因为:y=f(x)由方程x^(1+y)=y^(sinx)确定,所以对等式两边取对数:
(1+y)lnx=(sinx)lny, 等式两边对x求导:
y'lnx+(1+y)/x=(cosx)lny+(y'/y)sinx
y'[lnx-(sinx)/y]=(cosx)lny-(1+y)/x
y'=[(cosx)lny-(1+y)/x]/[lnx-(sinx)/y]
因为:y=f(x)由方程x^(1+y)=y^(sinx)确定,所以对等式两边取对数:
(1+y)lnx=(sinx)lny, 等式两边对x求导:
y'lnx+(1+y)/x=(cosx)lny+(y'/y)sinx
y'[lnx-(sinx)/y]=(cosx)lny-(1+y)/x
y'=[(cosx)lny-(1+y)/x]/[lnx-(sinx)/y]