设y=f(x)由方程x^(1+y)=y^(sinx)确定,求y'

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设y=f(x)由方程x^(1+y)=y^(sinx)确定,求y'

设y=f(x)由方程x^(1+y)=y^(sinx)确定,求y'
设y=f(x)由方程x^(1+y)=y^(sinx)确定,求y'

设y=f(x)由方程x^(1+y)=y^(sinx)确定,求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]

设F(x,y)=x^(1+y)-y^(sinx)
F对x求导(y看作常数),F'X=(1+y)*x^y-y^(sinx)lny*cosx
F对y求导(x看作常数),F'y=x^(1+y)lnx-sinx*y^(sinx-1)
y'= - F'X / F'y
= - [(1+y)*x^y-y^(sinx)lny*cosx]/[x^(1+y)lnx-sinx*y^(sinx-1)]