函数y=(x+cosx)⼀(x+sinx)在点x=π处得导数为

y=(x+cosx)/(x+sinx)
求导
y'=[(1-sinx)(x+sinx)-(x+cosx)(1+cosx)]/(x+sinx)²
带入x=π
y'=π/π²=1/π

y = (x + cosx)/(x + sinx)
dy/dx = [(x + sinx) * d/dx (x + cosx) - (x + cosx) * d/dx (x + sinx)]/(x + sinx)²
= [(x + sinx) * (1 - sinx) - (x + cosx) * (1 + cosx)]/(x + sinx)²
= (x - xsinx + sinx - sin²x - x - xcosx - cosx - cos²x)/(x + sinx)²
= [sinx - cosx - x(sinx + cosx) - 1]/(x + sinx)²

sin(π) = 0，cos(π) = -1
dy/dx|(x=π) = [0 + 1 - π(0 - 1) - 1]/(π + 0)² = 1/π

y'=[(1-sinx)(x+sinx)-(x+cosx)(1-cosx)]/(x+sinx)²
y'(π)=【π-（π-1）×（2）】/(π)²
=（2-π）/π²