求方程y=xysinx-cos(x-1)所确定的隐函数的导数dy⼀dx

2025-01-20 20:58:13
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回答1:

求导,y'=ysinx+x*(y'sinx+y*cosx)-sin(x-1)
移项
y'-xy'sinx=ysinx+xycosx-sin(x-1)
y'=(ysinx+xycosx-sin(x-1))/(1-xsinx)