-19957.gif) 解法二: y′=[f( )]′=f′( )·( )′ =f′( )· (x2+1) ·(x2+1)′ =f′( )· (x2+1) ·2x = f′( ) |