y(x)=sin(x²+1)
y'(x)=cos(x²+1)*(2x)=2xcos(x²+1)這是怎么求出來(lái)的？復(fù)雜導(dǎo)數(shù)我就是沒(méi)明白。。。。這個(gè)是復(fù)合函數(shù)u=g(y) y=f(x)則 u'=g'(y)*f'(x)在這里y'(x)=[sin(x²+1)]' × (x²+1)'=cos(x²+1) × (2x)=2xcos(x²+1)